Complex Numbers

$\frac{3-2 i \sin \theta}{1+2 i \sin \theta}=\frac{(3-2 i \sin \theta)}{(1+2 i \sin \theta)} \times \frac{(1-2 i \sin \theta)}{(1-2 i \sin \theta)}$

(by rationalising)

$ \begin{aligned} & =\frac{3-6 i \sin \theta-2 i \sin \theta-4 \sin ^2 \theta}{1+4 \sin ^2 \theta} \\ & =\frac{3-4 \sin ^2 \theta-8 i \sin \theta}{1+4 \sin ^2 \theta} \end{aligned} $

Since, $\frac{3-2 i \sin \theta}{1+2 i \sin \theta}$ is purely imaginary.

$ \begin{array}{ll} \therefore & \frac{3-4 \sin ^2 \theta}{1+4 \sin ^2 \theta}=0 \\ \Rightarrow & 3-4 \sin ^2 \theta=0 \Rightarrow 4 \sin ^2 \theta=3 \\ \Rightarrow & \sin ^2 \theta=\frac{3}{4} \Rightarrow \sin \theta=\frac{\sqrt{3}}{2} \\ \Rightarrow & \theta=n \pi \pm \frac{\pi}{3} \end{array} $

Given, $z=x+i y, x^2+y^2=1$ and

$z_1=z e^{i \theta}$

$ z_1=(x+i y) e^{i \theta} $

Taking (2n)th power of both sides of the equation

$ \begin{aligned} z_1^{2 n} & =(x+i y)^{2 n} e^{i 2 n \theta} \\ & =r^{2 n}(\cos (2 n \theta)+i \sin (2 n \theta)) \\ z_1^{2 n}+1 & =r^{2 n}(\cos (2 n \theta)+i \sin (2 n \theta))+1 \\ & =(\cos (2 n \theta)+i \sin (2 n \theta))+1 \\ & {\left[\because x^2+y^2=1\right] } \\ z_1^{2 n}-1 & =(\cos (2 n \theta)+i \sin (2 n \theta))-1 \\ \frac{z_1^{2 n}-1}{z_1^{2 n}+1} & =\frac{(\cos (2 n \theta)+i \sin (2 n \theta))-1}{(\cos (2 n \theta)+i \sin (2 n \theta))+1} \\ & =\frac{1-2 \sin ^2(n \theta)+i \sin (2 n \theta)-1}{2 \cos ^2(n \theta)-1+i \sin (2 n \theta)+1} \\ & =\frac{-2 \sin ^2(n \theta)+2 i \sin (n \theta) \cos (n \theta)}{2 \cos ^2(n \theta)+2 i \sin (n \theta) \cos (n \theta)} \\ & =\frac{[i \sin (n \theta)+\cos (n \theta)] i \sin (n \theta)}{[\cos (n \theta)+i \sin (n \theta)] \cos (n \theta)} \\ & =i \tan n \theta=i \tan \left(n\left(\theta+\tan { }^{-1} \frac{y}{x}\right)\right) \end{aligned} $

We have, $\frac{z+i}{z-1}=\frac{x+i y+i}{x+i y-1}$

$ \begin{aligned} & =\frac{x+i(1+y)}{(x-1)+i y} \times \frac{(x-1)-i y}{(x-1)-i y} \\ & =\frac{x(x-1)+i(1+y)(x-1)-i x y+y(1+y)}{(x-1)^2+y^2} \\ & =\frac{x^2-x+y^2+i(x+x y-y-1-x y)}{(x-1)^2+y^2} \\ & =\frac{x^2+y^2-x+y}{(x-1)^2+y^2}+i \frac{(x-y-1)}{(x-1)^2+y^2} \end{aligned} $

Since, $\frac{z+i}{z-1}$ is purely imaginary.

