Complex Numbers and Quadratic Equations

\( \alpha+\beta=3 \)
\( \frac{\alpha}{2}+2 \beta=-3\)

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 \(=-m\)

\( =(2 \alpha+\beta+2 \mathrm{r})\left(\alpha-2 \beta-\frac{\mathrm{r}}{2}\right) \)
\( \Rightarrow-\mathrm{m}=(-27)(21) \)
\( \Rightarrow \mathrm{m}=567\)

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 \(\frac{\alpha}{2}+\beta=\sqrt{\frac{9}{4}+4}\)

\(\alpha+\beta=3\)

$$\begin{gathered} \arg \left(z_2\right)=\arg \left(z_1\right)+\pi /6 \\ \arg \left(z_2\right)-\arg \left(z_1\right)=\pi /6\Rightarrow \arg \left(\frac{z_1}{z_2}\right)=\pi /6 \\ \text{ Area of }∆OAB=\frac{1}{2}\left|z_1\right|\left|z_2\right|\sin \pi /6 \\ =\frac{1}{2}\left|z_1\right|\sqrt{3}\left|z_1\right|\sin \frac{\mathrm{\pi }}{6}=\frac{1}{2}|z_1{|}^2\sqrt{3}\frac{1}{2}=\frac{11\sqrt{3}}{4} \end{gathered}$$

\(\begin{aligned}
&\text { Sol. }\\
&\begin{aligned}
& z_1=|1| e^{i \frac{\pi}{4}}=\frac{1}{\sqrt{2}}+i \cdot \frac{1}{\sqrt{2}} \\
& z_2=|1| e^{-(0)}= 1+0 i \\
& z_3=|1| e^{-i \frac{\pi}{4}}=\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}} \\
& \bar{z}_1 z_2=\left(\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}}\right)(1) \\
& \bar{z}_2 z_3=1\left(\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}}\right) \\
& \bar{z}_3 z_1=\left(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}\right)\left(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}\right) \\
& \Rightarrow \bar{z}_1 z_2+\bar{z}_2 z_3+\bar{z}_3 z_1=\left(\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}}\right)+\left(\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}}\right) \\
& +\left(\frac{1}{2}-\frac{1}{2}\right)+2 i\left(\frac{1}{2}\right) \\
& =\sqrt{2}-\sqrt{2} i+i \\
& \Rightarrow\left|\bar{z}_1 z_2+\bar{z}_2 z_3+\bar{z}_3 z_1\right|^2=|\sqrt{2}+i(-\sqrt{2}+1)|^2 \\
& =\left(\sqrt{(\sqrt{2})^2+(1-\sqrt{2})^2}\right)^2 \\
& =5-2 \sqrt{2} \\
& (a, b)=(5,-2)
\end{aligned}
\end{aligned}\)

\(z_1+z_2=5\)

\(z_1^3+z_2^3=20+15i\)

\(\left|z_1^4+z_2^4\right|=?\)

\(z_1^3+z_2^3=(z_1+z_2)^3-3z_1z_2(z_1+z_2)\)

\(z_1^3+z_2^3=125-15z_1z_2\)

\(20+15i=125-15z_1z_2\)

\(\Rightarrow z_1z_2=7-i\)

Now,

\((z_1+z_2)^2=5^2\)

\(z_1^2+z_2^2+2z_1z_2=25\)

\(z_1^2+z_2^2=25-2(7-i)\)

\(=11+2i\)

\((z_1^2+z_2^2)^2=(11+2i)^2\)

\(z_1^4+z_2^4+2(7-i)^2=117+44i\)

\(z_1^4+z_2^4=21+72i\)

\(\left|z_1^4+z_2^4\right|=\sqrt{21^2+72^2}\)

\(=\sqrt{441+5184}\)

\(=\sqrt{5625}\)

\(=75\)

$$\begin{gathered} \text{Given}, 2z^2-3z-2i=0 .......\left(i\right) \\ \text{Divide equation by z, we get,} \\ 2\left(z-\frac{i}{z}\right)=3 \\ \text{ As }\alpha ,\beta \text{ are roots of (i) } \\ \alpha -\frac{i}{\alpha }=\frac{3}{2} \\ \text{on squaring both sides, we get} \\ \Rightarrow {\alpha }^2-\frac{1}{{\alpha }^2}=\frac{9}{4}+2i \\ \text{ Again squaring both sides , we get} \\ \Rightarrow {\alpha }^4+\frac{1}{{\alpha }^4}=\frac{49}{16}+9i .........\left(ii\right) \\ \text{Similarly, for another root β} \\ {\beta }^4+\frac{1}{{\beta }^4}=\frac{49}{16}+9i ...........\left(ii\right) \\ Now, \\ \frac{{\alpha }^{19}+{\beta }^{19}+{\alpha }^{11}+{\beta }^{11}}{{\alpha }^{15}+{\beta }^{15}}=\frac{{\alpha }^{19}+{\alpha }^{11}+{\beta }^{19}+{\beta }^{11}}{{\alpha }^{15}+{\beta }^{15}} \\ =\frac{{\alpha }^{15}\left({\alpha }^4+\frac{1}{{\alpha }^4}\right)+{\beta }^{15}\left({\beta }^4+\frac{1}{{\beta }^4}\right)}{{\alpha }^{15}+{\beta }^{15}} \\ =\frac{49}{16}+9i \\ Re\left(\frac{{\alpha }^{19}+{\beta }^{19}+{\alpha }^{11}+{\beta }^{11}}{{\alpha }^{15}+{\beta }^{15}}\right)=\frac{49}{16} \\ lm\left(\frac{{\alpha }^{19}+{\beta }^{19}+{\alpha }^{11}+{\beta }^{11}}{{\alpha }^{15}+{\beta }^{15}}\right)=9 \\ \Rightarrow 16Re\left(\frac{{\alpha }^{19}+{\beta }^{19}+{\alpha }^{11}+{\beta }^{11}}{{\alpha }^{15}+{\beta }^{15}}\right)·lm\left(\frac{{\alpha }^{19}+{\beta }^{19}+{\alpha }^{11}+{\beta }^{11}}{{\alpha }^{15}+{\beta }^{15}}\right) \\ =16\times \frac{49}{16}\times 9=441 \end{gathered}$$

\(|z-i|=|z+i|=|z-1|\)

\(ABC\) is a triangle. Hence, its circum-centre will be the only point whose distance from \(A,B,C\) will be same.

