Complex Numbers and Quadratic Equations
146 JEE Maths previous year questions on Complex Numbers and Quadratic Equations — options free on every question; 15 include the answer & explanation free, the rest unlock with PYQ Pass.
Let \(\alpha ,\beta\) be the roots of the equation \({x}^{2}-3x+r=0\), and, \(\frac{\alpha }{2},2\beta\) be the roots of the equation \({x}^{2}+3x+r=0\). If the roots of the equation \({x}^{2}+6x=m\) are \(2\alpha +\beta +2r\) and \(\alpha -2\beta -\frac{r}{2}\), then \(m\) is equal to:
[JEE Main 2026, 2 Apr (Shift 2)]
\(567\)
\( \alpha+\beta=3 \)
\( \frac{\alpha}{2}+2 \beta=-3\)
On solving, we get:
\(\alpha=6 ; \beta=-3\)
Product of roots \(=\alpha \beta=r \Rightarrow r=-18\)
Now, for \(\mathrm{x}^2+6 \mathrm{x}-\mathrm{m}=0\)
Product of roots \(=-m\)
\( =(2 \alpha+\beta+2 \mathrm{r})\left(\alpha-2 \beta-\frac{\mathrm{r}}{2}\right) \)
\( \Rightarrow-\mathrm{m}=(-27)(21) \)
\( \Rightarrow \mathrm{m}=567\)
If \(z=\frac{1}{2}-2 i\) is such that \(|z+1|=\alpha z+\beta(1+i), i=\sqrt{-1}\) and \(\alpha, \beta \in \mathbb{R}\), then \(\alpha+\beta\) is equal to
[JEE Main 2024, 29 Jan (Shift 1)]
3
Given
\(z=\frac{1}{2}-2 i\)
\(|z+1|=\alpha z+\beta(1+i)\)
\(\left|\frac{3}{2}-2 i\right|=\frac{\alpha}{2}-2 \alpha i+\beta+\beta i\)
\(\left|\frac{3}{2}-2 i\right|=\left(\frac{\alpha}{2}+\beta\right)+(\beta-2 \alpha) i\)
On comparing
\(\beta=2 \alpha\) and \(\frac{\alpha}{2}+\beta=\sqrt{\frac{9}{4}+4}\)
\(\alpha+\beta=3\)
\({z}_{1}=\sqrt{3}+2\sqrt{2}i\&\sqrt{3}\left|{Z}_{1}\right|=\left|{Z}_{2}\right|\text{ and }\arg \left({z}_{2}\right)=\arg \left({z}_{1}\right)+\frac{\pi }{6}\text{ then area }\)\(\text{ of triangle with vertices }{z}_{1},{z}_{2}\text{ and origin. }\) (28 Jan, Shift I, Memory Based)
\(\frac{11\sqrt{3}}{4}\)
\(\arg \left({z}_{2}\right)=\arg \left({z}_{1}\right)+\pi /6\\ \arg \left({z}_{2}\right)-\arg \left({z}_{1}\right)=\pi /6\Rightarrow \arg \left(\frac{{z}_{1}}{{z}_{2}}\right)=\pi /6\\ \text{ Area of }∆OAB=\frac{1}{2}\left|{z}_{1}\right|\left|{z}_{2}\right|\sin \pi /6\\ =\frac{1}{2}|{z}_{1}|\sqrt{3}|{z}_{1}|\sin \frac{\pi }{6}=\frac{1}{2}|{z}_{1}{|}^{2}\sqrt{3}\frac{1}{2}=\frac{11\sqrt{3}}{4}\)
Let \(\left|z_{i}\right|=1\) for \(i=1,2,3\) satisfying \(\left|\bar{z}_1 z_2+\bar{z}_2 z_3+\bar{z}_3 z_1\right|^2=a+b \sqrt{2}\), where \(a, b\) an rational numbers such that \(\arg \left(z_1\right)=\frac{\pi}{4}, \arg \left(z_2\right)=0\) and \(\arg \left(z_3\right)=\frac{-\pi}{4}\), then find \((a, b)\)
\((5,-2)\)
\(\begin{aligned}&\text { Sol. }\\&& z_1=|1| e^{i \frac{\pi}{4}}=\frac{1}{\sqrt{2}}+i \cdot \frac{1}{\sqrt{2}} \\& z_2=|1| e^{-(0)}= 1+0 i \\& z_3=|1| e^{-i \frac{\pi}{4}}=\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}} \\& \bar{z}_1 z_2=\left(\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}}\right)(1) \\& \bar{z}_2 z_3=1\left(\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}}\right) \\& \bar{z}_3 z_1=\left(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}\right)\left(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}\right) \\& \Rightarrow \bar{z}_1 z_2+\bar{z}_2 z_3+\bar{z}_3 z_1=\left(\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}}\right)+\left(\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}}\right) \\& +\left(\frac{1}{2}-\frac{1}{2}\right)+2 i\left(\frac{1}{2}\right) \\& =\sqrt{2}-\sqrt{2} i+i \\& \Rightarrow\left|\bar{z}_1 z_2+\bar{z}_2 z_3+\bar{z}_3 z_1\right|^2=|\sqrt{2}+i(-\sqrt{2}+1)|^2 \\& =\left(\sqrt{(\sqrt{2})^2+(1-\sqrt{2})^2}\right)^2 \\& =5-2 \sqrt{2} \\& (a, b)=(5,-2)\end{aligned}\)
Let \({z}_{1}\) and \({z}_{2}\) be two complex numbers such that \({z}_{1}+{z}_{2}=5\) and \({z}_{1}^{3}+{z}_{2}^{3}=20+15i\). Then, \(\left|{z}_{1}^{4}+{z}_{2}^{4}\right|\) equals
[JEE Main 2024, 31 Jan (Shift 2)]
75
\(z_1+z_2=5\)
\(z_1^3+z_2^3=20+15i\)
\(\left|z_1^4+z_2^4\right|=?\)
\(z_1^3+z_2^3=(z_1+z_2)^3-3z_1z_2(z_1+z_2)\)
\(z_1^3+z_2^3=125-15z_1z_2\)
\(20+15i=125-15z_1z_2\)
\(\Rightarrow z_1z_2=7-i\)
Now,
\((z_1+z_2)^2=5^2\)
\(z_1^2+z_2^2+2z_1z_2=25\)
\(z_1^2+z_2^2=25-2(7-i)\)
\(=11+2i\)
\((z_1^2+z_2^2)^2=(11+2i)^2\)
\(z_1^4+z_2^4+2(7-i)^2=117+44i\)
\(z_1^4+z_2^4=21+72i\)
\(\left|z_1^4+z_2^4\right|=\sqrt{21^2+72^2}\)
\(=\sqrt{441+5184}\)
\(=\sqrt{5625}\)
\(=75\)
If \(\alpha\) and \(\beta\) are the roots of the equation \(2{z}^{2}-3z-2i=0,\text{ where }i=\sqrt{-1}\) then \(\text{ 16. }Re\left(\frac{{\alpha }^{19}+{\beta }^{19}+{\alpha }^{11}+{\beta }^{11}}{{\alpha }^{15}+{\beta }^{15}}\right)\cdot Im\left(\frac{{\alpha }^{19}+{\beta }^{19}+{\alpha }^{11}+{\beta }^{11}}{{\alpha }^{15}+{\beta }^{15}}\right)\) is equal to
[JEE Main 2025, 24 Jan (Shift 1)]
441
\(\text{Given},2{z}^{2}-3z-2i=0.......\left(i\right)\\ \text{Divide equation by z, we get,}\\ 2\left(z-\frac{i}{z}\right)=3\\ \text{ As }\alpha ,\beta \text{ are roots of (i) }\\ \alpha -\frac{i}{\alpha }=\frac{3}{2}\\ \text{on squaring both sides, we get}\\ \Rightarrow {\alpha }^{2}-\frac{1}{{\alpha }^{2}}=\frac{9}{4}+2i\\ \text{ Again squaring both sides , we get}\\ \Rightarrow {\alpha }^{4}+\frac{1}{{\alpha }^{4}}=\frac{49}{16}+9i.........\left(ii\right)\\ \text{Similarly, for another root β}\\ {\beta }^{4}+\frac{1}{{\beta }^{4}}=\frac{49}{16}+9i...........\left(ii\right)\\ Now,\\ \frac{{\alpha }^{19}+{\beta }^{19}+{\alpha }^{11}+{\beta }^{11}}{{\alpha }^{15}+{\beta }^{15}}=\frac{{\alpha }^{19}+{\alpha }^{11}+{\beta }^{19}+{\beta }^{11}}{{\alpha }^{15}+{\beta }^{15}}\\ =\frac{{\alpha }^{15}\left({\alpha }^{4}+\frac{1}{{\alpha }^{4}}\right)+{\beta }^{15}\left({\beta }^{4}+\frac{1}{{\beta }^{4}}\right)}{{\alpha }^{15}+{\beta }^{15}}\\ =\frac{49}{16}+9i\\ Re\left(\frac{{\alpha }^{19}+{\beta }^{19}+{\alpha }^{11}+{\beta }^{11}}{{\alpha }^{15}+{\beta }^{15}}\right)=\frac{49}{16}\\ lm\left(\frac{{\alpha }^{19}+{\beta }^{19}+{\alpha }^{11}+{\beta }^{11}}{{\alpha }^{15}+{\beta }^{15}}\right)=9\\ \Rightarrow 16Re\left(\frac{{\alpha }^{19}+{\beta }^{19}+{\alpha }^{11}+{\beta }^{11}}{{\alpha }^{15}+{\beta }^{15}}\right)\cdot lm\left(\frac{{\alpha }^{19}+{\beta }^{19}+{\alpha }^{11}+{\beta }^{11}}{{\alpha }^{15}+{\beta }^{15}}\right)\\ =16\times \frac{49}{16}\times 9=441\)
If \(S=\{z \in C:|z-i|=|z+i|=|z-1|\}\), then, \(n(S)\) is :
[JEE Main 2024, 27 Jan (Shift 1)]
1
\(|z-i|=|z+i|=|z-1|\)
\(ABC\) is a triangle. Hence, its circum-centre will be the only point whose distance from \(A,B,C\) will be same.
