Complex Numbers

Given, $\mathrm{Z}_k=\cos \frac{2 k \pi}{10}+i \sin \frac{2 k \pi}{10}, k=1,2,3, \ldots, 9$

$\begin{aligned} & \text { (P) } Z_k \cdot Z_j=1 \\ & \Rightarrow\left(\cos \frac{2 k \pi}{10}+i \sin \frac{2 k \pi}{10}\right) \\ & \left(\cos \frac{2 j \pi}{10}+i \sin \frac{2 j \pi}{10}\right)=1 \end{aligned}$

$\begin{aligned} & \Rightarrow \cos \left(\frac{2 \pi(k+j)}{10}\right)+i \sin \frac{2 \pi(k+j)}{10}=1 \\ & \Rightarrow \cos \frac{2 \pi(k+j)}{10}=1 \text { and } \sin \frac{2 \pi(k+j)}{10}=0 \\ & \Rightarrow \quad \frac{2 \pi(k+j)}{10}=2 \lambda \pi \text { where } \lambda \in \text { Integer } \\ & \Rightarrow \quad k+j=10 \lambda \\ & \Rightarrow \quad j=10 \lambda-k \\ \end{aligned}$

For every value of $k$ there is a possible value of $j$

Hence, $P$ match with 1.

(Q) Given $Z_1 \cdot Z=Z_k$

$\begin{aligned} & \left(\cos \frac{2 \pi}{10}+i \sin \frac{2 \pi}{10}\right) \cdot z=\cos \frac{2 \pi k}{10}+i \sin \frac{2 \pi k}{10} \\ & z=\frac{\cos \frac{2 \pi k}{10}+i \sin \frac{2 \pi k}{10}}{\cos \frac{2 \pi}{10}+i \sin \frac{2 \pi}{10}} \\ & z=\cos \frac{2 \pi(k-1)}{10}+i \sin \frac{2 \pi(k-1)}{10} \end{aligned}$

For every value of $k$, there is a unique value of $z$ is possible.

$\therefore$ The statement $z_1 \cdot z=z_k$ has no solution z in the set of complex numbers is false.

Hence, Q match with 2.

(R) Given, $z_k=\cos \frac{2 \pi k}{10}+i \sin \frac{2 \pi k}{10}, k=1,2,3, \ldots 9$

Here, $z_1, z_2, z_3, \ldots z_9$ are the roots of $z^{10}-1=0$

We know $z^{10}-1=(z-1)$

$\begin{aligned} & \left(z^9+z^8+z^7+z^6+z^5+z^4+z^3+z^2+z+1\right) \\ & =(z-1)\left(z-z_1\right)\left(z-z_2\right) \ldots\left(z-z_9\right) \\ \Rightarrow & 1+z+z^2+\ldots+z^9=\left(z-z_1\right)\left(z-z_2\right) \\ & \left(z-z_3\right) \ldots\left(z-z_9\right) \\ \Rightarrow & \left|1+z+z^2+\cdots+z^9\right| \\ & =\left|\left(z-z_1\right)\left(z-z_2\right)\left(z-z_3\right) \ldots\left(z-z_9\right)\right| \\ \Rightarrow & \left|z-z_1\right| \cdot\left|z-z_2\right|\left|z-z_3\right| \ldots\left|z-z_9\right| \\ & =\left|1+z+z^2+\ldots+z^9\right| \end{aligned}$

Put $z=1$

$\begin{aligned} & \Rightarrow\left|1-z_1\right| \cdot\left|1-z_2\right|\left|1-z_3\right| \ldots\left|1-z_9\right|=10 \\ & \Rightarrow \frac{\left|1-z_1\right|\left|1-z_2\right|\left|1-z_3\right| \ldots\left|1-z_9\right|}{10}=1 \end{aligned}$

Hence, R match with 3.

(S) We know the sum of all roots of $z^{10}-1=0$ is zero.

