MATHS MCQ

$⇒$ $\omega + {\omega ^{\left( {\frac{{1/2}}{{1 - 3/4}}} \right)}}$
$ \Rightarrow $ $\omega + {\omega ^2} = - 1$

$[\because 1 + \omega  + {\omega ^2} = 0]$

$[\because \,\,\omega  + {\omega ^2} =  - 1]$ 
= ${( - 2\omega )^4} = 16{\omega ^4} = 16\omega $.

$ \Rightarrow $${z^{69}} = {i^{69}}.{\omega ^{69}} = i$

$(\because {\omega ^{3n}} = {i^{4n}} = 1)$

$⇒$  ${x^3}(x - 1) + 1(x - 1) = 0$
$x - 1 = 0$ or ${x^3} + 1 = 0$

$⇒$ $x = 1,\, - 1,\,\frac{{1 + \sqrt 3 i}}{2},\,\,\frac{{1 - \sqrt 3 i}}{2}$
so its real roots are $1$ and $-1.$

[Since $1 + \omega + {\omega ^2} = 0,\,{\omega ^3} = 1$]
$= 729.$

[Since $1 + \omega + {\omega ^2} = 0,\,{\omega ^3} = 1$]
$= 0.$

$ + \cos (m\theta  + n\phi ) - i\sin (m\theta  + n\phi )$
$ = 2\cos (m\theta + n\phi )$

$......+ (n + 1)\,(n\omega + 1)\,\,(n{\omega ^2} + 1)$
= $\sum\limits_{r = 1}^n {(r + 1)\,(r\omega + 1)\,\,(r{\omega ^2} + 1)} $
= $\sum\limits_{r = 1}^n {(r + 1)\,({r^2}{\omega ^3} + r\omega + r{\omega ^2} + 1)} $
= $\sum\limits_{r = 1}^n {(r + 1)\,({r^2} - r + 1)} $= $\sum\limits_{r = 1}^n {({r^3} - {r^2} + r + {r^2} - r + 1)} $
= $\sum\limits_{r = 1}^n {({r^3}) + \sum\limits_{r = 1}^n {(1} )} $ = ${\left[ {\frac{{n(n + 1)}}{2}} \right]^2} + n$.

 For the equation ${(x - 2)^3} + 27 = 0$

==> ${(x - 2)^3} = - 27 = - {3^3}$

==> $x - 2 = - 3{(1)^{1/3}} = - 3(1,\omega ,{\omega ^2}) = - 3, - 3\omega ,3{\omega ^2}$

==> $x = - 1,2 - 3\omega ,2 - 3{\omega ^2}$.

For minimum, $Z =\frac{1}{2}+\frac{1}{2} ;$ minimum value

$=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=2 \sqrt{2}$

Let $z=x+i y$ then $\bar{z}=x-i y$

$\Rightarrow \quad z^2=x^2-y^2+2 i x y$

and $\bar{z}^2=x^2-y^2-2 i x y$

$\therefore \quad 10 z \bar{z}-3\left(z^2+\bar{z}^2\right)+4 i\left(z^2-\bar{z}^2\right)=0$ $\Rightarrow \quad 10\left(x^2+y^2\right)-3 \cdot 2\left(x^2-y^2\right)+4 i \cdot(4 i x y)=0$

$\Rightarrow \quad \quad 10\left(x^2+y^2\right)-6 x^2+6 y^2-16 x y=0$

$\Rightarrow \quad 2 x^2+8 y^2-8 x y=0$

$\Rightarrow \quad x^2-4 x y+4 y^2=0 \Rightarrow \quad(x-2 y)^2=0$

Which represents a straight line.

lt is given that complex numbers $z=x+i y$ and $w=u+i v$ are on the unit circle such that $z^2+w^2=1$

So, $\quad\left(z^2\right)+\left(w^2\right)=1$

$\Rightarrow \frac{1}{z^2}+\frac{1}{w^2}=1 \Rightarrow z^2+w^2=z^2 w^2$

$\Rightarrow \quad z^2 w^2=1 \quad \ldots (ii)$

$\because$ The number of solution of equations $x^2+y^2=1$ and $x^2 y^2=1$ is eight

$\therefore$ Number of ordered pair $(z, w)$ is also eight.