So, $\frac{x^2+y^2-x+y}{(x-1)^2+y^2}=0$

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

Where, $x \neq 1, y \neq 0$

The given set $ S = \{ z \in \mathbb{C} \mid |z + 1 - i| = 1 \} $ can be analyzed as follows:

First, consider the expression $ |z + 1 - i| = 1 $.

Break down $ z $ in terms of real and imaginary components:

$ z = x + iy, $

where $ x $ and $ y $ are real numbers. Thus, we can substitute $ z $ in the expression:

$ |x + iy + 1 - i| = 1 $

Simplifying inside the absolute value, we have:

$ |(x + 1) + i(y - 1)| = 1 $

The expression $|(x + 1) + i(y - 1)|$ represents the distance from the point $(-1, 1)$ in the complex plane. Therefore, the set of all such points for which this distance is $1$ forms a circle centered at $(-1, 1)$ with a radius of $1$. Mathematically, this can be simplified to:

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

Squaring both sides, we get:

$ (x + 1)^2 + (y - 1)^2 = 1 $

Thus, $ S $ represents a circle with center at $(-1, 1)$ and radius $1$ unit.

To solve for the least positive and greatest negative integer values of $ k $ in the expression $\left(\frac{1-i}{1+i}\right)^k = -i$, follow these steps:

Start with the given equation:

$ \left(\frac{1-i}{1+i}\right)^k = -i $

Simplify the fraction $\frac{1-i}{1+i}$:

$ \frac{1-i}{1+i} \cdot \frac{1-i}{1-i} = \frac{(1-i)^2}{(1+i)(1-i)} $

Calculate the numerator and the denominator:

$ (1-i)^2 = 1 - 2i + i^2 = 1 - 2i - 1 = -2i $

$ (1+i)(1-i) = 1 + i - i - i^2 = 1 + 1 = 2 $

Substitute back into the fraction:

$ \frac{(1-i)^2}{(1+i)(1-i)} = \frac{-2i}{2} = -i $

Therefore, the expression simplifies to:

$ \left(-i\right)^k = -i $

For $(-i)^k = -i$, it implies:

$ k \equiv 1 \pmod{4} $

From this, determine the least positive integer $ m $ and greatest negative integer $ n $:

$ m = 1 $ (since $ k = 1 $ is the smallest positive integer solution)

$ n = -3 $ (since $ k = -3 $ satisfies the condition $-i$ in the sequence and is the largest negative solution)

Calculate $ m - n $:

$ m - n = 1 - (-3) = 4 $

Thus, $ m - n = 4 $.

$\ \frac{z-2 i}{z-2}$ is purely imaginary

Let $z=x+i y$

$ \therefore \frac{x+i y-2 i}{x+i y-2}=\frac{x+i(y-2)}{(x-2)+i y} \times \frac{(x-2)-i y}{(x-2)-i y} $

Real part of $\frac{z-2 i}{z-2}=0$

$ x(x-2)+y(y-2)=0 $

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

$\Rightarrow(x-1)^2+(y-1)^2=2$, which is equation of circle with radius $\sqrt{2}$ units.

Required area $=\frac{1}{2} \times 2 \times 2+\frac{\pi}{2}(\sqrt{2})^2=(2+\pi)$ sq units

$ \begin{aligned} & \text {We have, } \frac{(\cos a+i \sin a)^6}{(\sin b+i \cos b)^8} \\ & =\frac{(\cos a+i \sin a)^6}{\left(-i^2 \sin b+i \cos b\right)^8} \quad\left[\because 1=-i^2\right] \\ & =\frac{(\cos a+i \sin a)^6}{i^8(\cos b-i \sin b)^8} \\ & =\frac{(\cos a+i \sin a)^6}{(\cos b-i \sin b)^8} \\ & =\frac{\cos 6 a+i \sin 6 a}{\cos 8 b-i \sin 8 b} \\ & =\frac{(\cos 6 a+i \sin 6 a)(\cos 8 b+i \sin 8 b)}{(\cos 8 b-i \sin 8 b)(\cos 8 b+i \sin 8 b)} \\ & \therefore \text { Required real part } \\ & =\frac{\cos 6 a \cos 8 b-\sin ^8 6 a \sin 8 b}{\cos ^2 8 b+\sin ^2 8 b} \\ & =\cos (6 a+8 b) \end{aligned} $