So, \(n(S)=1\)

$$\begin{gathered} \frac{z+i}{z-i}=\frac{x+i(y+1)}{x+i(y-1)}\times \frac{x-i(y-1)}{x-i(y-1)} \\ =\text{purely real} \end{gathered}$$

$$= \frac {\left(x ^ {2} + y ^ {2} - 1\right) + i \left(x (y + 1) - x (y - 1)\right)}{x ^ {2} + (y - 1) ^ {2}}$$

Imaginary part is zero.

$$\begin{gathered} x(y+1)=x(y-1)\text{ implies }x=0\text{ and }y\text{ is} \\ \text{ infinitely many values} \end{gathered}$$

Therefore statement 1 is wrong.

$$\begin{gathered} \frac{z-1}{z+1}=\frac{x+iy-1}{x+iy+1} \\ =\frac{(x-1)+iy}{(x+1)+iy}\times \frac{(x+1)-iy}{(x+1)-iy} \end{gathered}$$

$$= \frac {\left(\left(x ^ {2} - 1\right) y ^ {2} + i \left(y (x + 1) - y (x - 1)\right)\right)}{\left(x + 1\right) ^ {2} + y ^ {2}}$$

Purely imaginary implies real part is zero.

$$\therefore x ^ {2} + y ^ {2} = 1$$

Lies on infinitely many points

Therefore statement "2" is true.

$x^2-\left(3-2i\right)x-\left(2i-2\right)=0$

$x=\frac{(3-2i)\pm \sqrt{(3-2i{)}^2-4(1\left)\right(-(2i-2))}}{2\left(1\right)}$

$x=\frac{3-2i\pm \sqrt{-3-4i}}{2}$

$=\frac{3-2\mathrm{i}\pm \sqrt{(1{)}^2+(2\mathrm{i}{)}^2-2\left(1\right)\left(2\mathrm{i}\right)}}{2}$

$=\frac{3-2\mathrm{i}\pm (1-2\mathrm{i})}{2}$

$x=\frac{3-2\mathrm{i}+1-2\mathrm{i}}{2} , \frac{3-2i-1+2i}{2}$

$x=2 – 2i, 1 + 0i$

So $\alpha \gamma +\beta \delta =2\left(1\right)+\left(-2\right)\left(0\right)=2$

$$\begin{gathered} \text{Given,} \\ {z}_1=e^{-i\pi /4}, z_2=1, z_3=e^{i\pi /4} \\ \text{Now,} \\ {\left|z_1{\overline{z}}_2+z_2{\overline{z}}_3+z_3{\overline{z}}_1\right|}^2 \\ ={\left|e^{-i\frac{\pi }{4}}\times 1+1\times e^{-i\frac{\pi }{4}}+e^{i\frac{\pi }{4}}\times e^{i\frac{\pi }{4}}\right|}^2 \\ ={\left|e^{-i\frac{\pi }{4}}+e^{-i\frac{\pi }{4}}+e^{i\frac{\pi }{2}}\right|}^2 \\ ={\left|2\left(\cos \frac{\mathrm{\pi }}{4}-i\sin \frac{\mathrm{\pi }}{4}\right)+\cos \frac{\mathrm{\pi }}{2}+i\sin \frac{\mathrm{\pi }}{2}\right|}^2 \\ =|\sqrt{2}-\sqrt{2}\mathrm{i}+\mathrm{i}{|}^2 \\ =(\sqrt{2}{)}^2+(1-\sqrt{2}{)}^2 \\ =5-2\sqrt{2} \\ \mathrm{\alpha }=5,\mathrm{\beta }=-2 \\ \Rightarrow {\mathrm{\alpha }}^2+{\mathrm{\beta }}^2=29 \end{gathered}$$

We are given the equation $\frac{2+k^2 z}{k+\bar{z}} = kz$ and the condition $|z| = 1$.

Substitute $\bar{z} = \frac{1}{z}$ into the given equation, we get

$\frac{2 + k^2 z}{k + \frac{1}{z}} = kz$

$z(2 + k^2 z) = kz(kz + 1)$

$2z + k^2 z^2 = k^2 z^2 + kz$

$2z = kz$

$2 = k \implies k = 2$

Substituting $k = 2$ into $k + ik^2$, we get, $2 + i(2^2) = 2 + 4i$.

This corresponds to the point $P(2, 4)$.

The given circle is $|z - (1 + 2i)| = 1$. 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 \(=d+r=\sqrt{5}+1\)

$\left|z_1\right|=\sqrt{3+8}=\sqrt{11}$

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

$\angle AOB=\left|\text{arg}\left(z_2\right)-\text{arg}\left(z_1\right)\right|=\frac{\mathrm{\pi }}{6}$

Using cosine rule for

$A{B}^2=O{A}^2+O{B}^2-2\left(OA\right)\left(OB\right)$

$\text{cos}\left(\frac{\mathrm{\pi }}{6}\right)=11+\frac{11}{3}-2\left(\sqrt{11}\right)\left(\sqrt{\frac{11}{3}}\right)\left(\frac{\sqrt{3}}{2}\right)$

$=11+\frac{11}{3}-11=\frac{11}{3}$

Thus, $AB=\sqrt{11/3}=OB$

The triangle is isosceles.

Check for obtuse angle:

$O{A}^2=11,$

$O{B}^2+A{B}^2=\frac{11}{3}+\frac{11}{3}=\frac{22}{3}\approx 7.33$

$\because O{A}^2>O{B}^2+A{B}^2$, the angle at B is obtuse.