We have,

$|z| =1$

$z^n =1$

$z^n =e^{i 2 k \pi} ; k \in I$

$k \in(0, n-1)$

All roots of unity lie on arc of circle.

$\therefore$ There are infinitely many roots of unity.

We have,

$a=\cos 1^{\circ}$ and $b=\sin 1^{\circ}$

$(\cos x+i \sin x)^n=\cos n x+i \sin n x$

$(\cos x-i \sin x)^n=\cos n x-i \sin n x$

$2 \cos n x=(\cos x+i \sin x)^n +(\cos x-i \sin x)^n$

Put $x=1^{\circ}$ and $n=60 =2 \cos 60^{\circ}=2\left(\cos ^{60} 1^{\circ}+{ }^6 C_2 \cos ^{58} 1\right.$

Change all $\sin 1^{\circ}$ to $\cos 1^{\circ}$ using the identity $\cos ^2 1^{\circ}=1-\sin ^2 1^{\circ}$ equation with root $\cos 1^{\circ}$ so it is algebraic.

Similarly, for $b=\sin 1^{\circ}$

We have,

$11 z^8+20 i z^7+10 i z-22 =0$

$z^8+\frac{20}{11} i z^7+\frac{10 i}{11} z-2 =0$

Root of equation are $z_1, z_2, z_3, z_4, z_5, \ldots, z_8$ $\sum_{i=1}^8 z i=\frac{20}{11} i$

$z_1 z_2 z_3 z_4 z_5 z_6 z_7 z_8=2$

$\mid z_1\left\|z_2\right\| z_3\left\|z_4\right\| z_4\left\|z_5\right\| z_6\left\|z_7\right\| z_8 \|=2$

$1<\left|z_i\right|<2$

$1+1+1<|z|^2+|z|+1<2^2+2+1$

$3<|z|^2+|z|+1<7$

$3 < S < 7$

Given, $z_1=\sqrt{3}+i$ and $z_2=\sqrt{3}-i$ are two adjacent vertices of $n$-sided regular polygon centred of origin.

$z_1 =\sqrt{3}+i$

$\arg \left(z_1\right) =\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi}{6}$

$z_2 =\sqrt{3}-i$

$\arg \left(z_2\right) =\tan ^{-1}\left(-\frac{1}{\sqrt{3}}\right)=-\frac{\pi}{6}$

$\angle z_2 O z_1 =\frac{\pi}{3}$

$\frac{n \pi}{3} =2 \pi$

$n =6$

As the vertices lie on a unit circle with centre at origin in a complex plane, it can be argued that the vertices are the roots of the equation $z^7-1=0$.

Considering this polynomial equation, it evident that the sum of roots, the sum of roots taken $2$ at a time and so on upto roots taken $6$ at a time are all $0$ as the corresponding coefficients in the polynomial equation are $0$ .

We have,

$w=\left(\frac{z-a}{1-\bar{a} z}\right),|a| < 1$

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

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

$\Rightarrow \quad w+a=z(1+w \bar{a})$

$\Rightarrow \quad z=\frac{w+a}{1+w \bar{a}}$

When $|z| < 1$

$\left|\frac{w+a}{1+w}\right| < 1$

$\Rightarrow \quad|w+a|<|1+w \bar{a}|$

$\Rightarrow(w+a)(\bar{w}+\bar{a}) < (1+w \bar{a})(1+\bar{w} a)$

$\Rightarrow \quad w \bar{w}+w \bar{a}+a \bar{w}+a \bar{a} < 1+\bar{w} a$

$+w \bar{a}+w \bar{w} a \bar{a}$

$\Rightarrow \quad|w|^2+|a|^2 < 1+|u|^2|\alpha|^2$

$\Rightarrow|w|^2|a|^2-|w|^2-|a|^2+1 > 0$

$\Rightarrow \quad\left(|w|^2-1\right)\left(|a|^2-1\right) > 0$

$\Rightarrow \quad|u|^2-1 < 0 \quad[\because|a| < 1]$

$\therefore \quad|w| < 1,|z| < 1$

Given,

$A_r=\left\{e^{j \pi r n}: n\right.$ is a natural number $\}$

$A_1=e^{i n \pi}$ which is finite.