To find $ a + ib = \frac{\sqrt{-5-12i} + \sqrt{5+12i}}{\sqrt{-8-6i}} $ and calculate $ 2a + b $:

Calculate $\sqrt{-5-12i}$ and $\sqrt{5+12i}$:

Since the real parts of $\sqrt{-5-12i}$ and $\sqrt{5+12i}$ are positive:

$ \sqrt{-5-12i} = \sqrt{\frac{13 - 5}{2}} + i \sqrt{\frac{13 + 5}{2}} = 2 - 3i $

$ \sqrt{5+12i} = \sqrt{\frac{13 + 5}{2}} + i \sqrt{\frac{13 - 5}{2}} = 3 + 2i $

Calculate $\sqrt{-8-6i}$:

Given that the real part is negative:

$ \sqrt{-8-6i} = -\sqrt{\frac{10 - 8}{2}} + i \sqrt{\frac{10 + 8}{2}} = -1 + 3i $

Find $a + ib$:

Using the calculations from the previous steps:

$ a+ib = \frac{(2 - 3i) + (3 + 2i)}{-1 + 3i} = \frac{5 - i}{-1 + 3i} $

Multiply the numerator and the denominator by the conjugate of the denominator:

$ = \frac{(5 - i)(-1 - 3i)}{(-1 + 3i)(-1 - 3i)} $

Simplify:

$ = \frac{-5 + i - 15i - 3}{1 + 9} = \frac{-8 - 14i}{10} $

$ = -\frac{4}{5} - \frac{7i}{5} $

Thus, $ a = -\frac{4}{5} $ and $ b = -\frac{7}{5} $.

Calculate $2a + b$:

$ 2a + b = 2 \left(-\frac{4}{5}\right) + \left(-\frac{7}{5}\right) = -\frac{8}{5} - \frac{7}{5} = -3 $

Hence, the result is $-3$.

The given equation $ z\overline{z} + (4 - 3i)z + (4 + 3i)\overline{z} + c = 0 $ is analyzed to determine the real values of $ c $ that form a circle.

Key Steps

Identifying the Circle's Center:

$ z\overline{z} $ is the modulus squared of $ z $, and the terms $ (4 - 3i)z $ and $ (4 + 3i)\overline{z} $ suggest the transformation involves the complex number $ 4 - 3i $.

Thus, the center of the circle is at $ 4 - 3i $.

Finding the Radius:

To find the radius, use the formula that incorporates the modulus of the center:

$ \sqrt{(4)^2 + (-3)^2 - c} = \sqrt{25 - c} $

The radius needs to be non-negative for the equation to represent a circle:

$ \sqrt{25 - c} \geq 0 $

which simplifies to:

$ 25 - c \geq 0 \quad \Rightarrow \quad c \leq 25 $

Conclusion

The real values of $ c $ for which the equation represents a circle are in the interval:

$ (-\infty, 25] $

Given the equation $ z^3 + \overline{z} = 0 $, where $ z = x + iy $ is a complex number, we can find the distinct solutions as follows:

Rewrite the Equation:

Start with:

$ z^3 + \overline{z} = 0 $

This implies:

$ z^3 = -\overline{z} $

Equate Magnitudes:

Taking the magnitude of both sides, we have:

$ |z^3| = |-\overline{z}| $

This simplifies to:

$ |z|^3 = |\overline{z}| $

Since $ |\overline{z}| = |z| $, the equation becomes:

$ |z|^3 = |z| $

Solve for Magnitude:

This equation leads to:

$ |z|(|z|^2 - 1) = 0 $

Therefore, either $ |z| = 0 $ or $ |z|^2 = 1 $.