ABO is an obtuse angled isosceles triangle.

\(\frac{z^2+3i}{z-2+i}=2+3i\)

\(z^2+3i=(z-2+i)(2+3i)\)

\(z^2+3i=2z-4+2i+3iz-6i-3\)

\(z^2+3i=(2z-7)+i(3z-4)\)

\(z^2-(2+3i)z+(7+7i)=0\)

This is a quadratic in \(z\).

\(z_1+z_2=2+3i\)

\(z_1z_2=7+7i\)

\(z_1^2+z_2^2=(z_1+z_2)^2-2z_1z_2\)

\(=(2+3i)^2-2(7+7i)\)

\(=4-9+12i-14-14i\)

\(=-19-2i\)

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 \(P D: \sqrt{1^2+\left(5+\frac{4}{3}\right)^2}=\sqrt{1+\frac{361}{9}}=\sqrt{\frac{370}{9}}\)

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

\(|z^2|=|z|^2=1\)

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

\(|z^2|=|z|^2=1\)

\(S_1:|z|=1,\ \dfrac{z-i}{z+i}=\dfrac{\bar z+i}{\bar z-i}\)

\(\Rightarrow (z-i)(\bar z-i)=(z+i)(\bar z+i)\)

\(\Rightarrow |z|^2-i(z+\bar z)-1=|z|^2+i(z+\bar z)-1\)

\(\Rightarrow z+\bar z=0\)

\(\Rightarrow 2\operatorname{Re}(z)=0\)

Since \(|z|=1\), we get \(z=\pm i\).

But \(z=-i\) is not allowed because \(z+i=0\).

So, \(S_1=\{i\}\).

\(S_2:\dfrac{z-1}{z+1}+\dfrac{\bar z-1}{\bar z+1}=0\)

\(\Rightarrow (z-1)(\bar z+1)+(z+1)(\bar z-1)=0\)

\(\Rightarrow |z|^2+z-\bar z-1+|z|^2-z+\bar z-1=0\)

\(\Rightarrow 2|z|^2-2=0\)

\(\Rightarrow |z|^2=1\)

\(\Rightarrow |z|=1\)

\(S_1:|z|=1,\ \dfrac{z-i}{z+i}=\dfrac{\bar z+i}{\bar z-i}\)

\(\Rightarrow (z-i)(\bar z-i)=(z+i)(\bar z+i)\)

\(\Rightarrow |z|^2-i(z+\bar z)-1=|z|^2+i(z+\bar z)-1\)

\(\Rightarrow z+\bar z=0\)

\(\Rightarrow 2\operatorname{Re}(z)=0\)

Since \(|z|=1\), we get \(z=\pm i\).

But \(z=-i\) is not allowed because \(z+i=0\).

So, \(S_1=\{i\}\).

\(S_2:\dfrac{z-1}{z+1}+\dfrac{\bar z-1}{\bar z+1}=0\)

\(\Rightarrow (z-1)(\bar z+1)+(z+1)(\bar z-1)=0\)

\(\Rightarrow |z|^2+z-\bar z-1+|z|^2-z+\bar z-1=0\)

\(\Rightarrow 2|z|^2-2=0\)

\(\Rightarrow |z|^2=1\)

\(\Rightarrow |z|=1\)

\( \left.n x^2+x(2+4+6 \ldots+2 n)+(1 \cdot 3+\ldots+(n-1))(n+1)\right)=4 n \)
\( n x^2+n(n+1) x+\frac{n(n-1)(2 n+5)}{6}=4 n\)
\( x^2+(n+1) x+\frac{(n-1)(2 n+5)}{6}=4\)

Since the roots are \(\alpha\) and \(\alpha+2\), the difference of roots is \(2\)

For a quadratic equation, difference of roots \(=\frac{\sqrt D}{a}\).

Here \(a=1\), so \(\sqrt D=2\), hence \(D=4\)

Now,

\(D=(n+1)^2-4\left(\frac{(n-1)(2n+5)}{6}-4\right)\)

\(=n^2+2n+17-\frac{2(2n^2+3n-5)}{3}\)

\(=\frac{3n^2+6n+51-4n^2-6n+10}{3}\)

\(=\frac{61-n^2}{3}=20-\frac{n^2-1}{3}\)

\(\Rightarrow 20-\frac{n^2-1}{3}=4\)

\(\frac{n^2-1}{3}=16\)

\(n^2=49\)

Since \(n\in\mathbb N\), \(n=7\)

So, equation is:

\(x^2+8 x+\frac{6 \times 19}{6}-4=0\)
\( \mathrm{x}^2+8 \mathrm{x}+15=0 \)
\( \mathrm{x}=-3,-5 \)
\( \alpha=-5, \alpha+2=-3 \)
\( \alpha+\mathrm{n}=7-5=2\)

\(
\begin{aligned}
&z_1+z_2+z_3=3z_0\\
&\left(z_1+z_2+z_3\right)^2=9z_0^2\\
&\Rightarrow
z_1^2+z_2^2+z_3^2
+2\left(z_1z_2+z_2z_3+z_3z_1\right)
=9z_0^2
\end{aligned}
\)

\(
\begin{aligned}
&\sum_{k=1}^{3}(z_k-z_0)^2\\
&=(z_1-z_0)^2+(z_2-z_0)^2+(z_3-z_0)^2\\
&=z_1^2+z_2^2+z_3^2
+3z_0^2
-2(z_1+z_2+z_3)z_0\\
&=z_1^2+z_2^2+z_3^2
+3z_0^2
-6z_0^2\\
&=z_1^2+z_2^2+z_3^2-3z_0^2
\end{aligned}
\)