$\begin{aligned} & \Rightarrow 1+z_1+z_2+z_3+\ldots+z_9=0 \\ & \Rightarrow 1+\left(\cos \frac{2 \pi}{10}+i \sin \frac{2 \pi}{10}\right) \\ & +\left(\cos \left(\frac{4 \pi}{10}\right)+i \sin \left(\frac{4 \pi}{10}\right)\right) \\ & +\left(\cos \frac{6 \pi}{10}+i \sin \frac{6 \pi}{10}\right) \\ & +\ldots+\cos \frac{18 \pi}{10}+i \sin \frac{18 \pi}{10}=0 \\ \end{aligned}$

$\begin{aligned} & \begin{array}{l} \Rightarrow\left(1+\cos \frac{2 \pi}{10}+\cos \frac{4 \pi}{10}+\cdots+\cos \frac{18 \pi}{10}\right) \\ +i\left(\sin \frac{2 \pi}{10}+\sin \frac{4 \pi}{10}+\ldots+\sin \frac{18 \pi}{10}\right) \\ =0+i 0 \end{array} \\ & \Rightarrow 1+\cos \frac{2 \pi}{10}+\cos \frac{4 \pi}{10}+\cos \frac{6 \pi}{10}+\ldots+\cos \frac{18 \pi}{10}=0 \\ & \Rightarrow \sum_{k=1}^9 \cos \frac{2 k \pi}{10}=-1 \\ & \Rightarrow 1-\sum_{k=1}^9 \cos \frac{2 k \pi}{10}=1-(-1)=2 \end{aligned}$

Hint:

(i) Recall $(\cos \theta+i \sin \theta)(\cos \phi+i \sin \phi)=\cos (\theta+\phi)+i \sin (\theta+\phi)$

(ii) $\cos \theta=0$ when $\theta=2 n \pi$, where $n$ is an integer. and $\sin \theta=0$ when $\theta=n \pi$, where $n$ is an integer.

(iii) $1, z_1, z_2, z_3, \ldots z_9$ are the roots of $z^{10}-1=0$.

(iv) $z^{10}-1=(z-1)\left(1+z+z^2+\ldots+z^9\right)$

$=(z-1)\left(z-z_1\right)\left(z-z_2\right) \ldots\left(z-z_9\right)$

We know $|1-3 i-z|=|z-(1-3 i)|$ implies distance of $z$ from $(1,-3)$.

Let $P=(1,-3)$

$\min _{z \in \mathrm{S}}|1-3 i-z|=\mathrm{PN}$

$\begin{aligned} & \Rightarrow \min _{z \in S}|1-3 i-z|=\frac{|\sqrt{3} \cdot 1-3|}{\sqrt{(\sqrt{3})^2+1^2}} \\ & \Rightarrow \min _{z \in S}|1-3 i-z|=\frac{3-\sqrt{3}}{2} \end{aligned}$

Hints:

(i) $\left|z_1-z_2\right|$ implies distance between $z_1$ and $z_2$

(ii) The minimum distance of a point $\left(x_1, y_1\right)$ from the line $a x+b y+c=0$ is $\frac{\left|a x_1+b y_1+c\right|}{\sqrt{a^2+b^2}}$.

Given, $S_1=\{z \in C:|z|<4\}$ i.e. $S_1$ lies inside the circle of centre $(0,0)$ and radius 4.

Given, $S_3=\{z \in C: \operatorname{Re}(z)>0\}$ i.e. $S_3$ lies right side of the line $x=0$.