So, \(n(S)=1\)
Among the statements
(S1) : The set \(\left\{\mathrm{z}\in \mathrm{ℂ}-{-\mathrm{i}}:|\mathrm{z}|=1\right.\) and \(\frac{\mathrm{z}-\mathrm{i}}{\mathrm{z}+\mathrm{i}}\) is purely real} contains exactly two elements, and
(S2) : The set \(\left\{\mathrm{z}\in \mathrm{ℂ}-{-1}:|\mathrm{z}|=1\right.\) and \(\frac{\mathrm{z}-1}{\mathrm{z}+1}\) is purely imaginary} contains infinitely many elements.
[JEE Main 2025, 7 Apr (Shift 1)]
only (S2) is correct
\(\frac{z+i}{z−i}=\frac{x+i(y+1)}{x+i(y−1)}\times \frac{x−i(y−1)}{x−i(y−1)}\\ =\text{purely real}\)
\(=\frac{({x}^{2}+{y}^{2}−1)+i(x(y+1)−x(y−1))}{{x}^{2}+(y−1{)}^{2}}\)
Imaginary part is zero.
\(x(y+1)=x(y−1)\text{ implies }x=0\text{ and }y\text{ is}\\ \text{ infinitely many values}\)
Therefore statement 1 is wrong.
\(\frac{z−1}{z+1}=\frac{x+iy−1}{x+iy+1}\\ =\frac{(x−1)+iy}{(x+1)+iy}\times \frac{(x+1)−iy}{(x+1)−iy}\)
\(=\frac{(({x}^{2}−1){y}^{2}+i(y(x+1)−y(x−1)))}{{(x+1)}^{2}+{y}^{2}}\)
Purely imaginary implies real part is zero.
\(∴{x}^{2}+{y}^{2}=1\)
Lies on infinitely many points
Therefore statement "2" is true.
If \(\alpha+i \beta\) and \(\gamma+i \delta\) are the roots of \(x^2-(3-2 i) x-(2 i-2)=0, i=\sqrt{-1}\), then \(\alpha \gamma+\beta \delta\) is equal to :
[JEE Main 2025, 28 Jan (Shift 2)]
2
\({x}^{2}-\left(3-2i\right)x-\left(2i-2\right)=0\)
\(x=\frac{(3-2i)\pm \sqrt{(3-2i{)}^{2}-4(1)(-(2i-2))}}{2(1)}\)
\(x=\frac{3-2i\pm \sqrt{-3-4i}}{2}\)
\(=\frac{3-2\mathrm{i}\pm \sqrt{(1{)}^{2}+(2\mathrm{i}{)}^{2}-2(1)(2\mathrm{i})}}{2}\)
\(=\frac{3-2\mathrm{i}\pm (1-2\mathrm{i})}{2}\)
\(x=\frac{3-2\mathrm{i}+1-2\mathrm{i}}{2},\frac{3-2i-1+2i}{2}\)
\(x=2–2i,1+0i\)
So \(\alpha \gamma +\beta \delta =2\left(1\right)+\left(-2\right)\left(0\right)=2\)
Let \({z}_{1},{z}_{2}\text{and }{z}_{3}\) be three complex numbers on the circle \(|z|=1\text{ with }\arg \left({z}_{1}\right)=\frac{-\pi }{4},\arg \left({z}_{2}\right)=0\) and \(\text{ }\arg \left({z}_{3}\right)=\frac{\pi }{4}.\) If \({\left|{\mathrm{z}}_{1}{\overset{¯}{\mathrm{z}}}_{2}+{\mathrm{z}}_{2}{\overset{¯}{\mathrm{z}}}_{3}+{\mathrm{z}}_{3}{\overset{¯}{\mathrm{z}}}_{1}\right|}^{2}=\alpha +\beta \sqrt{2},\alpha ,\beta \in \mathrm{Z}\) then the value of \({\alpha }^{2}+{\beta }^{2}\) is
[JEE Main 2025, 22 Jan (Shift 1)]
29
\(\text{Given,}\\ {z}_{1}={e}^{-i\pi /4},{z}_{2}=1,{z}_{3}={e}^{i\pi /4}\\ \text{Now,}\\ {\left|{z}_{1}{\overset{¯}{z}}_{2}+{z}_{2}{\overset{¯}{z}}_{3}+{z}_{3}{\overset{¯}{z}}_{1}\right|}^{2}\\ ={\left|{e}^{-i\frac{\pi }{4}}\times 1+1\times {e}^{-i\frac{\pi }{4}}+{e}^{i\frac{\pi }{4}}\times {e}^{i\frac{\pi }{4}}\right|}^{2}\\ ={\left|{e}^{-i\frac{\pi }{4}}+{e}^{-i\frac{\pi }{4}}+{e}^{i\frac{\pi }{2}}\right|}^{2}\\ ={\left|2\left(\cos \frac{\pi }{4}-i\sin \frac{\pi }{4}\right)+\cos \frac{\pi }{2}+i\sin \frac{\pi }{2}\right|}^{2}\\ =|\sqrt{2}-\sqrt{2}\mathrm{i}+\mathrm{i}{|}^{2}\\ =(\sqrt{2}{)}^{2}+(1-\sqrt{2}{)}^{2}\\ =5-2\sqrt{2}\\ \alpha =5,\beta =-2\\ \Rightarrow {\alpha }^{2}+{\beta }^{2}=29\)
Let \(z\) be a complex number such that \(|z|=1\). If \(\frac{2+{\mathrm{k}}^{2}\mathrm{z}}{\mathrm{k}+\overset{¯}{\mathrm{z}}}=\mathrm{kz},\mathrm{k}\in R\), then the maximum distance of \(\mathrm{k}+{\mathrm{ik}}^{2}\) from the circle \(|\mathrm{z}-(1+2\mathrm{i})|=1\) is:
[JEE Main 2025, 2 Apr (Shift 1)]
\(\sqrt{5}+1\)
We are given the equation \(\frac{2+{k}^{2}z}{k+\overset{ˉ}{z}}=kz\) and the condition \(\mathrm{∣}z\mathrm{∣}=1\).
Substitute \(\overset{ˉ}{z}=\frac{1}{z}\) into the given equation, we get
\(\frac{2+{k}^{2}z}{k+\frac{1}{z}}=kz\)
\(z(2+{k}^{2}z)=kz(kz+1)\)
\(2z+{k}^{2}{z}^{2}={k}^{2}{z}^{2}+kz\)
\(2z=kz\)
\(2=k\text{ }⟹\text{ }k=2\)
Substituting \(k=2\) into \(k+i{k}^{2}\), we get, \(2+i({2}^{2})=2+4i\).