\(
\begin{aligned}
&\text{If }
z_1^2+z_2^2+z_3^2=3z_0^2,
\text{ then}\\
&\sum_{k=1}^{3}(z_k-z_0)^2\\
&=3z_0^2-3z_0^2\\
&=0.
\end{aligned}
\)

\begin{aligned}
& Z=\frac{1+i \cos \theta}{1-2 i \cos \theta} \\
& Z=-\bar{Z} \Rightarrow \frac{1+i \cos \theta}{1-2 i \cos \theta}=-\left(\frac{\overline{1+i \cos \theta}}{1-2 i \cos \theta}\right) \\
& (1+i \cos \theta)(\overline{1-2 i \cos \theta})=-(1-2 i \cos \theta)(\overline{1+i \cos \theta}) \\
& (1+i \cos \theta)(1+2 i \cos \theta)=-(1-2 i \cos \theta)(1-i \cos \theta) \\
& 1+3 i \cos \theta-2 \cos ^2 \theta=-\left(1-3 i \cos \theta-2 \cos ^2 \theta\right) \\
& 2-4 \cos ^2 \theta=0 \\
& \Rightarrow \cos ^2 \theta=\frac{1}{2} \Rightarrow \theta=-\frac{\pi}{4},-\frac{3 \pi}{4}, \frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4} \\
& \text { sum }=3 \pi
\end{aligned}

\(z=x+iy\)

\(\overline{z}=x-iy\)

\(\overline{z}^{\,2}=x^2-y^2-2ixy\)

\(\Rightarrow x^2-y^2-2ixy+\sqrt{x^2+y^2}=0\)

\(\Rightarrow x=0\)

\(-y^2+|y|=0\)

\(|y|=|y|^2\)

or

\(\Rightarrow y=0\)

\(x^2+|x|=0\)

\(\Rightarrow x=0\)

\(y=0,\pm 1\)

\(\Rightarrow i,-i\) are roots.

\(\Rightarrow \alpha=i-i=0\)

\(\beta=i(-i)=1\)

Now,

\(4(0+1)=4\)

\(z=x+iy\)

\(\overline{z}=x-iy\)

\(\overline{z}^{\,2}=x^2-y^2-2ixy\)

\(\Rightarrow x^2-y^2-2ixy+\sqrt{x^2+y^2}=0\)

\(\Rightarrow x=0\)

\(-y^2+|y|=0\)

\(|y|=|y|^2\)

or

\(\Rightarrow y=0\)

\(x^2+|x|=0\)

\(\Rightarrow x=0\)

\(y=0,\pm 1\)

\(\Rightarrow i,-i\) are roots.

\(\Rightarrow \alpha=i-i=0\)

\(\beta=i(-i)=1\)

Now,

\(4(0+1)=4\)

Given equation $z^2+az+b=0$

Then $z_1+z_2=-a$ and $z_1\times z_2=b$

$z_1$ and $z_2$ and $z_3$ from an equilateral triangle.

$\Rightarrow z_1^2+z_2^2+z_3^2=z_1\times z_2+z_2\times z_3+z_1\times z_3$

$\Rightarrow z_1^2+z_2^2=z_1\times z_2$ $\left\{\because z_3=0\right\}$

$\Rightarrow {\left(z_1+z_2\right)}^2-2z_1\times z_2=z_1\times z_2$

$\Rightarrow {\left(z_1+z_2\right)}^2=2\left(z_1\times z_2\right)+z_1\times z_2$

$\Rightarrow {\left(z_1+z_2\right)}^2=3z_1\times z_2\Rightarrow a^2=3b$

\(\frac{|\bar{z}-i|}{|2 \bar{z}+i|}=\frac{1}{3}\)

on squaring both sides, we get

\(\frac{|\bar{z}-i|^2}{|2 \bar{z}+i|^2}=\frac{1}{9}\)

\(9|\bar{z}-i|^2=|2 \bar{z}+i|^2\)

Let \(z=x+i y\). Then:

\(9|(x-i y)-i|^2=|2(x-i y)+i|^2\)

\(9\left(x^2+(y+1)^2\right)=(2 x)^2+(2 y-1)^2\)

\(9\left(x^2+y^2+2 y+1\right)=4 x^2+4 y^2-4 y+1\)

\(x^2+y^2+\frac{22}{5} y+\frac{8}{5}=0\)

This is the equation of a circle with center \(C\left(0,-\frac{11}{5}\right)\)

The area of the triangle with vertices \((0,0),\left(0,-\frac{11}{5}\right)\), and \((\alpha, 0)\) is given by:

$\left|\frac{1}{2}\left|\begin{array}{ccc}0& 0& 1\\ 0& -11/5& 1\\ \alpha & 0& 1\end{array}\right|\right|=11$

\(\alpha^2=100\)

$$\begin{gathered} x^2-(3-2i)x-(2i-2)=0 \\ {x}^2-2x-x+2+2i(x-1)=0 \\ (x-2)(x-1)+2i(x-1)=0 \\ (x-1)[x-2+2i]=0 \\ x=1 or x =2-2i \\ x=1+0i or x = 2-2i \\ \mathrm{Case} 1: \\ \alpha =1, \beta =0, \gamma =2, \delta =-2 \\ \alpha \gamma +\beta \delta =2 \\ \mathrm{Case} 2: \\ \alpha =2, \beta =-2, \gamma =\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right., \delta =0 \end{gathered}$$

this case is rejected because $\gamma \in I$

Given equation is \(z^2+4 z-(1+12 i)=0\)

Using quadratic formula, \(z=\frac{-4 \pm \sqrt{16+4(1+12 i)}}{2}\)

So \(z=\frac{-4 \pm \sqrt{20+48 i}}{2}\)

Since \(20+48 i=4(5+12 i), z=-2 \pm \sqrt{5+12 i}\)

Now \((3+2 i)^2=9+12 i+4 i^2=5+12 i\),

so \(\sqrt{5+12 i}=3+2 i\)

Therefore \(z_1=-2+(3+2 i)=1+2 i\)

and \(z_2=-2-(3+2 i)=-5-2 i\)

Now \(\left|z_1\right|^2=1^2+2^2=5\)

and \(\left|z_2\right|^2=(-5)^2+(-2)^2=29\)

Hence \(\left|z_1\right|^2+\left|z_2\right|^2=5+29=34\)

Let \(z=x+iy\).