Also given $S_2=\left\{z \in C: \operatorname{Im}\left(\frac{z-1+\sqrt{3} i}{1-\sqrt{3} i}\right)>0\right\}$

Put $z=x+i y$ in $\operatorname{Im}\left(\frac{z-1+\sqrt{3} i}{1-\sqrt{3} i}\right)>0$

$\begin{aligned} & \Rightarrow \operatorname{Im}\left(\frac{x-1+i(y+\sqrt{3})}{1-\sqrt{3} i}\right)>0 \\ & \Rightarrow \operatorname{Im}\left(\frac{(x-1)+i(y+\sqrt{3})}{1-\sqrt{3} i} \times \frac{1+\sqrt{3} i}{1+\sqrt{3} i}\right)>0 \\ & \Rightarrow \operatorname{Im}\left[\begin{array}{l} \frac{(x-1)-\sqrt{3}(y+\sqrt{3})}{4}+ \\ i\left(\frac{\sqrt{3}(x-1)+(y+\sqrt{3})}{4}\right) \end{array}\right]>0 \\ \end{aligned}$

$\begin{aligned} & \Rightarrow \quad \frac{\sqrt{3} x+y}{4}>0 \\ & \Rightarrow \sqrt{3} x+y>0 \end{aligned}$

$\text { So, } S_2 \text { lies above the line } \sqrt{3} x+y=0$

$\text { Given, } S=S_1 \cap S_2 \cap S_3$

Here, $\angle \mathrm{AOC}=\frac{\pi}{2}+\frac{\pi}{3}=\frac{5 \pi}{6}$

Area of $S=\frac{1}{2}\left(\frac{5 \pi}{6}\right) \cdot 4^2=\frac{20 \pi}{3}$ Sq. Units

Hints:

(i) $|z| < a$ implies $z$ lies inside the circle of centre $(0,0)$ and radius $a$

(ii) $\operatorname{Re}(z)>0$ implies $z$ lies on the right side of the line $x=0$ i.e. $y$-axis.

(iii) Put $z=x+i y$ in the expression $\operatorname{Im}\left(\frac{z-1+\sqrt{3} i}{1-\sqrt{3} i}\right)>0$

(iv) If a circular arc $A B$ form $\theta$ angle at the centre of circle of radius R, then the area of this circular section is $\frac{1}{2} \theta \cdot r^2$

If $z=x+i y$, then $\left(x-x_0\right)^2+\left(y-y_0\right)^2=r^2$ and $\left(x-x_0\right)^2+\left(y-y_0\right)^2=4 r^2$ can be written as $\left|Z-Z_0\right|^2=r^2$ and $\left|Z-Z_0\right|^2=4 r^2$ respectively where, $\mathrm{Z}_0=x_0+i y_0$ (given)

$\Rightarrow\left|Z-Z_0\right|^2=r^2$ and $\left|Z-Z_0\right|^2=4 r^2$

$\Rightarrow\left(Z-Z_0\right)\left(\bar{Z}-\bar{Z}_0\right)=r^2$ and $\left(Z-Z_0\right)\left(\bar{Z}-\bar{Z}_0\right)=4 r^2\left(|Z|^2=Z \bar{Z}\right)$

$\begin{aligned} & \text { Given, } \alpha \text { and } \frac{1}{\bar{\alpha}} \text { line on circles }\left(x-x_0\right)^2+\left(y-y_0\right)^2 \\ & =r^2 \text { and }\left(x-x_0\right)^2+\left(y-y_0\right)^2=4 r^2 \text { respectively } \\ & \begin{array}{l} \therefore \quad\left(\alpha-Z_0\right)\left(\bar{\alpha}-\bar{Z}_0\right)=r^2 \text { and } \\ \quad\left(\frac{1}{\bar{\alpha}}-Z_0\right)\left(\frac{1}{\alpha}-\bar{Z}_0\right)=4 r^2 \end{array} \end{aligned}$