This corresponds to the point \(P(2,4)\).
The given circle is \(\mathrm{∣}z−(1+2i)\mathrm{∣}=1\). This is a circle with:
Center (\(C\)): \(1+2i\), which is the point \((1,2)\).
Radius (\(r\)): \(1\).
The distance \(d\) between the point \(P(2,4)\) and the center of the circle \(C(1,2)\) is:
\(d=\sqrt{(2−1{)}^{2}+(4−2{)}^{2}}=\sqrt{{1}^{2}+{2}^{2}}=\sqrt{1+4}=\sqrt{5}\)
Maximum distance \(=d+r=\sqrt{5}+1\)
Let \(O\) be the origin, the point \(A\) be \({z}_{1}=\sqrt{3}+2\sqrt{2}i\), the point \(B({z}_{2})\) be such that \(\sqrt{3}\left|z_2\right|=\left|z_1\right|\) and \(\arg \left(z_2\right)=\arg \left(z_1\right)+\frac{\pi}{6}\). Then
\(ABO\) is an obtuse angled isosceles triangle
\(\left|{z}_{1}\right|=\sqrt{3+8}=\sqrt{11}\)
Then, \(\left|{z}_{2}\right|=\sqrt{11}/\sqrt{3}=\sqrt{11/3}\)
In \(△ABO,\) (where is the origin),
sides are \(OA=\sqrt{11}\)
and \(OB=\sqrt{11/3}\)
The angle
\(∠AOB=\left|\text{arg}\left({z}_{2}\right)−\text{arg}\left({z}_{1}\right)\right|=\frac{\pi }{6}\)
Using cosine rule for
\(A{B}^{2}=O{A}^{2}+O{B}^{2}−2\left(OA\right)\left(OB\right)\)
\(\text{cos}\left(\frac{\pi }{6}\right)=11+\frac{11}{3}−2\left(\sqrt{11}\right)\left(\sqrt{\frac{11}{3}}\right)\left(\frac{\sqrt{3}}{2}\right)\)
\(=11+\frac{11}{3}−11=\frac{11}{3}\)
Thus, \(AB=\sqrt{11/3}=OB\)
The triangle is isosceles.
Check for obtuse angle:
\(O{A}^{2}=11,\)
\(O{B}^{2}+A{B}^{2}=\frac{11}{3}+\frac{11}{3}=\frac{22}{3}\approx 7.33\)
\(∵O{A}^{2}>O{B}^{2}+A{B}^{2}\), the angle at B is obtuse.
ABO is an obtuse angled isosceles triangle.
Let \(\mathrm{z}\in \mathrm{C}\) be such that \(\frac{{\mathrm{z}}^{2}+3\mathrm{i}}{\mathrm{z}-2+\mathrm{i}}=2+3\mathrm{i}\). Then the sum of all possible values of \({z}^{2}\) is
[JEE Main 2025, 3 Apr (Shift 1)]
\(-19-2\mathrm{i}\)
\(\frac{z^2+3i}{z-2+i}=2+3i\)
\(z^2+3i=(z-2+i)(2+3i)\)
\(z^2+3i=2z-4+2i+3iz-6i-3\)
\(z^2+3i=(2z-7)+i(3z-4)\)
\(z^2-(2+3i)z+(7+7i)=0\)
This is a quadratic in \(z\).
\(z_1+z_2=2+3i\)
\(z_1z_2=7+7i\)
\(z_1^2+z_2^2=(z_1+z_2)^2-2z_1z_2\)
\(=(2+3i)^2-2(7+7i)\)
\(=4-9+12i-14-14i\)
\(=-19-2i\)
\(\text { If }\left|\frac{z}{z+i}=2\right| \text { represents a circle with centre } P \text { then distance of } P \text { from } D \text { is (where } D:(1,5) \text { ) }\)
\(\sqrt{\frac{370}{9}}\)
Let \(z=x+i y\)
\(\begin{aligned}& |z|=2|z+i| \\& \sqrt{x^2+y^2}=2 \sqrt{x^2+(y+1)^2} \\& x^2+y^2=4\left(x^2+(y+1)^2\right) \\& C: 3 x^2+3 y^2+8 y+4=0 \\& \therefore \quad P\left(0, \frac{-4}{3}\right)\end{aligned}\)
Now \(P D: \sqrt{1^2+\left(5+\frac{4}{3}\right)^2}=\sqrt{1+\frac{361}{9}}=\sqrt{\frac{370}{9}}\)
If \(z=x+ i y, x y \neq 0\), satisfies the equation \(z^2+ i \bar{z}=0\), then \(\left| z ^2\right|\) is equal to :
[JEE Main 2024, 30 Jan (Shift 1)]
1
Given \(z=x+iy\)
\(\overline{z}=x-iy\)
Substituting into
\(z^2+i\overline{z}=0\)
\((x+iy)^2+i(x-iy)=0\)
\(x^2-y^2+2ixy+ix+y=0\)
Comparing real and imaginary parts,
\(x^2-y^2+y=0\) and \(2xy+x=0\)
From \(x(2y+1)=0\)
and \(x\ne0\) \(2y+1=0\)
\(y=-\dfrac12\)
Substituting into
\(x^2-y^2+y=0\)
\(x^2-\dfrac14-\dfrac12=0\)
\(x^2=\dfrac34\)
Hence \(|z|^2=x^2+y^2\)
\(=\dfrac34+\dfrac14=1\)
Therefore
\(|z^2|=|z|^2=1\)
If \(z=x+ i y, x y \neq 0\), satisfies the equation \(z^2+ i \bar{z}=0\), then \(\left| z ^2\right|\) is equal to :
[JEE Main 2024, 30 Jan (Shift 1)]
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Among the statements
(S1) : The set \(\left\{\mathrm{z}\in \mathrm{ℂ}-{-\mathrm{i}}:|\mathrm{z}|=1\right.\) and \(\frac{\mathrm{z}-\mathrm{i}}{\mathrm{z}+\mathrm{i}}\) is purely real} contains exactly two elements, and
(S2) : The set \(\left\{\mathrm{z}\in \mathrm{ℂ}-{-1}:|\mathrm{z}|=1\right.\) and \(\frac{\mathrm{z}-1}{\mathrm{z}+1}\) is purely imaginary} contains infinitely many elements.
[JEE Main 2025, 7 Apr (Shift 1)]
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Let \(\alpha ,\alpha +2,\alpha \in \mathrm{ℤ}\), be the roots of the quadratic equation \(x(x+2)+(x+1)(x+3)+(x+2)(x+4)+\ldots\)\(+(x+n-1)(x+n+1)=4 n\) for some \(n \in \mathbb{N}\). Then \(n+\alpha\) is equal to:
[JEE Main 2026, 2 Apr (Shift 1)]
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If \({\mathrm{z}}_{1},{\mathrm{z}}_{2},{\mathrm{z}}_{3}\in \mathrm{C}\) are the vertices of an equilateral triangle, whose centroid is \({z}_{0}\), then \(\sum _{k=1}^{3}{\left({z}_{k}-{z}_{0}\right)}^{2}\) is equal to
[JEE Main 2025, 3 Apr (Shift 2)]
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The sum of all possible values of \(\theta \in [-\pi ,2\pi ]\), for which \(\frac{1+i\cos \theta }{1-2i\cos \theta }\) is purely imaginary, is equal to :
[JEE Main 2024, 8 Apr (Shift 2)]
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Let \(\alpha\) and \(\beta\) be the sum and the product of all the non-zero solutions of the equation \((\overset{¯}{z}{)}^{2}+|z|=0\), \(z\in C\). Then \(4\left({\alpha }^{2}+{\beta }^{2}\right)\) is equal to :
[JEE Main 2024, 04 Apr (Shift 1)]
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Let \(\alpha\) and \(\beta\) be the sum and the product of all the non-zero solutions of the equation \((\overset{¯}{z}{)}^{2}+|z|=0\), \(z\in C\). Then \(4\left({\alpha }^{2}+{\beta }^{2}\right)\) is equal to :
[JEE Main 2024, 04 Apr (Shift 1)]
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Let \({z}_{1}\) and \({z}_{2}\) be two roots of the equation \({z}^{2}+az+b=0,\) being complex. Further assume that the origin, \({z}_{1}\) and \({z}_{2}\) form an equilateral triangle. Then
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Let \(\left|\frac{\overset{¯}{\mathrm{z}}-\mathrm{i}}{2\overset{¯}{\mathrm{z}}+\mathrm{i}}\right|=\frac{1}{3},\mathrm{z}\in \mathrm{ℂ},\) be the equation of a circle with center at C. If the area of the triangle, whose vertices are at the points \((0,0), \mathrm{C}\) and \((\alpha, 0)\) is 11 square units, then \( \alpha^2\) equals
[JEE Main 2025, 23 Jan (Shift 1)]
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If \({x}^{2}-(3-2i)x-(2i-2)=0\) has roots \(\alpha +i\beta\) and \(\alpha \gamma +i\delta\) find the value of \(\alpha \gamma +\beta \delta\).