Then,

\((x-1)^2+y^2=1\) \(\tag{1}\)

and

\((\sqrt2-1)(2x)-i(2iy)=2\sqrt2\)

\(\Rightarrow (\sqrt2-1)x+y=\sqrt2\) \(\tag{2}\)

Solving \((1)\) and \((2)\), we get

Either \(x=1\) or \(x=\dfrac{1}{2-\sqrt2}\) \(\tag{3}\)

On solving \((3)\) with \((2)\), we get

For \(x=1\),

\(y=1\)

\(\Rightarrow z_2=1+i\)

and for

\(x=\dfrac{1}{2-\sqrt2}\),

\(y=\sqrt2-\dfrac{1}{\sqrt2}\)

\(\Rightarrow z_1=\left(1+\dfrac{1}{\sqrt2}\right)+\dfrac{i}{\sqrt2}\)

Now,

\(\left|\sqrt2z_1-z_2\right|^2=\left|\left(1+\dfrac{1}{\sqrt2}\right)\sqrt2+i-(1+i)\right|^2\)

\(=(\sqrt2)^2\)

\(=2\)

$$\begin{gathered} \arg \left(z_2\right)=\arg \left(z_1\right)+\pi /6 \\ \arg \left(z_2\right)-\arg \left(z_1\right)=\pi /6\Rightarrow \arg \left(\frac{z_1}{z_2}\right)=\pi /6 \\ \text{ Area of }∆OAB=\frac{1}{2}\left|z_1\right|\left|z_2\right|\sin \pi /6 \\ =\frac{1}{2}\left|z_1\right|\sqrt{3}\left|z_1\right|\sin \frac{\mathrm{\pi }}{6}=\frac{1}{2}|z_1{|}^2\sqrt{3}\frac{1}{2}=\frac{11\sqrt{3}}{4} \end{gathered}$$

\(|z-(4+8 i)|=\sqrt{10}\)

It is a circle with center \((4,8)\), radius \(=\sqrt{10}\)

\(|z-(3+5 i)|+\mid z-(5+11 i)=4 \sqrt{5}\)

It is an ellipse with focii \(S(3,5)\) and \(S^{\prime}(5,11)\)

\(S S^{\prime}=\sqrt{4+36} \Rightarrow \sqrt{40}=2 \sqrt{10}\)

Clearly \(4 \sqrt{5}>S S^{\prime}\)

Center of ellipse \(\left(\frac{3+5}{2}, \frac{5+11}{2}\right) \equiv(4,8)\), coincides with center of circle

Now, \(2 a=4 \sqrt{5}\), \(c=\sqrt{10}\)

and \(b^2=20-10=10\)

\(b=\sqrt{10}=\) radius of circle

Hence ellipse and circle will intersect at end points of minor axis
So, \(2\) points of intersection

$$\begin{gathered} \frac{|\overline{z}-i|}{|2\overline{z}+i|}=\frac{1}{3} \\ \text{on squaring both sides, we get} \\ \frac{|\overline{z}-i{|}^2}{|2\overline{z}+i{|}^2}=\frac{1}{9} \end{gathered}$$

$$\begin{gathered} 9|\overline{z}-i{|}^2=|2\overline{z}+i{|}^2 \\ \text{ Let }z=x+iy\text{. Then: } \\ 9\left|\right(x-iy)-i{|}^2=|2(x-iy)+i{|}^2 \\ 9\left(x^2+(y+1{)}^2\right)=(2x{)}^2+(2y-1{)}^2 \\ 9\left(x^2+y^2+2y+1\right)=4x^2+4y^2-4y+1 \\ {x}^2+y^2+\frac{22}{5}y+\frac{8}{5}=0 \\ \text{ This is the equation of a circle} \\ \text{with center }C\left(0,-\frac{11}{5}\right) \\ \text{ The area of the triangle with vertices} \\ (0,0),\left(0,-\frac{11}{5}\right)\text{, and }(\alpha ,0)\text{ is given by: } \end{gathered}$$

$$\begin{gathered} \left|\frac{1}{2}\left|\begin{array}{ccc}0& 0& 1\\ 0& -11/5& 1\\ \alpha & 0& 1\end{array}\right|\right|=11 \\ {\alpha }^2=100 \end{gathered}$$

Given \(z=2-2i\tan\dfrac{5\pi}{8}\)

\(\tan\dfrac{5\pi}{8}=-\tan\dfrac{3\pi}{8}\)

\(z=2+2i\tan\dfrac{3\pi}{8}\)

Factorising

\(z=2\left(1+i\tan\dfrac{3\pi}{8}\right)\)

\(=2\left(\dfrac{\cos\frac{3\pi}{8}+i\sin\frac{3\pi}{8}}{\cos\frac{3\pi}{8}}\right)\)

\(=2\sec\dfrac{3\pi}{8}\left(\cos\dfrac{3\pi}{8}+i\sin\dfrac{3\pi}{8}\right)\)

Hence the modulus is

\(r=2\sec\dfrac{3\pi}{8}\)

and the amplitude is

\(\theta=\dfrac{3\pi}{8}\)

Therefore \((r,\theta)=\left(2\sec\dfrac{3\pi}{8},\dfrac{3\pi}{8}\right)\)

Given \(z=2-2i\tan\dfrac{5\pi}{8}\)

\(\tan\dfrac{5\pi}{8}=-\tan\dfrac{3\pi}{8}\)

\(z=2+2i\tan\dfrac{3\pi}{8}\)

Factorising

\(z=2\left(1+i\tan\dfrac{3\pi}{8}\right)\)

\(=2\left(\dfrac{\cos\frac{3\pi}{8}+i\sin\frac{3\pi}{8}}{\cos\frac{3\pi}{8}}\right)\)

\(=2\sec\dfrac{3\pi}{8}\left(\cos\dfrac{3\pi}{8}+i\sin\dfrac{3\pi}{8}\right)\)