$\begin{aligned} \Rightarrow & |\alpha|^2-\alpha \bar{Z}_0-\bar{\alpha} Z_0+\left|Z_0\right|^2=r^2 \text { and } \\ & \frac{1}{|\alpha|^2}-\frac{1}{\bar{\alpha}} \bar{Z}_0-\frac{1}{\alpha} Z_0+\left|\bar{Z}_0\right|^2=4 r^2 \\ \Rightarrow & \alpha Z_0+\bar{\alpha} Z_0=|\alpha|^2+\left|Z_0\right|^2-r^2 \text { and } \\ & \frac{\alpha \bar{Z}_0}{|\alpha|^2}+\frac{\bar{\alpha} Z_0}{|\alpha|^2}=\frac{1}{|\alpha|^2}+\left|Z_0\right|^2-4 r^2 \\ \Rightarrow & \alpha \bar{Z}_0+\bar{\alpha} Z_0=|\alpha|^2+\left|Z_0\right|^2-r^2 =|\alpha|^2\left(\frac{1}{\left|\alpha^2\right|}+\left|Z_0^2\right|-4 r^2\right) \end{aligned}$

$\Rightarrow|\alpha|^2+\left|Z_0\right|^2-r^2=1+|\alpha|^2\left|Z_0\right|^2-4|\alpha|^2 r^2 \quad\text{... (i)}$

Given, $2\left|Z_0\right|^2=r^2+2$

Put $\left|Z_0\right|^2=\frac{r^2}{2}+1$ in the equation (i)

$\begin{aligned} & \Rightarrow|\alpha|^2+\frac{r^2}{2}+1-r^2=1+|\alpha|^2\left(\frac{r^2}{2}+1\right)-4|\alpha|^2 r^2 \\ & \Rightarrow|\alpha|^2-\frac{r^2}{2}=\frac{1}{2}|\alpha|^2 r^2+|\alpha|^2-4|\alpha|^2 r^2 \end{aligned}$

$\begin{array}{ll} \Rightarrow & -\frac{r^2}{2}=\frac{1}{2}|\alpha|^2 r^2-4|\alpha|^2 r^2 \\ \Rightarrow & -\frac{1}{2}=\frac{1}{2}(\alpha)^2-4|\alpha|^2 \\ \Rightarrow & \frac{-1}{2}=-\frac{7}{2}|\alpha|^2 \\ \Rightarrow & |\alpha|^2=\frac{1}{7} \\ \Rightarrow & |\alpha|=\frac{1}{\sqrt{7}} \end{array}$

Hints :

(i) Apply the property $Z \cdot \bar{Z}=|\bar{Z}|^2$

(ii) If $Z=(x+i y)$ and $Z_0=\left(x_0+i y_0\right)$ then $\left(x-x_0\right)^2+\left(y-y_0\right)^2=r^2$ and $\left(x-x_0\right)^2+ \left(y-y_0\right)^2=4 r^2$ can be written as $\left|Z-Z_0\right|^2=r^2$ and $\left|Z-Z_0\right|^2=4 r^2$ respectively.

Given, $\quad a=z^2+z+1$

$\Rightarrow z^2+z+1-a=0$

Using Sridharacharya Rule

$\begin{aligned} & \qquad z=\frac{-1 \pm \sqrt{1^2-4(1-a)}}{2} \\ & \Rightarrow \quad z=\frac{-1 \pm \sqrt{1-4+4 a}}{2} \\ & \Rightarrow \quad z=\frac{-1 \pm \sqrt{4 a-3}}{2} \end{aligned}$

Now, we can observe that if $\mathrm{a}=\frac{3}{4}$, then $z= \frac{-1}{2}$ which will become purely real.

Hence, if $a \neq \frac{3}{4}$, otherwise $z$ will be purely real.

(i) Make quadratic equation in $z$ and apply Sridharacharya rule.

(ii) Analyse the value of '$a$' for which '$z$' becomes purely real.

(A) z is equidistant from the points $i|z|$ and $ - i|z|$, whose perpendicular bisector is ${\mathop{\rm Im}\nolimits} (z) = 0$.

(B) Sum of distance of z from (4, 0) and ($-$4, 0) is a constant 10, hence locus of z is ellipse with semi-major axis 5 and focus at ($\pm$ 4, 0), ae = 4.