(where \(\alpha ,\beta ,\gamma ,\delta \in I\))
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Let \(z_1, z_2 \in \mathrm{C}\) be the distinct solutions of the equation \(z^2+4 z-(1+ 12 i)=0\). Then, \(\left|z_1\right|^2+\left|z_2\right|^2\) is equal to:
[JEE Main 2026, 5 Apr (Shift 2)]
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Let \(S =\{ z \in C :|z-1|=1\) and \((\sqrt{2}-1)(z+\bar{z})-i(z-\bar{z})=2 \sqrt{2}\}\). Let \(z_1, z_2 \in S\) be such that \(\left|z_1\right|=\max _{Z \in S}|z|\) and \(\left|z_2\right|=\min _{z \in S}|z|\). Then \(\left|\sqrt{2} z_1-z_2\right|^2\) equals :
[JEE Main 2024, 1 Feb (Shift 1)]
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\({z}_{1}=\sqrt{3}+2\sqrt{2}i\&\sqrt{3}\left|{Z}_{1}\right|=\left|{Z}_{2}\right|\text{ and }\arg \left({z}_{2}\right)=\arg \left({z}_{1}\right)+\frac{\pi }{6}\text{ then area }\)\(\text{ of triangle with vertices }{z}_{1},{z}_{2}\text{ and origin. }\) (28 Jan, Shift I, Memory Based)
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The number of values of \(z\in C\), satisfying the equations \(|z-(4+8i)|=\sqrt{10}\) and \(|z-(3+5i)|+|z-(5+11i)|=4\sqrt{5}\), is:
[JEE Main 2026, 8 Apr (Shift 2)]
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Let \(\left|\frac{\overset{¯}{\mathrm{z}}-\mathrm{i}}{2\overset{¯}{\mathrm{z}}+\mathrm{i}}\right|=\frac{1}{3},\mathrm{z}\in \mathrm{ℂ},\) be the equation of a circle with center at C. If the area of the triangle, whose vertices are at the points \((0,0),\mathrm{C}\text{ and }(\alpha ,0)\text{ is }11\) square units, then \( \alpha^2\) equals
[JEE Main 2025, 23 Jan (Shift 1)]
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Let \(r\) and \(\theta\) respectively be the modulus and amplitude of the complex number \(z=2-i\left(2 \tan \frac{5 \pi}{8}\right)\), then \(( r , \theta)\) is equal to
[JEE Main 2024, 29 Jan (Shift 2)]
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Let \(r\) and \(\theta\) respectively be the modulus and amplitude of the complex number \(z=2-i\left(2 \tan \frac{5 \pi}{8}\right)\), then \(( r , \theta)\) is equal to
[JEE Main 2024, 29 Jan (Shift 2)]
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If the set \(R=\{(a, b): a+5 b=42, a, b \in N \}\) has \(m\) elements and \(\sum_{n=1}^m\left(1-i^{n !}\right)=x+i y\), where \(i=\sqrt{-1}\), then the value of \(m+x+y\) is
[JEE Main 2024, 8 Apr (Shift 1)]
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If the set \(R=\{(a, b): a+5 b=42, a, b \in N \}\) has \(m\) elements and \(\sum_{n=1}^m\left(1-i^{n !}\right)=x+i y\), where \(i=\sqrt{-1}\), then the value of \(m+x+y\) is
[JEE Main 2024, 8 Apr (Shift 1)]
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Let \(\left|z_{i}\right|=1\) for \(i=1,2,3\) satisfying \(\left|\bar{z}_1 z_2+\bar{z}_2 z_3+\bar{z}_3 z_1\right|^2=a+b \sqrt{2}\), where \(a, b\) an rational numbers such that \(\arg \left(z_1\right)=\frac{\pi}{4}, \arg \left(z_2\right)=0\) and \(\arg \left(z_3\right)=\frac{-\pi}{4}\), then find \((a, b)\)
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The number of complex numbers \(z\) , satisfying \(|\mathrm{z}|=1\) and \(\left|\frac{\mathrm{z}}{\overline{\mathrm{z}}}+\frac{\overline{\mathrm{z}}}{\mathrm{z}}\right|=1\), is :
[JEE Main 2025, 23 Jan (Shift 2)]
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Let \({z}_{1},{z}_{2}\text{and }{z}_{3}\) be three complex numbers on the circle \(|z|=1\text{ with }\arg \left({z}_{1}\right)=\frac{-\pi }{4},\arg \left({z}_{2}\right)=0\) and \(\arg \left({z}_{3}\right)=\frac{\pi }{4}.\)
If \({\left|{\mathrm{z}}_{1}{\overset{¯}{\mathrm{z}}}_{2}+{\mathrm{z}}_{2}{\overset{¯}{\mathrm{z}}}_{3}+{\mathrm{z}}_{3}{\overset{¯}{\mathrm{z}}}_{1}\right|}^{2}=\alpha +\beta \sqrt{2},\alpha ,\beta \in \mathrm{Z}\) then the value of \({\alpha }^{2}+{\beta }^{2}\) is
[JEE Main 2025, 22 Jan (Shift 1)]
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If \(z=\frac{1}{2}-2 i\) is such that \(|z+1|=\alpha z+\beta(1+i), i=\sqrt{-1}\) and \(\alpha, \beta \in \mathbb{R}\), then \(\alpha+\beta\) is equal to
[JEE Main 2024, 29 Jan (Shift 1)]
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If \(z=\frac{1}{2}-2 i\) is such that \(|z+1|=\alpha z+\beta(1+i), i=\sqrt{-1}\) and \(\alpha, \beta \in \mathbb{R}\), then \(\alpha+\beta\) is equal to
[JEE Main 2024, 29 Jan (Shift 1)]
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If \({z}_{1},{z}_{2}\) are two distinct complex number such that \(\left|\frac{{z}_{1}-2{z}_{2}}{\frac{1}{2}-{z}_{1}{\overset{¯}{z}}_{2}}\right|=2\), then
[JEE Main 2024, 6 Apr (Shift 2)]
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If \({z}_{1},{z}_{2}\) are two distinct complex number such that \(\left|\frac{{z}_{1}-2{z}_{2}}{\frac{1}{2}-{z}_{1}{\overset{¯}{z}}_{2}}\right|=2\), then
[JEE Main 2024, 6 Apr (Shift 2)]
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If \({z}_{1},{z}_{2}\) are two distinct complex number such that \(\left|\frac{{z}_{1}-2{z}_{2}}{\frac{1}{2}-{z}_{1}{\overset{¯}{z}}_{2}}\right|=2\), then
[JEE Main 2024, 6 Apr (Shift 2)]
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Let \(\text{S}={z\in ℂ:\left|\frac{z−6i}{z−2i}\right|=1\) and \(\left|\frac{z−8+2i}{z+2i}\right|=\frac{3}{5}}\). Then \(\sum _{z\in \text{ s}}|z{|}^{2}\) is equal to
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The points represented by the complex numbers \(1+i,-2+3 i, \frac{5}{3} i\) on the Argand plane are:
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The area (in sq. units) of the region \(S={z\in C:|z-1|\leq 2;(z+\overset{¯}{z})+i(z-\overset{¯}{z})\leq 2,Im(z)\geq 0}\) is
[JEE Main 2024, 4 Apr (Shift 2)]
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Let the set of all values of \(k\in R\) such that the equation \(z(\overset{¯}{z}+2+i)+k(2+3i)=0,z\in C\), has at least one solution, be the interval \([\alpha ,\beta ]\). Then \(9(\alpha +\beta )\) is equal to:
[JEE Main 2026, 6 Apr (Shift 1)]
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Let \(\left|\frac{\overset{¯}{\mathrm{z}}-\mathrm{i}}{2\overset{¯}{\mathrm{z}}+\mathrm{i}}\right|=\frac{1}{3},\mathrm{z}\in \mathrm{ℂ},\) be the equation of a circle with center at C. If the area of the triangle, whose vertices are at the points \((0,0),\mathrm{C}\text{ and }(\alpha ,0)\text{ is }11\) square units, then \( \alpha^2\) equals
[JEE Main 2025, 23 Jan (Shift 1)]
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If \({x}^{2}-(3-2i)x-(2i-2)=0\) has roots \(\alpha +i\beta\) and \(\alpha \gamma +i\delta\) find the value of \(\alpha \gamma +\beta \delta\).