$A_{0.3}=e^{j 0 \cdot 3 n \pi}$ is also finite

$A \frac{1}{\pi}=e^{n i}$ is infinite set

$\therefore$ Option $(d)$ is correct.

We have, $\left|z^3+z^{-3}\right| \leq 2$

Now, $\left|z^3+\frac{1}{z^3}\right| \leq 2$

$\Rightarrow \quad\left|z^3+\frac{1}{z^3}\right| \leq\left|z^3\right|+\frac{1}{\left|z^3\right|}$

$\left.\Rightarrow \frac{|z|^3+\frac{1}{|z|^3}}{2} \geq \sqrt{|z|^3 \frac{1}{|z|^3}} \quad \because \text { AM } \geq GM \right]$

$\Rightarrow|z|^3+\frac{1}{|z|^3} \geq 2 \Rightarrow|z|^3+\frac{1}{|z|^3}=2$

$\therefore \quad|z|=1$

$\therefore \text { Maximum value of }\left|z+\frac{1}{z}\right| \leq|z|+\frac{1}{|z|}=2$

We have, $\left(1+x+x^2\right)^{2014}$

$=a_0+a_1 x+a_2 x^2+\ldots+a_{4028} x^{4028}$

Put $x=-1$

$1=a_0-a_1+a_2-a_3+a_4-\ldots+a_{4028} \ldots$ (i)

Put $x=-\omega$

$\left(1-\omega+\omega^2\right)^{2014} =(2 \omega)^{2014}$

$\quad=a_0-a_1 \omega+a_2 \omega^2-a_3+\ldots \ldots \text { (ii) }$

$\text { Put }$ $x =-\omega^2$

$\left(1-\omega^2+\omega\right)^{2014} =\left(2 \omega^2\right)^{2014}$

$\quad=a_0-\omega_1 \omega^2+a_2(1)-\alpha_3+a_4 \omega^2+\ldots \ldots \text { (iii) }$

On adding Eqs. $(i), (ii)$ and $(iii)$, we get

$1+(2 \omega)^{2014}+\left(2 \omega^2\right)^{2014}$

$\quad=3\left(a_0-a_3+a_6+\ldots\right)$

$\Rightarrow a_0-a_3+a_6+\ldots=\frac{1+2^{2014} \omega+2^{2014} \omega}{3}$

$\Rightarrow A=\frac{1-2^{2014}}{3}-$

$\Rightarrow |A|=\frac{2^{2014}-1}{3}$

and Eq.$(i)$ $+ Eq$. $(ii)$ $\omega+ Eq$. $(iii)$ $\omega^2$,

we get

$\frac{1+2^{2014} \omega^{2014} \omega+2^{2014}\left(\omega^2\right)^{2014} \omega^2}{3}$

$=a_2-a_5+a_8+\ldots$

$\Rightarrow \quad C=\frac{1+2^{2014} \omega^{2015}+\omega^{2014} \omega^{4030}}{3}$

$\Rightarrow \quad C=\frac{1-2^{2014}}{3}$

$\Rightarrow|C|=\frac{2^{2014}-1}{-}=|A|$ and Similarly,

Eq. $(i)$ $+$ Eq.$(ii)$ $\omega^2+$ Eq.$(iii)$ $\omega$, we get

$\frac{1+2^{2014} \cdot \omega^{2014} \omega^2+2^{2014}\left((\omega)^2\right)^{2014} \cdot \omega}{3}$

$=a_1-a_4+a_7+\ldots$

$\Rightarrow \quad B=\frac{1+2^{2014} \omega^{2016}+2^{2014} \omega^{4029}}{3}$

$\Rightarrow \quad B=\frac{1+2^{2014}+2^{2014}}{3}=\frac{1+2^{2015}}{3}$

$\Rightarrow|B| > |A|=|C|$

We have, $\left|e^{\sum \limits_{k=0}^n c_k \omega^k}\right|$, where $\omega$ is cube root of unity.