Consider Each Case:

If $ |z| = 0 $, then $ z = 0 $. This gives one distinct solution: $ z = 0 $.

If $ |z|^2 = 1 $, then $ |z| = 1 $ and consequently $ z\overline{z} = 1 $. This implies:

$ z = \frac{1}{z} $

Substitute Back:

Substituting $\overline{z} = \frac{1}{z}$ into the original equation:

$ z^3 + \frac{1}{z} = 0 $

Multiply through by $ z $ to clear the fraction:

$ z^4 + 1 = 0 $

Solve $ z^4 + 1 = 0 $:

This equation has four roots:

$ z = e^{i(\pi/4)}, e^{i(3\pi/4)}, e^{i(5\pi/4)}, e^{i(7\pi/4)} $

Count Distinct Solutions:

Including $ z = 0 $, we have:

One solution for $ z = 0 $

Four solutions from $ z^4 + 1 = 0 $

Therefore, the number of distinct solutions to the equation is $ \boxed{5} $.

$\begin{aligned} & i^{18}-3 i^7+i^2\left(1+i^4\right)(i)^{22} \\ & =i^{4 \times 4+2}-3 i^{4+3}+-1(1+1)\left(i^2\right)^{11} \\ & =i^2-3 i^3+2 \quad\left[\because i^2=-1 \text { and } i^4=1\right] \\ & =-1+3 i+2=1+3 i \quad \end{aligned}$

Conjugate of $\cos x-i \sin 2 x$ is $\cos x+i \sin 2 x$

So, $\sin x+i \cos 2 x=\cos x+i \sin 2 x$

Comparing real and imaginary parts,

$\sin x=\cos x \text { or } \tan x=1$

and $\cos 2 x=\sin 2 x$ or $\tan 2 x=1$

But for same value of $x$, both $\tan x$ and $\tan 2 x$ can not. Hence, no solution is possible.

$\begin{aligned} \text { Let } z & =x+i y \\ |z| & =\sqrt{x^2+y^2} \end{aligned}$

Now, $|z|^2=\operatorname{Re}(z)$

$\begin{aligned} x^2+y^2 & =x \\ x^2+y^2-x & =0 \\ g=1 / 2, f & =0 \end{aligned}$

So, centre of circle $\left(\frac{1}{2}, 0\right)$.

Multiplicative inverse of complex number $(\sin \theta, \cos \theta)$ is

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

Thus, multiplicative inverse of $(\sin \theta, \cos \theta)$ is $(\sin \theta,-\cos \theta)$.

$\begin{aligned} & \sum_\limits{k=0}^{440} i^k=x+i y \\ & \Rightarrow \quad x+i y=1+i+i^2+i^3+i^4+\ldots .+i^{440}=-1 \\ & \therefore \quad x=-1, y=0 \\ & \text { Then, } x^{100}+x^{99} \cdot y+x^{242} \cdot y^2+x^{97} \cdot y^3 \\ & \Rightarrow \quad(-1)^{100}+0=1 \end{aligned}$

$ \begin{aligned} & e^{i \theta}=\operatorname{cis} \theta=\cos \theta+i \sin \theta \\ & \sum_{n=0}^{\infty} \frac{\cos (n \theta)}{2^n} \\ & =\operatorname{Re}\left[\sum_{n=0}^{\infty} \frac{\cos n \theta}{2^n}\right] \\ & =\operatorname{Re}\left[\sum_{n=0}^{\infty}\left(\frac{\cos \theta}{2}\right)^n\right] \text { [Using De-moivre theorem] } \\ & =\operatorname{Re}\left[\sum_{n=0}^{\infty}\left(\frac{e^{i \theta}}{2}\right)^n\right] \\ & =\operatorname{Re}\left[1+\frac{e^{i \theta}}{2}+\left(\frac{e^{i \theta}}{2}\right)^2+\ldots . \infty\right]\\ &1+\frac{e^{i \theta}}{2}+\left(\frac{e^{i \theta}}{2}\right)^2+\ldots . . \text { is an G P } \\ & =\operatorname{Re}\left[\frac{1}{1-\frac{e^{i \theta}}{2}}\right]=\operatorname{Re}\left[\frac{2}{2-e^{i \theta}}\right] \end{aligned}$