(where \(\alpha ,\beta ,\gamma ,\delta \in I\))
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If \(\alpha\) and \(\beta\) are the roots of the equation \(2{z}^{2}-3z-2i=0,\text{ where }i=\sqrt{-1}\), then \(16\cdot Re\left(\frac{{\alpha }^{19}+{\beta }^{19}+{\alpha }^{11}+{\beta }^{11}}{{\alpha }^{15}+{\beta }^{15}}\right)\cdot Im\left(\frac{{\alpha }^{19}+{\beta }^{19}+{\alpha }^{11}+{\beta }^{11}}{{\alpha }^{15}+{\beta }^{15}}\right)\) is equal to
[JEE Main 2025, 24 Jan (Shift 1)]
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The points represented by the complex numbers \(1+i,-2+3 i, \frac{5}{3} i\) on the Argand plane are:
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Let z be a complex number such that \(|z|=1\). If \(\frac{2+{\mathrm{k}}^{2}\mathrm{z}}{\mathrm{k}+\overset{¯}{\mathrm{z}}}=\mathrm{kz},\mathrm{k}\in R\), then the maximum distance of \(k+i{k}^{2}\) from the circle \(|\mathrm{z}-(1+2\mathrm{i})|=1\) is:
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\(\text { If }\left|\frac{z}{z+i}=2\right| \text { represents a circle with centre } P \text { then distance of } P \text { from } D \text { is (where } D:(1,5) \text { ) }\)
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For a non-zero complex number \(z\), let \(\arg (z)\) denote the principal argument of \(z\), with \(-\pi<\arg (z) \leq \pi\). Let \(\omega\) be the cube root of unity for which \(0<\arg (\omega)<\pi\). Let \(\alpha=\arg \left(\sum_{n=1}^{2025}(-\omega)^n\right)\) . Then the value of \(\frac{3 \alpha}{\pi}\) is_________.
[JEE Advanced 2025]
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Let \(S=\left\{z\in C:{z}^{2}+4z+16=0\right\}\). Then \(\sum _{z\in S}|z+\sqrt{3}i{|}^{2}\) is equal to
[JEE Main 2026, 4 Apr (Shift 2)]
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Let \(x\) and \(y\) be real numbers such that \(50\left(\frac{2x}{1+3i}-\frac{y}{1-2i}\right)=31+17i\), \(i=\sqrt{-1}\). Then the value of \(10(x-3y)\) is:
[JEE Main 2026, 2 Apr (Shift 1)]
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Consider the following two statements:
Statement I: For any two non-zero complex numbers \({z}_{1},{z}_{2},\left(\left|{z}_{1}\right|+\left|{z}_{2}\right|\right)\left|\frac{{z}_{1}}{\left|{z}_{1}\right|}+\frac{{z}_{2}}{\left|{z}_{2}\right|}\right|\leq 2\left(\left|{z}_{1}\right|+\left|{z}_{2}\right|\right)\), and
Statement II : If \(x, y, z\) are three distinct complex numbers and \(a, b, c\) are three positive real numbers such that\(\frac{a}{|y-z|}=\frac{b}{|z-x|}=\frac{c}{|x-y|}\), then \(\frac{{a}^{2}}{y-z}+\frac{{b}^{2}}{z-x}+\frac{{c}^{2}}{x-y}=1.\)
Between the above two statements,
[JEE Main 2024, 5 Apr (Shift 1)]
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Consider the following two statements:
Statement I: For any two non-zero complex numbers \({z}_{1},{z}_{2},\left(\left|{z}_{1}\right|+\left|{z}_{2}\right|\right)\left|\frac{{z}_{1}}{\left|{z}_{1}\right|}+\frac{{z}_{2}}{\left|{z}_{2}\right|}\right|\leq 2\left(\left|{z}_{1}\right|+\left|{z}_{2}\right|\right)\), and
Statement II : If \(x, y, z\) are three distinct complex numbers and \(a, b, c\) are three positive real numbers such that\(\frac{a}{|y-z|}=\frac{b}{|z-x|}=\frac{c}{|x-y|}\), then \(\frac{{a}^{2}}{y-z}+\frac{{b}^{2}}{z-x}+\frac{{c}^{2}}{x-y}=1.\)
Between the above two statements,
[JEE Main 2024, 5 Apr (Shift 1)]
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If the locus of \(z\in C\), such that \(Re\left(\frac{z-1}{2z+\mathrm{i}}\right)+Re\left(\frac{\overset{¯}{z}-1}{2\overset{¯}{z}-\mathrm{i}}\right)=2,\) is a circle of radius r and center \((a,b)\) then \(\frac{15ab}{{r}^{2}}\) is equal to:
[JEE Main 2025, 7 Apr (Shift 2)]
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Let \(\mathrm{A}=\) \(\left\{\theta \in [0,2\pi ]:1+10Re\left(\frac{2\mathrm{cosθ}+\mathrm{isinθ}}{\mathrm{cosθ}-3\mathrm{isinθ}}\right)=0\right\}.\) Then \(\sum _{\theta \in \mathrm{A}}{\theta }^{2}\) is equal to
[JEE Main 2025, 8 Apr (Shift 1)]
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Let \(z\) be a complex number such that \(|z+2|=|z-2|\) and \(\text{arg}\left(\frac{z+3}{z-i}\right)=\frac{\pi }{4}\) then \(|z{|}^{2}\) is equal to:
[JEE Main 2026, 4 Apr (Shift 1)]
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If \(z\) is a complex number, then the number of common roots of the equations \(z^{1985}+z^{100}+1=0\) and \(z^3+2 z^2+2 z+1=0\), is equal to
[JEE Main 2024, 30 Jan (Shift 2)]
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If \(z=\frac{\sqrt{3}}{2}+\frac{i}{2},i=\sqrt{−1}\), then \(\left(z^{201}-i\right)^8\) is equal to
[JEE Main 2026, 23 Jan (Shift 2)]
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Let \(z\) be a complex number such that \(|z+2|=1\) and \(\operatorname{Im}\left(\frac{z+1}{z+2}\right)=\frac{1}{5}\). Then the value of \(|\operatorname{Re}(\overline{z+2})|\) is
[JEE Main 2024, 8 Apr (Shift 1)]
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Let \(z\) be a complex number such that \(|z+2|=1\) and \(\operatorname{Im}\left(\frac{z+1}{z+2}\right)=\frac{1}{5}\). Then the value of \(|\operatorname{Re}(\overline{z+2})|\) is
[JEE Main 2024, 8 Apr (Shift 1)]
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Let the product of \({\omega }_{1}=(8+\mathrm{i})\mathrm{sinθ}+(7+4\mathrm{i})\mathrm{cosθ}\) and \({\omega }_{2}=(1+8\mathrm{i})\mathrm{sinθ}+(4+7\mathrm{i})\mathrm{cosθ}\) be \(\alpha +\mathrm{iβ}\), \(\mathrm{i}=\sqrt{-1}\). Let \(p\) and \(q\)be the maximum and the minimum values of \(\alpha +\beta\) respectively. Then \(p+q\) is
[JEE Main 2025, 4 Apr (Shift 2)]
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If \(z=\frac{\sqrt{3}}{2}+\frac{i}{2},i=\sqrt{−1}\), then \(\left(z^{201}-i\right)^8\) is equal to
[JEE Main 2026, 23 Jan (Shift 2)]
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Let \(\mathrm{A}=\) \(\left\{\theta \in [0,2\pi ]:1+10Re\left(\frac{2\mathrm{cosθ}+\mathrm{isinθ}}{\mathrm{cosθ}-3\mathrm{isinθ}}\right)=0\right\}.\) Then \(\sum _{\theta \in \mathrm{A}}{\theta }^{2}\) is equal to
[JEE Main 2025, 8 Apr (Shift 1)]
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If \(\alpha+i \beta\) and \(\gamma+i \delta\) are the roots of \(x^2-(3-2 i) x-(2 i-2)=0, i=\sqrt{-1}\), then \(\alpha \gamma+\beta \delta\) is equal to :
[JEE Main 2025, 28 Jan (Shift 2)]
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Let \(S=\left\{z\in C:{z}^{2}+\sqrt{6}iz-3=0\right\}\). Then \(\sum _{z\in S}{z}^{8}\) is equal to:
[JEE Main 2026, 6 Apr (Shift 2)]
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Let \(a, b \in C\). Let \(\alpha, \beta\) be the roots of the equation \(x^2+a x+b=0\). If \(\beta-\alpha=\sqrt{11 } \) and \(\beta^2-\alpha^2\) \(=3 i \sqrt{11}\), then \(\left(\beta^3-\alpha^3\right)^2\) is equal to:
[JEE Main 2026, 5 Apr (Shift 1)]
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For \(a \in C\), let \(A=\{z \in C: \operatorname{Re}(a+\bar{z})>\operatorname{Im}(\bar{a}+z)\}\) and \(B=\{z \in C: \operatorname{Re}(a+\bar{z})<\operatorname{Im}(\bar{a}+z)\}\). Then among the two statements:
(S1): If \(\operatorname{Re}(A), \operatorname{Im}(A)>0\), then the set \(A\) contains all the real numbers.