$\sum \limits_{k=0}^n{ }^n C_k \omega{ }^k={ }^n C_0+{ }^n C_1 \omega+{ }^n C_2 \omega{ }^2+{ }_{\left.\ldots+{ }^n C_n \omega\right)^n}$

$=(1+\omega)^n$

$=\left(-\omega^2\right)^n \quad\left[\because 1+w+w^2=0\right]$

$=(-1)^n \omega^{2 n} \quad$

$\left|e^{\sum \limits_{k=0}^n C_k \omega^k}\right|=\left|e^{(-1)^n \omega^{2 n}}\right|=\left|e^{\left(-\omega^2\right)^n}\right|$

$=\left|e^{\left(-\cos \frac{4 \pi}{3}-i \sin \frac{4 \pi}{3}\right)^n \mid}\right|$

$=\left|e^{\left(\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}\right)^n}\right|$

$=e^{\cos \frac{n \pi}{3}+i \sin \frac{n \pi}{3}}$

$=\left|e^{\cos \frac{n \pi}{3}}\right| \cdot\left|e^{i \sin \frac{n \pi}{3}}\right|$

$=\left|e^{\cos \frac{n \pi}{3}}\right| \quad\left[\because\left|e^{i \theta}\right|=1\right]$

$=e^1, e^{1 / 2}, e^{-1 / 2}, e^{-1}$

only four values possible.

We have,

$\left|a+b \omega+c \omega^2\right|, a, b, c \in\{-1,1\}$

For maximum put $a=1, b=-1, c=-1$

$\left|1-\omega-\omega^2\right|$

$\left|1-\left(\omega+\omega^2\right)\right|=\mid 1+ 1 \mid=2$

${\left[\because \omega+\omega^2=-1\right] }$

Given, $1, \omega, \omega^2$ are cube roots of unity.

Let the roots of polynomials are

$ \alpha = 2 \omega^2, \quad \beta = 3 + 4 \omega $
$ \gamma = 3 + 4 \omega^2, \quad \delta = 5 - \omega + \omega^2 $
$ = 5 - (\omega + \omega^2) = 5 + 1 = 6 $

$\left[\because 1+\omega+\omega^2=0\right]$

If $\alpha=2 \omega^2$ is a root the $2 \omega$ has to be also root.

$\therefore$ Total roots of polynomial $=5$.

Hence, minimum degree of polynomial  $=5$.

$ \left|z^2\right|=|i \bar{z}|$

$ \left|z^2\right|=|z|$

$ |z|^2-|z|=0$

$ |z|(|z|-1)=0$

$ |z|=0 \text { (not acceptable) } $

$ \therefore|z|=1 $

$ \therefore|z|^2=1$

$ \text { Let } z+2=\cos \theta+i \sin \theta $

$ \frac{1}{z+2}=\cos \theta-i \sin \theta $

$ \Rightarrow \frac{z+1}{z+2}=1-\frac{1}{z+2}=1-(\cos \theta-i \sin \theta) $

$ =(1-\cos \theta)+\operatorname{isin} \theta $

$ \operatorname{Im}\left(\frac{z+1}{z+2}\right)=\sin \theta, \sin \theta=\frac{1}{5} $

$ \cos \theta= \pm \sqrt{1-\frac{1}{25}}= \pm \frac{2 \sqrt{6}}{5} $

$ |\operatorname{Re}(\overline{z+2})|=\frac{2 \sqrt{6}}{5}$

$ r=\sqrt{x^2+y^2}  \theta=\tan ^{-1} \frac{y}{x} $

$ r=\sqrt{(2)^2+\left(2 \tan \frac{5 \pi}{8}\right)^2} $

$ =\left|2 \sec \frac{5 \pi}{8}\right|=\left|2 \sec \left(\pi-\frac{3 \pi}{8}\right)\right|$

$ =2 \sec \frac{3 \pi}{8} $

$  \theta=\tan ^{-1}\left(\left.\frac{-2 \tan \frac{5 \pi}{8}}{2} \right)\,\right. $

$ =\tan ^{-1}\left(\tan \left(\pi-\frac{5 \pi}{8}\right)\right) $

$ =\frac{3 \pi}{8}$

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

So $n(S)=1$

Min. value of $\left|z+\frac{3}{2}+2 i\right|$  is actually zero.