$\therefore$ $e = {4 \over 5}$

(C) $|z| \le |w| + \left| {{1 \over w}} \right| = {5 \over 2} < 3$

(D) $|z| \le |w| + \left| {{1 \over w}} \right| = 2$

$\therefore$ ${\mathop{\rm Re}\nolimits} (z) \le |z| \le 2$

We have

$\overline z {z^3} + z{\overline z ^3} = 350$

Substituting $z = x + iy$, we get

$({x^2} + {y^2})({x^2} - {y^2}) = 175$

$({x^2} + {y^2})({x^2} - {y^2}) = 5 \times 5 \times 7$

${x^2} + {y^2} = 25$

${x^2} - {y^2} = 7$

whose solutions are $x = \pm 4$ and $y = \pm 3;x,y \in I$.

Therefore, that is, area is found as $8 \times 6 = 48$ sq. unit.

We have

$X = \sin \theta + \sin 3\theta \, + \,...\, + \,\sin 29\theta $

$2(\sin \theta )X = 1 - \cos 2\theta + \cos 2\theta - \cos 4\theta \, + \,...\, + \,\cos 28\theta - \cos 30\theta $

$X = {{1 - \cos 30\theta } \over {2\sin \theta }} = {1 \over {4\sin 2^\circ }}$

given $-$ $\alpha$ particle

In the direction of, or, Now rotation about origin through angle of means multiply by

${z_0} = 1 + 2i = (1,2) = ({x_0},{y_0})$

${z_1} = ({x_0} + 5,{y_0} + 3)$

$ = (6,5) = 6 + 5i$

$\Rightarrow 2$ in the direction of $i = j$

${x_1} = 2\cos 45^\circ $

${y_1} = 2\sin 45^\circ $

${z_2}(7 + 6i)$

Now, rotation about origin through an angle of 2$\pi$ means multiply z$_2$ by $ \to {e^{i\pi /2}}$

$\cos 90^\circ + i\sin 90^\circ = i$

${z_3} = i(7 + 6i)$

$ = - 6 + 7i$

$|z + 1 - i{|^2} + |z - 5 - i{|^2}$

The points ($-1,1$) and (5, 1) are the extremities of the diameter of the given circle of radius 3

Hence, PA$^2$ + PB$^2$ = AB$^2$ = 36

Since, $|w - (2 + i)| < 3$

$|w| - |2 + i| < 3$

$ - 3 + \sqrt 5 < |w| < 3 + \sqrt 5 $

$ - 3 - \sqrt 5 < |w| < 3 - \sqrt 5 $

Also, $|z - (2 + i)| = 3$

$ - 3 + \sqrt 5 \le |z| \le 3 + \sqrt 5 $

$\therefore$ $ - 3 < |z| - |w| + 3 < 9$

In the Cartesian coordinates sets A, B and C defined the regions given by

$A:y\ge1,B:(x-2)+(y-1)^2=9$

B and C being a circle and a straight line intersect in two points, out of which only one satisfies $y > -1$.

Thus, the no. of elements in the set A $\cap$ B $\cap$ C is 1.

Given, $|z|=1$ and $z \neq \pm 1$

To find : values of $\frac{z}{1-z^{2}}$ lies on ?

Any complex number $z$ can be written as :

Since, $|z|=1$

$z=r(\cos \theta+i \sin \theta), \text { where } r=|z| $

Then, $z=(\cos \theta+i \sin \theta)$

Consider, $\frac{z}{1-z^{2}}$

$\begin{aligned}& =\frac{\cos \theta+i \sin \theta}{1-\left(\cos ^{2} \theta+i^{2} \sin ^{2} \theta+2 i \sin \theta \cos \theta\right)} \\& =\frac{\cos \theta+i \sin \theta}{1-\cos ^{2} \theta+\sin ^{2} \theta-2 i \sin \theta \cos \theta} \\& =\frac{\cos \theta+i \sin \theta}{\sin ^{2} \theta+\sin ^{2} \theta-2 i \sin \theta \cos \theta} \\ & =\frac{\cos \theta+i \sin \theta}{2 \sin ^{2} \theta-2 i \sin \theta \cos \theta} \\& =\frac{\cos \theta+i \sin \theta}{2 \sin \theta(\sin \theta-i \cos \theta)} \end{aligned}$