(S2): If \(\operatorname{Re}(A), \operatorname{Im}(A)<0\), then the set \(B\) contains all the real numbers,
[JEE Main 2023, 11 Apr (Shift 2)]
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The number of real solutions of the equation \((x-1)^2+(x-2)^2+(x-3)^2=0\) is
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Let \(p, q \in R\) and \((1-\sqrt{3} i)^{200}=2^{199}(p+i q), i=\sqrt{-1}\) Then \(p+q+q^2\) and \(p-q+q^2\) are roots of the equation.
[JEE Main 2023, 24 Jan (Shift 1)]
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If \( z_{1}, z_{2} \) are complex numbers such that \( \operatorname{Re}\left(z_{1}\right)=\left|z_{1}-1\right|, \operatorname{Re}\left(z_{2}\right)=\left|z_{2}-1\right| \), and \( \arg \left(z_{1}-z_{2}\right)=\frac{\pi} { 6} \), then \( \operatorname{Im}\left(z_{1}+z_{2}\right) \) is equal to
[JEE Main 2020, 3 Sep (Shift 2)]
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Let \(a, b\) be two real numbers such that \(a b<0\). If the complex number \(\frac{1+a i}{b+i}\) is of unit modulus and \(a+i b\) lies on the circle \(|z-1|=|2 z|\), then a possible value of \(\frac{1+[a]}{4 b}\), where \([t]\) is greatest integer function, is:
[JEE Main 2023, 1 Feb (Shift 2)]
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The value of \(\left(\frac{-1+i \sqrt{3}}{1-i}\right)^{30}\) is:
[JEE Main 2020, 5 Sep (Shift 2)]
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If the equation \(a|z|^2+\overline{\bar{a} z+\alpha \bar{z}}+d=0\) represents a circle where \(a, d\) are real constants, then which of the following condition is correct?
[JEE Main 2021, 18 Mar (Shift 1)]
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If the center and radius of the circle \(\left|\frac{z-2}{z-3}\right|=2\) are respectively \((\alpha, \beta)\) and \(\gamma\), then \(3(\alpha+\beta+\gamma)\) is equal to
[JEE Main 2023, 1 Feb (Shift 1)]
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Let \(w_1\) be the point obtained by the rotation of \(z_1=5+4 i\) about the origin through a right angle in the anticlockwise direction, and \(w_2\) be the point obtained by the rotation of \(z_2=3+5 i\) about the origin through a right angle in the clockwise direction. Then the principal argument of \(w_1-w_2\) is equal to
[JEE Main 2023, 11 Apr (Shift 1)]
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Let a complex number \( z,|z| \neq 1 \), satisfy \( \log _{\frac{1}{\sqrt{2}}}\left(\frac{|z|+11}{(|z|-1)^{2}}\right) \leq 2 \). Then the largest value of \( |z| \) is equal to
[JEE Main 2021, 16 Mar (Shift 1)]
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Let \(z\) be a complex number such that \(\left|\frac{z-2 i}{z+i}\right|=2, z \neq-i\). Then \(z\) lies on the circle of radius 2 and centre
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\(\mathbf{i}^{57}+\frac{1}{i^{25}}\), when simplified has the value
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If \(S=\left\{z \in C: \frac{z-i}{z+2 i} \in R\right\}\), then
[JEE Main 2021, 27 Aug (Shift 1)]
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\(\text{ Let }C\text{ be the set of all complex numbers. Let }\\ {S}_{1}=\left\{z\in C||z-3-{\left.2i\right|}^{2}=8\right\},\\ {S}_{2}={z\in C∣Re(z)\geq 5}\text{ and }\\ {S}_{3}={z\in C||z-\overset{¯}{z}∣\geq 8}.\\ \text{ Then the number of element in }{S}_{1}\cap {S}_{2}\cap {S}_{3}\text{ is equal to: }\)
[JEE Main 2021, 27 Jul (Shift 1)]
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For all \(z \in C\) on the curve \(C_1:|z|=4\), let the locus of the point \(z+\frac{1}{z}\) be the curve \(C _2\). Then
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Let \( u=\frac{2 z+i}{z-k i}, \mathrm{z}=x+i y \) and \( k>0 \). If the curve represented by \( \operatorname{Re}(u)+\operatorname{Im}(u)=1 \) intersects the \( y \)-axis at the point \( P \) and \( Q \) where \( P Q=5 \), then the value of \( k\) is
[JEE Main 2020, 4 Sep (Shift 1)]
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Let \(S=\left\{Z \in C: \bar{z}=i\left(z^2+\operatorname{Re}(\bar{z})\right)\right\}\). Then \(\sum_{z \in S}|z|^2\) is equal to
[JEE Main 2023, 13 Apr (Shift 2)]
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Let \({z}_{1}=2+3i\) and \({z}_{2}=3+4i.\) The set S = \({z\in C:|z–{z}_{1}{|}^{2}-|z–{z}_{2}{|}^{2}=|{z}_{1}–{z}_{2}{|}^{2}}\) represents a
[JEE Main 2023, 25 Jan (Shift 1)]
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The value of \( \left(\frac{1+\sin \frac{2 \pi}{9}+i \cos \frac{2 \pi}{9}}{1+\sin \frac{2 \pi}{9}-i \cos \frac{2 \pi}{9}}\right)^{3} \) is
[JEE Main 2020, 2 Sep (Shift 1)]
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For two non-zero complex number \(z_1\) and \(z_2\), if \(\operatorname{Re}\left(z_1 z_2\right)\) \(=0\) and \(\operatorname{Re}\left(z_1+z_2\right)=0\), then which of the following are possible?