Hence the modulus is

\(r=2\sec\dfrac{3\pi}{8}\)

and the amplitude is

\(\theta=\dfrac{3\pi}{8}\)

Therefore \((r,\theta)=\left(2\sec\dfrac{3\pi}{8},\dfrac{3\pi}{8}\right)\)

\(R=\{(a, b): a+5 b=42\}\)
Then \( R=\{(2,8),(7,7),(12,6),(17,5),(22,4),(27,3),(32,2),(37,1)\}\)
$\Rightarrow m=8$
and \(\sum_{n=1}^{m=8}\left(1-i^{n!}\right)=x+i y \)
\(\therefore \sum_{n=1}^8\left(1-i^{n!}\right)=8-\left(i+i^2+i^6+1+1+1+1+1\right) \)
\( =5-i \)
\( \therefore x=5, y=-1 \)
\( x+y+m=5-1+8=12\)

\(R=\{(a, b): a+5 b=42\}\)
Then \( R=\{(2,8),(7,7),(12,6),(17,5),(22,4),(27,3),(32,2),(37,1)\}\)
$\Rightarrow m=8$
and \(\sum_{n=1}^{m=8}\left(1-i^{n!}\right)=x+i y \)
\(\therefore \sum_{n=1}^8\left(1-i^{n!}\right)=8-\left(i+i^2+i^6+1+1+1+1+1\right) \)
\( =5-i \)
\( \therefore x=5, y=-1 \)
\( x+y+m=5-1+8=12\)

\(\begin{aligned}
&\text { Sol. }\\
&\begin{aligned}
& z_1=|1| e^{i \frac{\pi}{4}}=\frac{1}{\sqrt{2}}+i \cdot \frac{1}{\sqrt{2}} \\
& z_2=|1| e^{-(0)}= 1+0 i \\
& z_3=|1| e^{-i \frac{\pi}{4}}=\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}} \\
& \bar{z}_1 z_2=\left(\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}}\right)(1) \\
& \bar{z}_2 z_3=1\left(\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}}\right) \\
& \bar{z}_3 z_1=\left(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}\right)\left(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}\right) \\
& \Rightarrow \bar{z}_1 z_2+\bar{z}_2 z_3+\bar{z}_3 z_1=\left(\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}}\right)+\left(\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}}\right) \\
& +\left(\frac{1}{2}-\frac{1}{2}\right)+2 i\left(\frac{1}{2}\right) \\
& =\sqrt{2}-\sqrt{2} i+i \\
& \Rightarrow\left|\bar{z}_1 z_2+\bar{z}_2 z_3+\bar{z}_3 z_1\right|^2=|\sqrt{2}+i(-\sqrt{2}+1)|^2 \\
& =\left(\sqrt{(\sqrt{2})^2+(1-\sqrt{2})^2}\right)^2 \\
& =5-2 \sqrt{2} \\
& (a, b)=(5,-2)
\end{aligned}
\end{aligned}\)

\begin{aligned}
&\begin{aligned}
& \mathbf{Z}=e^{i \theta} \\
& \frac{\mathrm{Z}}{\overline{\mathrm{Z}}}=\mathrm{e}^{\mathrm{i} 2 \theta} \\
& \left|\frac{\mathrm{Z}}{\overline{\mathrm{Z}}}+\frac{\overline{\mathrm{Z}}}{\mathrm{Z}}\right|=1 \Rightarrow\left|\mathrm{e}^{\mathrm{i} 2 \theta}+\mathrm{e}^{-12 \theta}\right|=1 \Rightarrow|\cos 2 \theta|=\frac{1}{2}
\end{aligned}\\
&8 \text { solution in }[0,2 \pi]
\end{aligned}

Given,

\(z_1=e^{-i \pi / 4}, z_2=1, z_3=e^{i \pi / 4}\)

Now,

\(\left|z_1 \bar{z}_2+z_2 \bar{z}_3+z_3 \bar{z}_1\right|^2\)

\(=\left|e^{-i \frac{\pi}{4}} \times 1+1 \times e^{-i \frac{\pi}{4}}+e^{i \frac{\pi}{4}} \times e^{i \frac{\pi}{4}}\right|^2\)

$={\left|e^{-i\frac{\pi }{4}}+e^{-i\frac{\pi }{4}}+e^{i\frac{\pi }{2}}\right|}^2$

$={\left|2\left(\cos \frac{\mathrm{\pi }}{4}-i\sin \frac{\mathrm{\pi }}{4}\right)+\cos \frac{\mathrm{\pi }}{2}+i\sin \frac{\mathrm{\pi }}{2}\right|}^2$

\(=|\sqrt{2}-\sqrt{2}i+i|^2\)

\(=(\sqrt{2})^2+(1-\sqrt{2})^2\)

\(=5-2 \sqrt{2}\)

\(\alpha=5, \beta=-2\)

\(\Rightarrow \alpha^2+\beta^2=29\)

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 \(\frac{\alpha}{2}+\beta=\sqrt{\frac{9}{4}+4}\)

\(\alpha+\beta=3\)

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 \(\frac{\alpha}{2}+\beta=\sqrt{\frac{9}{4}+4}\)

\(\alpha+\beta=3\)

\(\left|\frac{z_1-2 z_2}{\frac{1}{2}-z_1 \bar{z}_2}\right|=2\)

$\Rightarrow \frac{z_1-2z_2}{\frac{1}{2}-z_1{\overline{z}}_2}\times \frac{{\overline{z}}_1-2{\overline{z}}_2}{\frac{1}{2}-{\overline{z}}_1{z}_2}=4$
$$\begin{gathered} \Rightarrow {\left|z_1\right|}^2-2z_1{\overline{z}}_2-2{\overline{z}}_1{z}_2+4{\left|z_2\right|}^2 \\ =4\left(\frac{1}{4}-\frac{{\overline{z}}_1{z}_2}{2}-\frac{z_1{\overline{z}}_2}{2}+{\left|z_1\right|}^2{\left|z_2\right|}^2\right) \end{gathered}$$
$\Rightarrow {\left|z_1\right|}^2-1-4{\left|z_1\right|}^2{\left|z_2\right|}^2+4{\left|z_2\right|}^2=0$
$\Rightarrow \left({\left|z_1\right|}^2-1\right)\left({\left|2z_2\right|}^2-1\right)=0$