$ a=42-5 b, b=1, a=37 $

$ b=2, a=32 $

$ b=3, a=27 $

$ \vdots $

$ b=8, a=2 $

$ R \text { has "8" elements } \Rightarrow \mathrm{m}=8 $

$ \sum_{n=1}^8\left(1-i^{n !}\right)=x+i y $

$ \text { for } n \geq 4, i^{n !}=1 $

$ \Rightarrow(1-i)+\left(1-i^{2 !}\right)+\left(1-i^{3 !}\right) $

$ =1-I+2+1+1 $

$ =5-I=x+i y $

$ m+x+y=8+5-1=12$

$ 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$

$ |\mathrm{z}+1|=\alpha \mathrm{z}+\beta(1+\mathrm{i}) $

$ \left|\frac{3}{2}-2 \mathrm{i}\right|=\frac{\alpha}{2}-2 \alpha \mathrm{i}+\beta+\beta \mathrm{i} $

$ \left|\frac{3}{2}-2 \mathrm{i}\right|=\left(\frac{\alpha}{2}+\beta\right)+(\beta-2 \alpha) \mathrm{i} $

$ \beta=2 \alpha \text { and } \frac{\alpha}{2}+\beta=\sqrt{\frac{9}{4}+4} $

$ \alpha+\beta=3$

$z=\frac{\pi}{4}(1+i)^4\left[\frac{\sqrt{\pi}-\pi i-i-\sqrt{\pi}}{\pi+1}+\frac{\sqrt{\pi}-i-\pi i-\sqrt{\pi}}{1+\pi}\right]$

$=-\frac{\pi i}{2}\left(1+4 i+6 i^2+4 i^3+1\right)$

$=2 \pi i$

$\beta=\frac{2 \pi}{\frac{\pi}{2}}=4$

Distance from $(1,4)$ to $4 x-3 y=7$

Will be $\frac{15}{5}=3$

$z_1^3+z_2^3=20+15 i$

$z_1^3+z_2^3=\left(z_1+z_2\right)^3-3 z_1 z_2\left(z_1+z_2\right)$

$z_1^3+z_2^3=125-3 z_1 \cdot z_2(5)$

$\Rightarrow 20+15 i=125-15 z_1 z_2$

$\Rightarrow 3 z_1 z_2=25-4-3 i$

$\Rightarrow 3 z_1 z_2=21-3 i$

$\Rightarrow z_1 \cdot z_2=7-i$

$\Rightarrow\left(z_1+z_2\right)^2=25$

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

$\Rightarrow 11+2 i$

$\left(z_1^2+z_2^2\right)^2=121-4+44 i$

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

$\Rightarrow z_1^4+z_2^4=117+44 i-2(49-1-14 i)$

$\Rightarrow\left|z_1^4+z_2^4\right|=75$

Then $(x-1)^2+y^2=1 \rightarrow(1)$

$(\sqrt{2}-1)(2 x)-i(2 i y)=2 \sqrt{2}$

$\quad \Rightarrow(\sqrt{2}-1) x+y=\sqrt{2} \rightarrow(2)$

Solving $(1) and (2)$ we get

Either $\mathrm{x}=1$ or $x=\frac{1}{2-\sqrt{2}} \rightarrow$ $(3)$

On solving $(3)$ with $(2)$ we get

For $x=1 \Rightarrow y=1 \Rightarrow Z_2=1+i$  and  for

$x=\frac{1}{2-\sqrt{2}} \Rightarrow y=\sqrt{2}-\frac{1}{\sqrt{2}} \Rightarrow Z_1=\left(1+\frac{1}{\sqrt{2}}\right)+\frac{i}{\sqrt{2}}$