A. \(\operatorname{Im}\left(z_1\right)>0\) and \(\operatorname{Im}\left(z_2\right)>0\)
B. \(\operatorname{Im}\left(z_1\right)<0\) and \(\operatorname{Im}\left(z_2\right)>0\)
C. \(\operatorname{Im}\left(z_1\right)>0\) and \(\operatorname{Im}\left(z_2\right)<0\)
D. \(\operatorname{Im}\left(z_1\right)<0\) and \(\operatorname{Im}\left(z_2\right)<0\)
Choose the correct answer from the options given below:
[JEE Main 2023, 29 Jan (Shift 1)]
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Let \(z\) be a complex number such that \(\left|\frac{z-2 i}{z+i}\right|=2, z \neq-i\). Then \(z\) lies on the circle of radius 2 and centre
[JEE Main 2023, 25 Jan (Shift 2)]
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Let \(A=\left\{\theta \in\left(-\frac{\pi}{2}, \pi\right): \frac{3+2 i \sin \theta}{1-2 i \sin \theta}\right.\) is purely imaginary \(\}\). Then the sum of the elements in \(\mathrm{A}\) is
[JEE Main 2023, 8 Apr (Shift 2)]
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Let the lines \( (2-\mathrm{i}) z=(2+\mathrm{i}) \bar{z} \) and \( (2+\mathrm{i}) z+(\mathrm{i}-2) \bar{z}-4 \mathrm{i}=0\) here \( i^{2}=-1 \) be normal to a circle \( \mathrm{C} \). If the line \( i z+\bar{z}+1+i=0 \) is tangent to this circle \( C \), then its radius is:
[JEE Main 2021, 25 Feb (Shift 1)]
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If the set \(\left\{\mathrm{Re}\left(\frac{z−\overset{¯}{z}+z\overset{¯}{z}}{2−3z+5\overset{¯}{z}}\right):z\in C,\mathrm{Re}(z)=3\right\}\) is equal to the interval \((\alpha ,\beta ]\), then 24 \((\beta −\alpha )\) is equal to
[JEE Main 2023, 15 Apr (Shift 1)]
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If \( z \) be \( a \) complex number satisfying \( |\operatorname{Re}(z)|+|\operatorname{Im}(z)|=4 \), then \( |z| \) cannot be
[JEE Main 2020, 9 Jan (Shift 2)]
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The imaginary part of \( (3+2 \sqrt{-54})^{1 / 2}-(3-2 \sqrt{-54})^{1 / 2} \) can be:
[JEE Main 2020, 2 Sep (Shift 2)]
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Let \(z_1=2+3 i\) and \(z_2=3+4 i\). The set \(S =\left\{z \in C :\left|z-z_1\right|^2\right.\) \(\left.-\left|z-z_2\right|^2=\left|z_1-z_2\right|^2\right\}\) represents a
[JEE Main 2023, 25 Jan (Shift 1)]
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If the equation, \(x^2+b x+45=0(b \in R)\) has conjugate complex roots and they satisfy \(|z+1|=2 \sqrt{10}\), then
[JEE Main 2020, 8 Jan (Shift 1)]
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Let the complex number \(z=x+i y\) be such that \(\frac{2 z-3 i}{2 z+i}\) is purely imaginary. If \(x+y^2=0\), then \(y^4+y^2-y\) is equal to:
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If for \(z=\alpha+i \beta,|z+2|=z+4(1+i)\), then \(\alpha+\beta\) and \(\alpha \beta\) are the roots of the equation
[JEE Main 2023, 8 Apr (Shift 1)]
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If \((\sqrt{3}+i)^{100}=2^{99}(p+i q)\). Then \(p\) and \(q\) are roots of the equation:
[JEE Main 2021, 26 Aug (Shift 2)]
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Let α = 8 – 14i, \(A=\left\{z\in C:\frac{\alpha z-\overset{¯}{\alpha }\overset{¯}{z}}{{z}^{2}-{(\overset{¯}{z})}^{2}-112i}=1\right\}\) and \(B={z\in C:|z+3i|=4}\). Then \(\sum _{z\in A\cap B}(Rez-Imz)\) is equal to ______.
[JEE Main 2023, 29 Jan (Shift 2)]
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If \( \operatorname{Re}\left(\frac{z-1}{2 z+i}\right)=1 \), where \( z=x+i y \), then the point \( (x, y) \) lies on a
[JEE Main 2020, 7 Jan (Shift 1)]
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If the four complex numbers \( z, \bar{z}, \bar{z}-2 \operatorname{Re}(\bar{z}) \) and \( z-2 \operatorname{Re}(z) \) represent the vertices of a square of side 4 units in the Argand plane, then \( |z| \) is equal to
[JEE Main 2020, 5 Sep (Shift 1)]
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Let a complex number \( z,|z| \neq 1 \), satisfy \( \log _{\frac{1}{\sqrt{2}}}\left(\frac{|z|+11}{(|z|-1)^{2}}\right) \leq 2 \). Then, the largest value of \( |z| \) is equal to
[JEE Main 2021, 16 Mar (Shift 1)]
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Let \(\mathrm{C}\) be the set of all complex numbers. Let \(S_1=\{z \in C:|z-2| \leqslant 1\} \text { and } S_2=\{z \in C: z(1+i)+\bar{z}(1-i) \geqslant 4\}\). Then, the maximum value of \(\left|z-\frac{5}{2}\right|^2\) for \(z \in S_1 \cap S_2\) is equal to:
[JEE Main 2021, 27 Jul (Shift 2)]
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If \(\ z_1=\sqrt{2}\left[\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right] \) and \(\ z_2=\sqrt{3}\left[\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}\right] \), then \(\ \left|z_1 z_2\right| \) is equal to \(\ \sqrt{\mathbf{m}} \). Value of \(\ \mathbf{m} \) is
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Let \(\alpha, \beta\) be the roots of the equation \(x^2-\sqrt{2} x+2=0\). Then \(\alpha^{14}+\beta^{14}\) is equal to
[JEE Main 2023, 13 Apr (Shift 2)]
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If \( a \) and \( b \) are real numbers such that \( (2+\alpha)^{4}=a+b \alpha \) where \( \alpha=\frac{-1+i \sqrt{3}}{2} \) then \( a+b \) is equal to
[JEE Main 2020, 4 Sep (Shift 2)]
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If \(z\) and \(\omega\) are two complex numbers such that \(|z \omega|=1\) and \(\arg (z)-\arg (w)=\frac{3 \pi}{2}\), then \(\arg \left(\frac{1-2 \bar{z} \omega}{1+3 \bar{z} \omega}\right)\) is:
(Here \(\arg (z)\) denotes the principal argument of complex number \(z\) )
[JEE Main 2021, 20 Jul (Shift 1)]
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Let \( n \) denote the number of solutions of the equation \( z^{2}+3 \bar{z}=0 \), where \( z \) is a complex number. Then the value of \( \sum_{k=0}^{\infty} \frac{1}{n^{k}} \) is equal to
[JEE Main 2021, 22 Jul (Shift 2)]
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Let \( \mathrm{z} \) be complex number such that \( \left|\frac{\mathrm{z}-\mathrm{i}}{\mathrm{z}+2 \mathrm{i}}\right|=1 \) and \( |\mathrm{z}|=\frac{5}{2} \). Then the value of \( |\mathrm{z}+3 \mathrm{i}| \) is :
[JEE Main 2020, 9 Jan (Shift 1)]
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Let C be the set of all complex numbers. Let
\[\begin{aligned}& S_1=\{z \in C:|z-2| \leq 1\} \text { and }& S_2=\{z \in C: z(1+i)+\bar{z}(1-i) \geq 4\} .\end{aligned}\]
Then, the maximum value of \(\left|z-\frac{5}{2}\right|^2\) for \(z \in S_1 \cap S_2\) is equal to:
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\(\text{ Let }A=\left\{\theta \in (0,2\pi ):\frac{1+2i\sin \theta }{1-i\sin \theta }\text{ is purely imaginary }\right\}.\\ \text{Then the sum of the elements in }A\text{ is }\)
[JEE Main 2023, 08 Apr (Shift 2)]
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Let the complex number \(z=x+i y\) be such that \(\frac{2 z-3 i}{2 z+i}\) is purely imaginary. If \(x+y^2=0\), then \(y^4+y^2-y\) is equal to:
[JEE Main 2023, 10 Apr (Shift 1)]
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The value of \(\left(\frac{1+\sin \frac{2 \pi}{9}+i \cos \frac{2 \pi}{9}}{1+\sin \frac{2 \pi}{9}-i \cos \frac{2 \pi}{9}}\right)^3\) is:
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Let \(C\) be the set of all complex numbers. Let \(S_1=\{z \in C:|z-2| \leq 1\}\) and \(S_2=\{z \in C: z(1+i)+\bar{z}(1-i) \geq 4\}\).
Then, the maximum value of \(\left|z-\frac{5}{2}\right|^2\) for \(z \in S_1 \cap S_2\) is equal to:
[JEE Main 2021, 27 Jul (Shift 2)]
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If for \(z=\alpha+i \beta,|z+2|=z+4(1+i)\), then \(\alpha+\beta\) and \(\alpha \beta\) are the roots of the equation
[JEE Main 2023, 08 Apr (Shift 1)]
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If \(a, b, c\) are positive numbers, then least value of \((a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\) is
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If \(\ z \) is a complex number such that \(\ z^2=(\bar{z})^2 \), then
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If \(x, y\) and \(z\) are real numbers, then \(x^2+4 y^2+9 z^2-6 y z-3 z x-2 x y\) is always
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For two non-zero complex numbers \(z_1\) and \(z_2\), if \(\operatorname{Re}\left(z_1 z_2\right)\) \(=0\) and \(\operatorname{Re}\left(z_1+z_2\right)=0\), then which of the following are possible?