Either \(\left|z_1\right|^2-1=0\) or \(1-4\left|z_2\right|^2=0\)

\(\left|z_1\right|=1 \quad\) or \(\quad\left|z_2\right|=\frac{1}{2}\)

Therefore either \(\mathrm{Z}_1\) lies on circle of radius 1

or \(z_2\) lies on circle of radius \(\frac{1}{2}\)

\(\left|\frac{z_1-2 z_2}{\frac{1}{2}-z_1 \bar{z}_2}\right|=2\)

$\Rightarrow \frac{z_1-2z_2}{\frac{1}{2}-z_1{\overline{z}}_2}\times \frac{{\overline{z}}_1-2{\overline{z}}_2}{\frac{1}{2}-{\overline{z}}_1{z}_2}=4$
$$\begin{gathered} \Rightarrow {\left|z_1\right|}^2-2z_1{\overline{z}}_2-2{\overline{z}}_1{z}_2+4{\left|z_2\right|}^2 \\ =4\left(\frac{1}{4}-\frac{{\overline{z}}_1{z}_2}{2}-\frac{z_1{\overline{z}}_2}{2}+{\left|z_1\right|}^2{\left|z_2\right|}^2\right) \end{gathered}$$
$\Rightarrow {\left|z_1\right|}^2-1-4{\left|z_1\right|}^2{\left|z_2\right|}^2+4{\left|z_2\right|}^2=0$
$\Rightarrow \left({\left|z_1\right|}^2-1\right)\left({\left|2z_2\right|}^2-1\right)=0$

Either \(\left|z_1\right|^2-1=0\) or \(1-4\left|z_2\right|^2=0\)

\(\left|z_1\right|=1 \quad\) or \(\quad\left|z_2\right|=\frac{1}{2}\)

Therefore either \(\mathrm{Z}_1\) lies on circle of radius 1

or \(z_2\) lies on circle of radius \(\frac{1}{2}\)

\(\left|\frac{z_1-2 z_2}{\frac{1}{2}-z_1 \bar{z}_2}\right|=2\)

$\Rightarrow \frac{z_1-2z_2}{\frac{1}{2}-z_1{\overline{z}}_2}\times \frac{{\overline{z}}_1-2{\overline{z}}_2}{\frac{1}{2}-{\overline{z}}_1{z}_2}=4$
\(\Rightarrow\left|z_1\right|^2-2 z_1 \bar{z}_2-2 \bar{z}_1 z_2+4\left|z_2\right|^2=4\left(\frac{1}{4}-\frac{\bar{z}_1 z_2}{2}-\frac{z_1 \bar{z}_2}{2}+\left|z_1\right|^2\left|z_2\right|^2\right)\)
$\Rightarrow {\left|z_1\right|}^2-1-4{\left|z_1\right|}^2{\left|z_2\right|}^2+4{\left|z_2\right|}^2=0$
$\Rightarrow \left({\left|z_1\right|}^2-1\right)\left({\left|2z_2\right|}^2-1\right)=0$

Either \(\left|z_1\right|^2-1=0\) or \(1-4\left|z_2\right|^2=0\)

\(\left|z_1\right|=1 \quad\) or \(\quad\left|z_2\right|=\frac{1}{2}\)

Therefore either \(\mathrm{Z}_1\) lies on circle of radius 1 or \(z_2\) lies on circle of radius \(\frac{1}{2}\)

Given, $\left|\frac{z-6i}{z-2i}\right|=1$

$\Rightarrow |z-6i|=|z-2i|$

Let $z=x+iy$

\(\Rightarrow x^2+(y-6)^2=x^2+(y-2)^2\)

\(\Rightarrow y^2-12 y+36=y^2-4 y+4\)

\(\Rightarrow y=4\)

Now $\left|\frac{z-8+2i}{z+2i}\right|=\frac{3}{5}$

$\Rightarrow \left|\frac{(x-8)+i(y+2)}{x+i(y+2)}\right|=\frac{3}{5}$

\(\Rightarrow \left[(x-8)^2+(y+2)^2\right] 25=9\left[x^2+(y+2)^2\right]\)

$\Rightarrow x^2+y^2-25x+4y+104=0$

put \(y=4\), we get

\(x^2-25 x+136=0\)

\(\Rightarrow x=8,17\)

\(\Rightarrow z=17+4 i\) and \(8+4 i\)

$\Sigma |z{|}^2={\left(17\right)}^2+{\left(4\right)}^2+{\left(8\right)}^2+{\left(4\right)}^2=385$

Let

$$\begin{gathered} {\mathrm{z}}_1=1+\mathrm{i},{\mathrm{z}}_2=-2+3\mathrm{i} \\ \text{ and }{\mathrm{z}}_3=0+\frac{5}{3}i \end{gathered}$$

then,

$\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|=\left|\begin{array}{ccc}1& 1& 1\\ -2& 3& 1\\ 0& 5/3& 1\end{array}\right|$

= $1\left(3-\frac{5}{3}\right)+1\left(2\right)+\left(\frac{-10}{3}\right)=0$

Hence, Area of the triangle is zero. So they are collinear points.

Put \(z=x+iy\).