$\begin{aligned} & =\operatorname{Re}\left[\frac{2}{2-\cos \theta-i \sin \theta}\right] \\ & =\operatorname{Re}\left[\frac{2(2-\cos \theta-i \sin \theta)}{(2-\cos \theta)^2-i^2 \sin ^2 \theta}\right] \\ & =\operatorname{Re}\left[\frac{4-2 \cos \theta}{4+\cos ^2 \theta-4 \cos \theta+\sin ^2 \theta}\right. \\ & \left.+\frac{i \sin \theta}{4+\cos ^2 \theta-4 \cos \theta+\sin ^2 \theta}\right] \\ & =\operatorname{Re}\left[\frac{4-2 \cos \theta}{5-4 \cos \theta}+\frac{i \sin \theta}{5-4 \cos \theta}\right] \\ & =\frac{4-2 \cos \theta}{5-4 \cos \theta} \end{aligned}$

$\begin{array}{l} & \text { } i z^3+z^2-z+i=0 \\ \Rightarrow & i z^3-z+z^2+i=0 \\ \Rightarrow & i z^3+i^2 z+z^2+i=0 \\ \Rightarrow & i z\left(z^2+i\right)+\left(z^2+i\right)=0 \\ \Rightarrow & \left(z^2+i\right)(i z+1)=0 \\ \Rightarrow & \left(z^2+i\right)\left(i z-i^2\right)=0 \\ \Rightarrow & \left(z^2+i\right) i(z-i)=0 \\ \Rightarrow & z^2=-i \text { or } z=i \\ & z^2=1\left(\cos \frac{3 \pi}{2}+i \sin \frac{3 \pi}{2}\right) \\ \text { or } & z=1\left(\cos \frac{\pi}{2}+i \sin \frac{\pi}{2}\right) \end{array}$

We can see that all the roots have modulus as 1.

So, $|z|=1$

$\begin{array}{ll} \text {} & \frac{x-1}{3+i}+\frac{y-1}{3-i}=i \\ \Rightarrow & \frac{(x-1)(3-i)+(y-1)(3+i)}{(3+i)(3-i)}=i \\ \Rightarrow & \frac{3 x-i x-3+i+3 y+i y-3-i}{9-i^2}=i \\ \Rightarrow & 3 x+3 y-i x+i y-6=i[9-(-1)] \end{array}$

$\begin{array}{ll} \Rightarrow \quad 3 x+3 y-i x+i y-6=10 i \\ \Rightarrow \quad 3(x+y)+i(-x+y)=6+10 i \end{array}$

On comparing both the sides, we get

$3(x+y)=6 \text { and }-x+y=10$

$\begin{array}{lcrl} \Rightarrow & x+y & =2 \quad \text{... (i)}\\ \text { and } & -x+y & =10 \quad \text{.... (ii)} \end{array}$

On solving Eqs. (i) and (ii), we get

$2 y=12 \Rightarrow y=6 \text { and } x=-4$

$|1-i|^x=2^x$

$\begin{array}{l} \Rightarrow & \left(\sqrt{\Lambda)^2+(-1)^2}\right)^x=2^x \\ \Rightarrow & (\sqrt{2})^x =2^x \\ \Rightarrow & 2^{\frac{x}{2}} =2^x \\ \Rightarrow & \frac{x}{2} =x \\ \Rightarrow & 0 =\frac{x}{2} \\ \Rightarrow & x =0 \end{array}$

Number of integer solution is 1.