Now

$\left|\sqrt{2} z_1-z_2\right|^2$

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

$=(\sqrt{2})^2$

$=2$

$\bar{z}=x-i y $

$\bar{z}^2=x^2-y^2-2 i x y $

$\Rightarrow x^2-y^2-2 i x y+\sqrt{x^2+y^2}=0 $

$\Rightarrow x=0 \quad \text { or } $$ y=0 $

$-y^2+|y|=0 $ $ x^2+|x|=0 $

$|y|=|y|^2 $ $ \Rightarrow x=0 $

$y=0, \pm 1 $ $ \Rightarrow \alpha=i-i=0 $

$\Rightarrow i,-i $ $ \beta=i(-i)=1 $

are roots $ 4(0+1)=4$

are roots

$\left(\left|z_1\right|+\left|z_2\right|\right)\left|\frac{z_1}{\left|z_1\right|}+\frac{z_2}{\left|z_2\right|}\right|$

Since $\left|\frac{z_1}{\left|z_1\right|}+\frac{z_2}{\left|z_2\right|}\right| \leq\left|\frac{z_1}{\left|z_1\right|}\right|+\left|\frac{z_2}{\left|z_2\right|}\right|$

$ \left|\frac{z_1}{\left|z_1\right|}+\frac{z_2}{\left|z_2\right|}\right| \leq \frac{\left|z_1\right|}{\left|z_1\right|}+\frac{\left|z_2\right|}{\left|z_2\right|} $

$ \left|\frac{z_1}{\left|z_1\right|}+\frac{z_2}{\left|z_2\right|}\right| \leq 2$

$\left(\left|z_1\right|+\left|z_2\right|\right)\left(\left|\frac{z_1}{\left|z_1\right|}+\frac{z_2}{\left|z_2\right|}\right|\right) \leq 2\left(\left|z_1\right|+\left|z_2\right|\right)$

$\therefore$ statement $I$ is correct

For Statement $II$ :

$ \frac{a}{|y-z|}=\frac{b}{|z-x|}=\frac{c}{|x-y|} $

$ \frac{a^2}{|y-z|^2}=\frac{b^2}{|z-x|^2}=\frac{c^2}{|x-y|^2}=\lambda $

$ a^2=\lambda\left(|y-z|^2\right)=\lambda(y-z)(\bar{y}-\bar{z}) $

$ b^2=\lambda(z-x)(\bar{z}-\bar{x}) \text { and } c^2=\lambda(x-y)(\bar{x}-\bar{y}) $

$ \frac{a^2}{y-z}+\frac{b^2}{z-x}+\frac{c^2}{x-y}=\lambda(\bar{y}-\bar{z}+\bar{z}-\bar{x}+\bar{x}-\bar{y})=0$

Statement $II$ is false

$ \Rightarrow\left(\alpha-z_0\right)\left(\bar{\alpha}-\bar{z}_0\right)=4 $

$ \Rightarrow \alpha \bar{\alpha}-\alpha \bar{z}_0-z_0 \bar{\alpha}+\left|z_0\right|^2=4 $

$ \Rightarrow|\alpha|^2-\alpha \bar{z}_0-z_0 \bar{\alpha}=2 \ldots \ldots .$$...............(i)$

$ \left|z-z_0\right|^2=16$

$ \Rightarrow\left(\frac{1}{\bar{\alpha}}-z_0\right)\left(\frac{1}{\alpha}-\bar{z}_0\right)=16 $

$ \Rightarrow\left(1-\bar{\alpha} z_0\right)\left(1-\alpha \bar{z}_0\right)=16|\alpha|^2 $

$ \Rightarrow 1-\bar{\alpha} z_0-\alpha \bar{z}_0+|\alpha|^2\left|z_0\right|^2=16|\alpha|^2 $

$ \Rightarrow 1-\bar{\alpha} z_0-\alpha \bar{z}_0=14|\alpha|^2 \ldots \ldots \ldots$ .$.........(ii)$

From $(1)$ and $(2)$

$\Rightarrow 5|\alpha|^2=1 $

$\Rightarrow 100|\alpha|^2=20$