A. \(\operatorname{Im}\left(z_1\right)>0\) and \(\operatorname{Im}\left(z_2\right)>0\)
B. \(\operatorname{Im}\left(z_1\right)<0\) and \(\operatorname{Im}\left(z_2\right)>0\)
C. \(\operatorname{Im}\left(z_1\right)>0\) and \(\operatorname{Im}\left(z_2\right)<0\)
D. \(\operatorname{Im}\left(z_1\right)<0\) and \(\operatorname{Im}\left(z_2\right)<0\)
Choose the correct answer from the options given below:
[JEE Main 2023, 29 Jan (Shift 1)]
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Let \(A=\left\{\theta \in(0,2 \pi): \frac{1+2 i \sin \theta}{1-i \sin \theta}\right.\) is purely imaginary \(\}\). Then the sum of the elements in \(A\) is
[JEE Main 2019, 9 Jan (Shift 1)]
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Let \(S=\left\{z=x+iy:\frac{2z-3i}{4z+2i}\text{is a real number}\right\}\). Then which of the following is NOT correct?
[JEE Main 2023, 10 Apr (Shift 2)]
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The least value of \( |z| \) where \( z \) is complex number which satisfies the inequality \( \exp \left(\frac{(|z|+3)(|z|-1)}{|| z|+1|} \log _{e} 2\right) \geq \log _{\sqrt{2}}|5 \sqrt{7}+9 i|, i=\sqrt{-1} \), is equal to:
[JEE Main 2021, 16 Mar (Shift 2)]
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For all \(z \in C\) on the curve \(C_1:|z|=4\), let the locus of the point \(z+\frac{1}{z}\) be the curve \(C _2\). Then
[JEE Main 2023, 31 Jan (Shift 1)]
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The complex number \(z=\frac{i-1}{\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}}\) is equal to:
[JEE Main 2023, 31 Jan (Shift 2)]
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The value of \(\left(\frac{1+\sin \frac{2 \pi}{9}+i \cos \frac{2 \pi}{9}}{1+\sin \frac{2 \pi}{9}-i \cos \frac{2 \pi}{9}}\right)^3\) is
[JEE Main 2023, 24 Jan (Shift 2)]
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The modulus of the complex number \( z \) such that
\( |z+3-i|=1 \) and \( \arg (z)=\pi \) is equal to:
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The equation \(\arg \left(\frac{z-1}{z+1}\right)=\frac{\pi}{4}\) represents a circle with:
[JEE Main 2021, 26 Aug (Shift 1)]
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Let \(S={z\in C:\overset{¯}{z}=i({z}^{2}+\mathrm{Re}(\overset{¯}{z}))}\). Then \(\sum _{z\in S}|z{|}^{2}\) is equal to:
[JEE Main 2023, 13 Apr (Shift 2)]
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The region represented by \({z=x+iy\in C:|z|-Re(z)\leq 1}\) is also given by the inequality:
[JEE Main 2020, 6 Sep (Shift 1)]
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If \(\left|{z}_{1}\right|\text{ }=\text{ }\left|{z}_{2}\right|=\text{ }........\text{ }\left|{z}_{n}\right|=1,\) then the value of
\(\left|{z}_{1}+{z}_{2}+.......{z}_{n}\right|−\left|\frac{1}{{z}_{1}}+\frac{1}{{z}_{2}}+.....+\frac{1}{{z}_{n}}\right|\) is
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A complex number is given by \(z=\frac{1+2i}{1−{(1−i)}^{2}}\) ,what is the modulus of \(z\) ?
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The equation \( |z-i|=|z-1|, i=\sqrt{-1} \), represents
[JEE Main 2019, 12 Apr (Shift 1)]
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Let \(a\neq \pm b\) be two non-zero real numbers. Then the number of elements in the set \(X=\left\{z \in C: \operatorname{Re}\left(a z^2+b z\right)=a \text { and } \operatorname{Re}\left(b z^2+a z\right)=b\right\}\) is equal to
[JEE Main 2023, 6 Apr (Shift 2)]
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For two non-zero complex numbers z1 and z2, if Re(z1z2) = 0 and Re (z1 + z2) = 0, then which of the following are possible?
(i) Im (z1) > 0 and Im (z2) > 0
(ii) Im (z1) < 0 and Im (z2) > 0
(iii) Im (z1) > 0 and Im (z2) < 0
(iv) Im (z1) < 0 and Im (z2) < 0
Choose the correct answer from the options given below:
[JEE Main 2023, 29 Jan (Shift 1)]
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If \((\sqrt{3}+i)^{100}=2^{99}(p+i q)\). Then \(p\) and \(q\) are roots of the equation:
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If the set \(\left\{\operatorname{Re}\left(\frac{z-\bar{z}+z \bar{z}}{2-3 z+5 \bar{z}}\right): z \in C , \operatorname{Re}(z)=3\right\}\) is equal to the interval \((\alpha, \beta]\), then \(24(\beta-\alpha)\) is equal to
[JEE Main 2023, 15 Apr (Shift 1)]
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Let \( \alpha=\frac{-1+\mathrm{i} \sqrt{3}}{2} \). If \( \mathrm{a}=(1+\alpha) \sum_{\mathrm{k}=0}^{100} \alpha^{2 \mathrm{k}} \) and \( \mathrm{b}=\sum_{\mathrm{k}=0}^{100} \alpha^{3 \mathrm{k}} \), then \( \mathrm{a} \) and \( \mathrm{b} \) are the roots of the quadratic equation:
[JEE Main 2020, 8 Jan (Shift 2)]
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If \(\frac{3+i\sin \theta }{4-i\cos \theta },\theta \in [0,2\pi ]\)is a real number, then an argument of \(\sin \theta +i\cos \theta\) is :
[JEE Main 2020, 7 Jan (Shift 2)]
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If the equation \(a{\left|z\right|}^{2}+\overset{¯}{\overset{¯}{\alpha }z+\alpha \overset{¯}{z}}+d=0\) represents a circle where \(a, d\) are real constants, then which of the following condition is correct?
[JEE Main 2021, 18 Mar (Shift 1)]
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Let \( \theta_{1}, \theta_{2}, \ldots, \theta_{10} \) be positive valued angles (in radian) such that \( \theta_{1}+\theta_{2}+\ldots+\theta_{10}=2 \pi \). Define the complex numbers \( z_{1}=e^{i \theta_{1}}, z_{k}=z_{k-1} e^{i \theta_{k}} \) for \( k=2,3, \ldots, 10 \),
where \( i=\sqrt{-1} \). Consider the statements \( P \) and \( Q \) given below:
\(\begin{array}{l}P:\left|z_{2}-z_{1}\right|+\left|z_{3}-z_{2}\right|+\ldots+\left|z_{10}-z_{9}\right|+\left|z_{1}-z_{10}\right| \leq 2 \pi \\Q:\left|z_{2}^{2}-z_{1}^{2}\right|+\left|z_{3}^{2}-z_{2}^{2}\right|+\ldots+\left|z_{10}^{2}-z_{9}^{2}\right|+\left|z_{1}^{2}-z_{10}^{2}\right| \leq 4 \pi\end{array}\)
.Then,
[JEE Advanced 2021]
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The point \(P(a, b)\) undergoes the following three transformations successively:
(A) reflection about the line \(y=x\)
(B) translation through 2 units along the positive direction of \(x\)-axis.
(C) rotation through angle \(\frac{\pi}{4}\) about the origin in the anticlockwise direction.
If the co-ordinates of the final position of the point \(P\) are \(\left(-\frac{1}{\sqrt{2}}, \frac{7}{\sqrt{2}}\right)\) then the value of \(2 a+b\) is equal to:
[JEE Main 2021, 27 Jul (Shift 2)]
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If \( \alpha, \beta \in \mathrm{R} \) are such that \( 1-2 \mathrm{i} \) (here \( \mathrm{i}^{2}=-1 \) ) is a root of \( z^{2}+\alpha z+\beta=0 \), then \( (\alpha-\beta) \) is equal to :
[JEE Main 2021, 25 Feb (Shift 2)]
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Let \(z=x+iy\) be a non-zero complex number such that \({\mathrm{z}}^{2}=i|\mathrm{z}{|}^{2}\), where \(i=\sqrt{-1}\), then \(\mathrm{z}\) lies on the :
[JEE Main 2020, 6 Sep (Shift 2)]
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