$\begin{array}{left} \left|\begin{array}{aligned} a & b & c \\ b & c & a \\ c & a & b \end{array}\right|=a\left(b c-a^2\right)-b\left(b^2-a c\right)+c\left(a b-c^2\right) \\ \Rightarrow a b c-a^3-b^3+a b c+a b c-c^3=0 \\ \Rightarrow 3 a b c-a^3-b^3-c^3=0 \\ \Rightarrow a^3+b^3+c^3-3 a b c=0 \\ \Rightarrow (a+b+c)=0 \end{array}$

Since, $a=\left|Z_1\right|, b=\left|Z_2\right|, c=\left|Z_3\right|$

$\therefore \quad\left|Z_1\right|+\left|Z_2\right|+\left|Z_3\right|=0$

$\left|z_1+z_2\right|^2=\left|z_1\right|^2+\left|z_2\right|^2$ $\quad$ (given) ..... (i)

Let $z_1=x+i y, z_2=a+i b$.

Then

$\begin{aligned} & \frac{z_1}{z_2}=\frac{x+i y}{a+i b}=\frac{x+i y}{a+i b} \times \frac{a-i b}{a-i b} \\ & \frac{z_1}{z_2}=\frac{(a x+y b)+i(a y-x b)}{a^2+b^2} \quad \text{...... (ii)} \end{aligned}$

From Eq. (i), we get

$\begin{aligned} & \text { LHS }=\left|z_1+z_2\right|^2=|x+a+i(y+b)|^2 \\ & =(x+a)^2+(y+b)^2 \\ & =x^2+a^2+2 a x+y^2+b^2+2 b y \\ & \text { RHS }=|x+i y|^2+|a+i b|^2 \\ & =x^2+y^2+a^2+b^2 \\ & \because \text { LHS }=\text { RHS } \\ & \Rightarrow x^2+y^2+a^2+b^2+2(a x+b y) \\ & =x^2+y^2+a^2+b^2 \\ & \Rightarrow \quad 2(a x+b y)=0 \Rightarrow a x+b y=0 \\ \end{aligned}$

Put $a x+b y=0$ in Eq. (ii),

$\frac{z_1}{z_2}=\frac{i(a y-x b)}{a^2+b^2}$, which is purely imaginary.

$\begin{aligned} & \text { LHS }=\frac{3-4 i x}{3+4 i x} \times \frac{3-4 i x}{3-4 i x} \\ & =\frac{9+16 i^2 x^2-24 i x}{9-\left(16 i^2 x^2\right)}=\frac{9-16 x^2-24 i x}{9+16 x^2} \\ & =\frac{9-16 x^2}{9+16 x^2}-\frac{24 i x}{9+16 x^2}=\alpha-i \beta=\text { RHS } \\ & \therefore \quad \alpha=\frac{9-16 x^2}{9+16 x^2} \\ & \beta=\frac{24 x}{9+16 x^2} \\ & \Rightarrow \quad a^2+\beta^2=\frac{\left(9-16 x^2\right)^2+(24 x)^2}{\left(9+16 x^2\right)^2} \\ & =\frac{\left(9+16 x^2\right)^2}{\left(9+16 x^2\right)^2}=1 \\ & \therefore \quad \alpha^2+\beta^2=1 \\ & \end{aligned}$

$\begin{aligned}(1-i \sqrt{3})^9 & =2^9\left(\frac{1}{2}-i \frac{\sqrt{3}}{2}\right)^9 \\ & =2^9\left(\cos \frac{\pi}{3}-i \sin \frac{\pi}{3}\right)^9=2^9\left(e^{-i \pi / 3}\right)^9 \\ & =2^9\left(e^{-i 3 \pi}\right)=2^9(\cos 3 \pi-i \sin 3 \pi) \\ & =2^9\left((-1)^3-0\right)=-2^9\end{aligned}$

${\left( {\sqrt 2 {{(\sqrt 3 - 1)} \over {2\,.\,2}} + i\sqrt 2 {{(\sqrt 3 + 1)} \over {2\,.\,2}}} \right)^{2020}}$