$\begin{aligned}& =\frac{(\cos \theta+i \sin \theta)(\sin \theta+i \cos \theta)}{2 \sin \theta} \\ & {\left[\frac{1}{(\sin \theta-i \cos \theta)}=\sin \theta+i \cos \theta\right]} \\& =\frac{\sin \theta \cos \theta+i \cos ^{2} \theta+i \sin ^{2} \theta+i^{2} \sin \theta \cos \theta}{2 \sin \theta} \\ & =\frac{i\left(\cos ^{2} \theta+\sin ^{2} \theta\right)}{2 \sin \theta}=\frac{i}{2 \sin \theta}\end{aligned}$

Thus, $\frac{z}{1-z^{2}}$ lies on the imaginary axis that is the $\mathrm{Y}$-axis in argand plane.

$\overrightarrow {OP} = \overrightarrow {OA} + \overrightarrow {OP} $

$\overrightarrow {OP} = \overrightarrow {OA} + \overrightarrow {OB} $

$\overrightarrow {OP} = 3{e^{{{i\pi } \over 4}}} + 4{e^{i\left( {{\pi \over 2} + {\pi \over 4}} \right)}}$

$ = 3{e^{{{i\pi } \over 4}}} + 4{e^{{{i\pi } \over 2}}}.\,{e^{{{i\pi } \over 4}}}$

$ = 3{e^{{{i\pi } \over 4}}} + 4i{e^{{{i\pi } \over 4}}}$

$ = {e^{{{i\pi } \over 4}}}(3 + 4i)$

$w=\alpha+i \beta$

since, $\left(\frac{w-\bar{w} z}{1-z}\right)$ is purely real

then, $\quad \frac{w-\bar{w} z}{1-z}=\frac{\overline{w-\bar{w} z}}{\overline{1-z}}=\frac{\bar{w}-w \bar{z}}{1-\bar{z}}$

$\Rightarrow(w-\bar{w} z)(1-\bar{z})=(\bar{w}-w \bar{z})(1-z)$

$\Rightarrow(w-\bar{w} z)-w \bar{z}+\bar{w}|z|^{2}$

$=\bar{w}-w \bar{z}-\bar{w} z+w|z|^{2}$

$\Rightarrow \quad w+\bar{w}|z|^{2}=\bar{w}+w|z|^{2}$

$\Rightarrow \quad|z|^{2}(w-\bar{w})=w-\bar{w}$

$\Rightarrow \quad|z|^{2}=\frac{w-\bar{w}}{w-\bar{w}}$

$\Rightarrow \quad|z|^{2}=1$

Since, $\quad w-\bar{w}=\alpha+i \beta-\alpha+i \beta$

$=2 i \beta$ and $\beta \neq 1$

$\therefore \quad|z|=1$ but $|z| \neq 1$

Shortcut Method:

Since $z=\bar{z}$ then

$\begin{aligned} x+i y & =x-i y \\ \Rightarrow \quad x+i y-x+i y & =0 \\ \Rightarrow \quad 2 i y & =0 \\ \Rightarrow \text { Imaginary part } & =0 \end{aligned}$

Here, equation of $C_2:(x-1)^2+(y-1)^2=(\sqrt{2})^2$ and $C_1:(x-1)^2+(y-1)^2=(1)^2$

$\begin{aligned} & \therefore P(1+\cos \theta, 1+\sin \theta) \text { and } Q(1+\sqrt{2} \sin \theta, 1+\sqrt{2} \sin \theta) \\\\ & \therefore P A^2+P B^2+P C^2+P D^2\end{aligned}$

$\begin{aligned}=\{(1 & \left.+\cos \theta)^2+(1+\sin \theta)^2\right\} +\left\{(\cos \theta-1)^2+(1+\sin \theta)^2\right\} +\left\{(1+\cos \theta)^2+(1-\sin \theta)^2\right\} +\left\{(1-\cos \theta)^2+(1-\sin \theta)^2\right\} \\\\ =16 & \end{aligned}$

Similarly,

$ \begin{array}{ll} Q A^2+Q B^2+Q C^2+Q D^2 & =12 \\ \therefore \frac{\Sigma Q A^2}{\Sigma P A^2}=\frac{12}{16}=0.75 \end{array} $

Hence (a) is the correct answer.