\(|z-1|\leq2\)

\(\Rightarrow (x-1)^2+y^2\leq4\quad ...(1)\)

\((z+\bar z)+i(z-\bar z)\leq2\)

\(\Rightarrow 2x+i(2iy)\leq2\)

\(\Rightarrow 2x-2y\leq2\)

\(\Rightarrow x-y\leq1\quad ...(2)\)

\(\operatorname{Im}(z)\geq0\)

\(\Rightarrow y\geq0\quad ...(3)\)

Required area \(=\) Area of semi-circle \(-\) area of sector \(A\)

\(=\dfrac{1}{2}\pi(2)^2-\dfrac{\pi}{2}\)

\(=2\pi-\dfrac{\pi}{2}\)

\(=\dfrac{3\pi}{2}\)

Let \(z=x+i y\), then \(\bar{z}=x-i y\)

\((x+i y)((x-i y)+2+i)+k(2+3 i)=0\)

\(x^2+2 x+i(x-x y)+i(x y+2 y)-y(1-y)+2 k+3 k i=0\)

Real Part: \(x^2+2 x-y+y^2+2 k=0\)

\(\Rightarrow x^2+y^2+2 x-y+2 k=0\) \(\ldots(i)\)

Imaginary Part: \(x-x y+x y+2 y+3 k=0\)

\(\Rightarrow x+2 y+3 k=0\)

\(\Rightarrow x=-2 y-3 k\)

Substituting this into equation \((i)\):

\((-2 y-3 k)^2+y^2+2(-2 y-3 k)-y+2 k=0\)

\(5 y^2+y(12 k-5)+\left(9 k^2-4 k\right)=0\)

For \(y\) (and thus \(z\)) to have at least one real solution,

\(D=(12 k-5)^2-4(5)\left(9 k^2-4 k\right) \geq 0\)

\(36 k^2+40 k-25 \leq 0\)

\(\therefore \frac{-40-20 \sqrt{13}}{72} \leq k \leq \frac{-40+20 \sqrt{13}}{72}\)

So, \(9(\alpha+\beta)=9\left(-\frac{80}{72}\right)=-10\)

$$\begin{gathered} \frac{|\overline{z}-i|}{|2\overline{z}+i|}=\frac{1}{3} \\ \text{on squaring both sides, we get} \\ \frac{|\overline{z}-i{|}^2}{|2\overline{z}+i{|}^2}=\frac{1}{9} \end{gathered}$$

$$\begin{gathered} 9|\overline{z}-i{|}^2=|2\overline{z}+i{|}^2 \\ \text{ Let }z=x+iy\text{. Then: } \\ 9\left|\right(x-iy)-i{|}^2=|2(x-iy)+i{|}^2 \\ 9\left(x^2+(y+1{)}^2\right)=(2x{)}^2+(2y-1{)}^2 \\ 9\left(x^2+y^2+2y+1\right)=4x^2+4y^2-4y+1 \\ {x}^2+y^2+\frac{22}{5}y+\frac{8}{5}=0 \\ \text{ This is the equation of a circle} \\ \text{with center }C\left(0,-\frac{11}{5}\right) \\ \text{ The area of the triangle with vertices} \\ (0,0),\left(0,-\frac{11}{5}\right)\text{, and }(\alpha ,0)\text{ is given by: } \end{gathered}$$

$$\begin{gathered} \left|\frac{1}{2}\left|\begin{array}{ccc}0& 0& 1\\ 0& -11/5& 1\\ \alpha & 0& 1\end{array}\right|\right|=11 \\ {\alpha }^2=100 \end{gathered}$$

$$\begin{gathered} x^2-(3-2i)x-(2i-2)=0 \\ {x}^2-2x-x+2+2i(x-1)=0 \\ (x-2)(x-1)+2i(x-1)=0 \\ (x-1)[x-2+2i]=0 \\ x=1 or x =2-2i \\ x=1+0i or x = 2-2i \\ \mathrm{Case} 1: \\ \alpha =1, \beta =0, \gamma =2, \delta =-2 \\ \alpha \gamma +\beta \delta =2 \\ \mathrm{Case} 2: \\ \alpha =2, \beta =-2, \gamma =\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right., \delta =0 \end{gathered}$$

this case is rejected because $\gamma \in I$

Given, \(2 z^2-3 z-2 i=0\) .... (i)

Divide equation by \(z\), we get, \(2\left(z-\frac{i}{z}\right)=3\)
As \(\alpha, \beta\) are roots of eq(i)

\(\alpha-\frac{i}{\alpha}=\frac{3}{2}\)

on squaring both sides, we get

\(\alpha^2-\frac{1}{\alpha^2}=\frac{9}{4}+2 i\)

Again squaring both sides, we get

\(\Rightarrow \alpha^4+\frac{1}{\alpha^4}=\frac{49}{16}+9 i \)

Similarly, for another root \(\beta\)

\(\beta^4+\frac{1}{\beta^4}=\frac{49}{16}+9 i\)

Now

\(\frac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}=\frac{\alpha^{19}+\alpha^{11}+\beta^{19}+\beta^{11}}{\alpha^{15}+\beta^{15}}\)

\(=\frac{\alpha^{15}\left(\alpha^4+\frac{1}{\alpha^4}\right)+\beta^{15}\left(\beta^4+\frac{1}{\beta^4}\right)}{\alpha^{15}+\beta^{15}}\)

\(=\frac{49}{16}+9 i\)

\(\operatorname{Re}\left(\frac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}\right)=\frac{49}{16}\)

\(\operatorname{lm}\left(\frac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}\right)=9\)

\(\Rightarrow 16 \operatorname{Re}\left(\frac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}\right) \cdot \operatorname{Im}\left(\frac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}\right)\)

\(=16 \times \frac{49}{16} \times 9=441\)

Let

$$\begin{gathered} {\mathrm{z}}_1=1+\mathrm{i},{\mathrm{z}}_2=-2+3\mathrm{i} \\ \text{ and }{\mathrm{z}}_3=0+\frac{5}{3}i \end{gathered}$$

then,

$\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|=\left|\begin{array}{ccc}1& 1& 1\\ -2& 3& 1\\ 0& 5/3& 1\end{array}\right|$

= $1\left(3-\frac{5}{3}\right)+1\left(2\right)+\left(\frac{-10}{3}\right)=0$

Hence, Area of the triangle is zero. So they are collinear points.