$ = {\left( {{{\left( {{{\sqrt 3 } \over 2} - {1 \over 2}} \right)} \over {\sqrt 2 }} + i{{\left( {{{\sqrt 3 } \over 2} + {1 \over 2}} \right)} \over {\sqrt 2 }}} \right)^{2020}}$

$ = {\left( {{{{1 \over {\sqrt 2 }}\left( {{{\sqrt 3 } \over 2} - {1 \over 2}} \right)} \over 1} + {{i{1 \over {\sqrt 2 }}\left( {{{\sqrt 3 } \over 2} + {1 \over 2}} \right)} \over 1}} \right)^{2020}}$

$ = {\left( {\left( {{1 \over {\sqrt 2 }}\,.\,{{\sqrt 3 } \over 2} - {1 \over {\sqrt 2 }}\,.\,{1 \over 2}} \right) + i\left( {{1 \over {\sqrt 2 }}\,.\,{{\sqrt 3 } \over 2} + \left( {{1 \over {\sqrt 2 }}\,.\,{1 \over 2}} \right)} \right)} \right)^{2020}}$

$ = (\cos 45^\circ \cos 30^\circ - \sin 45^\circ \sin 30^\circ ) + i(\sin 45^\circ \cos 30^\circ + \cos 45^\circ \sin 30^\circ ){)^{2020}}$

$ = {(\cos 75^\circ + i\sin 75^\circ )^{2020}}$

$ = {\left( {\cos {{5\pi } \over {12}} + i\sin {{5\pi } \over {12}}} \right)^{2020}}$

$ = \left( {\cos {{5\pi } \over {12}} \times 2020 + i\sin {{5\pi } \over {12}} \times 2020} \right)$ [De-moivre]

$ = \left( {\cos {{2525\pi } \over 3} + i\sin {{2525\pi } \over 3}} \right)$

$ = (\cos (2525 \times 60)^\circ + i\sin (2525 \times 60)^\circ $

$ = (\cos (n\pi - 60)^\circ + i\sin (n\pi - 60)^\circ )$, where n = 842

$ = ({( - 1)^n}\cos 60^\circ + i( - \sin 60^\circ ))$

$ = (\cos 60^\circ - i\sin 60^\circ )$ [$\because$ n = even]

$ = {1 \over 2} - i{{\sqrt 3 } \over 2}$

$\begin{aligned} & \text { } z_1=2+3 i, z_2=3+2 i \\ & \Rightarrow\left[\begin{array}{cc}z_1 & z_2 \\ -\bar{z}_2 & \bar{z}_1\end{array}\right]\left[\begin{array}{cc}\bar{z}_1 & -z_2 \\ \bar{z}_2 & z_1\end{array}\right] \\ & =\left[\begin{array}{cc}\left|z_1\right|^2+\left|z_2\right|^2 & 0 \\ 0 & \left|z_1\right|^2+\left|z_2\right|^2\end{array}\right] \\ & \Rightarrow \quad\left|z_1\right|^2=4+9=13 \\ & \Rightarrow \quad\left|z_2\right|^2=9+4=13=\left[\begin{array}{cc}26 & 0 \\ 0 & 26\end{array}\right]=26 I \\ & \end{aligned}$

$\begin{array}{rlrl} & \text { } & (1+i)(1+3 i)(1+7 i) & =x+i y \\ \Rightarrow & & (-2+4 i)(1+7 i) & =x+i y \\ \Rightarrow & & -30-10 i & =x+i y \\ \Rightarrow & & |x+i y| & =\sqrt{30^2+10^2} \\ & & =\sqrt{1000}=10 \sqrt{10}\end{array}$