$|z-1|=\sqrt{2}$

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

$\therefore$ Centre $=(1,0)$

i.e., $1+i \cdot 0\left(\right.$ say $\left.z_{1}\right)$

$\begin{aligned} z_{2} & =2+\sqrt{3} i \\ \text { radius } & =\sqrt{2} \end{aligned}$

To obtain B, OA is rotated with $90^{\circ}$

$\begin{aligned} \frac{z_{3}-z_{1}}{z_{2}-z_{1}} & =\left|\frac{z_{3}-z_{1}}{z_{2}-z_{1}}\right| e^{i \pi / 2} \\ \Rightarrow \quad \frac{z_{3}-z_{1}}{z_{2}-z_{1}} & =\left|\frac{z_{3}-z_{1}}{z_{2}-z_{1}}\right|\left(\cos \frac{\pi}{2}+i \sin \frac{\pi}{2}\right) \\ \Rightarrow \quad \frac{z_{3}-z_{1}}{z_{2}-z_{1}} & =\frac{\mathrm{OB}}{\mathrm{OA}}(0+i) \\ \Rightarrow \quad \frac{z_{3}-z_{1}}{z_{2}-z_{1}} & =1 . i\quad \because (\mathrm{OA=OB (radius))} \\ \Rightarrow \quad\left(z_{3}-z_{1}\right) & =i\left(z_{2}-z_{1}\right) \\ \Rightarrow \quad z_{3}-1 & =i(2+\sqrt{3} i-1) \\ \Rightarrow \quad z_{3} & =1+i+\sqrt{3} i^{2} \\ & =1-\sqrt{3}+i \end{aligned}$

Similarly, we can calculate Co-ordinate D

$\begin{aligned} \frac{z_{5}-z_{1}}{z_{2}-z_{1}} & =\left|\frac{z_{5}-z_{1}}{z_{2}-z_{1}}\right| e^{i\left(\frac{-\pi}{2}\right)} \\ & =\frac{\mathrm{OD}}{\mathrm{OA}} \cos \left(\frac{-\pi}{2}\right)+i\left(-\sin \frac{\pi}{2}\right) \\ & =0-i \\ \Rightarrow \quad z_{5}-z_{1} & =\left(z_{2}-z_{1}\right)(-i) \\ \Rightarrow \quad z_{5} & =z_{1}+(1+i \sqrt{3})(-i) \\ & =1+(-i)+\sqrt{3} \\ & =(1+\sqrt{3})-i \end{aligned}$

Similarly, we can calculate co-ordinate C

$\begin{aligned} \frac{z_{4}-z_{1}}{z_{2}-z_{1}} & =\left|\frac{z_{4}-z_{1}}{z_{2}-z_{1}}\right| e^{i \pi} \\ \Rightarrow \quad \frac{z_{4}-z_{1}}{z_{2}-z_{1}} & =\frac{\mathrm{OC}}{\mathrm{OA}}(-1+0)\end{aligned}$

$\begin{aligned} \Rightarrow \quad z_{4}-z_{1} & =\left(z_{2}-z_{1}\right)(-1) \\ & =(-1)(1+\sqrt{3} i)=-1-\sqrt{3} i \\ \Rightarrow \quad z_{4} & =z_{1}-1-\sqrt{3} i \\ & =1-1-\sqrt{3} i=-\sqrt{3} i \end{aligned}$