$\because 1, \alpha_1, \alpha_2, \alpha_3, \alpha_4$ are the roots of $z^5-1=0$

$\begin{gathered} \Rightarrow \quad(z-1)\left(z-\alpha_1\right)\left(z-\alpha_2\right)\left(z-\alpha_3\right)\left(x-\alpha_4\right)=z^5-1 \\ \Rightarrow \quad(\omega-1)\left(\omega-\alpha_1\right)\left(\omega-\alpha_2\right)\left(\omega-d_3\right)\left(\omega-\alpha_4\right)=\omega^5-1 \\ (\omega-1)\left(\omega-\alpha_1\right)\left(\omega-\alpha_2\right)\left(\omega-\alpha_3\right)\left(\omega-\alpha_4\right)+\omega \\ =\left(\omega^5-1\right)+\omega=\omega^2-1+\omega \\ =-1-1=-2 \end{gathered}$

$\begin{aligned} & \text { } a > 0, z=x+i y \\ & \log _{\cos ^2{ }_\theta}|z-a| > \log _{\cos ^2{ }_\theta}|z-a i| \\ & 0 < \cos ^2 \theta < 1 \\ & \text { So, }|z-a| < |z-a i| \\ & \Rightarrow \quad(x-a)^2+y^2 < x^2+(y-a)^2 \\ & \Rightarrow x^2+a^2-2 a x+y^2 < x^2+y^2+a^2-2 a y \\ & \Rightarrow-2 a x < -2 a y \\ & \Rightarrow \quad x > y\end{aligned}$

$i x^2-2(i+1) x+(2-i)=0$ ..... (i)

Let the other roots be $\alpha$.

$\begin{aligned} & {[x-(2-i)][x-\alpha]=0} \\ & \Rightarrow x^2-(2-i+\alpha) x-(\alpha(2-i))=0 \\ & \text { \{comparing with Eq. (i) } \\ & \Rightarrow i x^2-(2 i+1+\alpha i) x-[\alpha(2 i+1)]=0 \\ & \Rightarrow \quad 2 i+1+\alpha i=2 i+2 \\ & \Rightarrow \\ & \alpha i=1 \\ & \Rightarrow \quad \alpha=\frac{1}{i}=-i \\ & \end{aligned}$

$\because|z-2|=|z-1|$

Let $z=x+i y$, then

$\begin{array}{ll} \Rightarrow & |x+i y-2|^2=|x+i y-1|^2 \\ \Rightarrow & (x-2)^2+y^2=(x-1)^2+y^2 \\ \Rightarrow & x^2-4 x+4=x^2-2 x+1 \\ \Rightarrow & -2 x+4 x=4-1 \Rightarrow x=3 / 2 \end{array}$

Which is parallel to $Y$-axis.

${\left( {{{1 + i} \over {1 - i}}} \right)^m} = 1 \Rightarrow {\left[ {{{(1 + i)(1 + i)} \over {(1 - i)(1 + i)}}} \right]^m} = 1$

$ \Rightarrow \left( {{{1 - 1 + 2i} \over {1 + 1}}} \right) = 1$

$\Rightarrow i^m=1$

$\Rightarrow m$ is multiple of $4\left[\because i^{4 n}=1, \forall n \in N\right]$

Option (a) is not a multiple of 4 .

$\therefore \quad m \neq 1934$

$(\sin \theta-i \cos \theta)^3=(-i)^3(\cos \theta+i \sin \theta)^3$

$=(-i)^3(\cos 3 \theta+i \sin 3 \theta)$ [by De-Moivre theorem]

Real part of $(\cos 4+i \sin 4+1)^{2020}=$ Real part of $\left(2 \cos ^2 2+i 2 \sin 2 \cos 2\right)^{2020}$

$\begin{aligned} & =\text { Real part of } 2^{2020} \cos ^{2020}(2)(\cos 2+i \sin 2)^{2020} \\ & =\text { Real part of } 2^{2020} \cos ^{2020}(2)(\cos 4040+i \sin 4040) \\ & =2^{2020} \cos ^{2020} 2 \cdot \cos 4040 \quad[z+\bar{z}=2 \operatorname{Re}(z)] \end{aligned}$