Complex Numbers

If $\omega$ is the cube root of unity.

$ \frac{a+b \omega+c \omega^2}{c+a \omega+b \omega^2}+\frac{a+b \omega+c \omega^2}{b+c \omega+a \omega^2}=? $

Now, we write as

$ \begin{aligned} & A=a+b \omega+c \omega^2 \\ & B=b+c \omega+a \omega^2 \\ & C=c+a \omega+b \omega^2 \end{aligned} $

Find common denominator for the fractions.

$ \frac{A}{C}+\frac{A}{B}=\frac{A \cdot B+A \cdot C}{B \cdot C}=\frac{A(B+C)}{B C} $

Now, we know that according to given condition.

$ \begin{gathered} \omega^3=1 \\ 1+\omega+\omega^2=0 \\ B+C=\left(b+c \omega+a \omega^2\right)+\left(c+a \omega+b \omega^2\right) \\ b+c \omega+a \omega^2+c+a \omega+b \omega^2 \\ =\left(b+\omega^2 b\right)+(c \omega+c)+\left(a \omega^2+a \omega\right) \\ =b\left(1+\omega^2\right)+c(1+\omega)+a\left(\omega^2+\omega\right) \\ \quad\left(\because \text { using } 1+w+w^2=0\right) \\ 1+\omega^2=-\omega, 1+\omega=-\omega^2 \end{gathered} $

$ \begin{aligned} B+C & =b(-\omega)+C\left(-\omega^2\right)+a(-1) \\ & =-a-b \omega-c \omega^2 \end{aligned} $

Thus, $B+C=-A$

Because, $A=a+b \omega+c \omega^2$

Substituting this into the expression

$ \frac{A(B+C)}{B C}=\frac{A(-A)}{B C}=\frac{-A^2}{B C} $

The denominator $(B C)$ does not change the overall simplification that each term involves $\bar{A}, B$ and $C$ with symmetrical properties.

Thus, $\frac{A}{C}+\frac{A}{B}=-1$

$ \begin{aligned} &\text { Given, }(3+i) \text { is a root of }\\ &\begin{aligned} & x^2+a x+b=0 \\ &(3+i)^2+a(3+i)+b=0 \\ & 9-1+6 i+a(3+i)+b=0 \\ & 8+6 i+3 a+a i+b=0 \\ &(8+3 a+b)+i(6+a)=0 \\ & 3 a+b=-8 \\ & a+6=0 \Rightarrow a=-6 \end{aligned} \end{aligned} $

Given, $z_1=10+6 i, z_2=4+6 i$

and $\arg \left(\frac{z-z_1}{z-z_2}\right)=\frac{\pi}{4}$

Let $z=x+i y$

Now, $\frac{z-z_1}{z-z_2}=\frac{(x+i y)-(10+6 i)}{(x+i y)-(4+6 i)}$

$ \begin{aligned} & =\frac{(x-10)+i(y-6)}{(x-4)+i(y-6)} \\ & =\frac{(x-10)+i(y-6)}{(x-4)+i(y-6)} \times \frac{(x-4)-i(y-6)}{(x-4)-i(y-6)} \\ & =\frac{(x-10)(x-4)+(y-6)^2+i[-(x-10)(y-6)+(y-6)(x-4)]}{(x-4)^2+(y-6)^2} \end{aligned} $

Since, $\arg \left(\frac{z-z_1}{z-z_2}\right)=\frac{\pi}{4}$

i.e. $\tan \theta=1$

$ \therefore \frac{(y-6)(x-4)-(x-10)(y-6)}{(x-10)(x-4)+(y-6)^2}=1 $

$ \Rightarrow \frac{x y-6 x-4 y+24-x y+6 x+10 y-60}{x^2-14 x+40+y^2-12 y+36}=1 $

$ \Rightarrow 6 y-36=x^2+y^2-12 y-14 x+76 $

$ \Rightarrow x^2+y^2-18 y-14 x+112=0 $

$ \Rightarrow \quad(x-7)^2+(y-9)^2=18 $

$ \Rightarrow|z-(7+9 i)|=3 \sqrt{2} $

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

(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} $.

To solve the problem, let's break it down step by step.

Step 1 : Find $w_{1}$

Given $z_{1} = 5 + 4i$.

When you rotate $z_{1}$ by $90^{\circ}$ anticlockwise about the origin, the real part becomes negative of the imaginary part of $z_{1}$, and the imaginary part becomes the real part of $z_{1}$.

Therefore, $w_{1}$ becomes :

$w_{1} = -4 + 5i$

Step 2 : Find $w_{2}$

Given $z_{2} = 3 + 5i$.

When you rotate $z_{2}$ by $90^{\circ}$ clockwise about the origin, the real part becomes the imaginary part of $z_{2}$, and the imaginary part becomes negative of the real part of $z_{2}$.

Therefore, $w_{2}$ becomes :

$w_{2} = 5 - 3i$

Step 3 : Calculate $w_{1} - w_{2}$

$w_{1} - w_{2} = (-4 + 5i) - (5 - 3i)$

$w_{1} - w_{2} = -9 + 8i$

Step 4 : Find the principal argument

To find the principal argument, we need to compute the tangent inverse of the ratio of the imaginary part to the real part.

Argument = $\tan^{-1}\left(\frac{\text{Imaginary part}}{\text{Real part}}\right)$

Argument = $\tan^{-1}\left(\frac{8}{-9}\right)$

Argument = $-\tan^{-1}\left(\frac{8}{9}\right)$

Since it's in the third quadrant, the principal argument is :

Argument = $\pi + (-\tan^{-1}\left(\frac{8}{9}\right))$

Argument = $\pi - \tan^{-1}\left(\frac{8}{9}\right)$

So, the correct option is :

Option C : $\pi-\tan^{-1} \frac{8}{9}$

Let, ${z_1} = {x_1} + i{y_1}$

and ${z_2} = {x_2} + i{y_2}$

$\therefore$ ${z_1}{z_2} = {x_1}{x_2} - {y_1}{y_2} + i({x_1}{y_2} + {x_2}{y_1})$

Given, ${\mathop{\rm Re}\nolimits} ({z_1} + {z_2}) = 0$

$ \Rightarrow {x_1} + {x_2} = 0$ ...... (1)

also given, ${\mathop{\rm Re}\nolimits} ({z_1}{z_2}) = 0$

$ \Rightarrow {x_1}{x_2} - {y_1}{y_2} = 0$

$ \Rightarrow {x_1}{x_2} = {y_1}{y_2}$

$ \Rightarrow {y_1}{y_2} = - x_1^2$ [$\because$ ${x_2} = - {x_1}$]

So, multiplication of imaginary part's of z₁ and z₂ is negative. It means sign of y₁ and y₂ are opposite of each other.

$ \therefore $ B and C are correct.

${\left( {1 - \sqrt 3 i} \right)^{200}}$

$ = {\left[ {2\left( {{1 \over 2} - {{\sqrt 3 } \over 2}i} \right)} \right]^{200}}$

$ = {2^{200}}{\left( {\cos {\pi \over 3} - i\sin {\pi \over 3}} \right)^{200}}$

$ = {2^{200}}\left( {\cos {{200\pi } \over 3} - i\sin {{200\pi } \over 3}} \right)$

$ = {2^{200}}\left( {\cos \left( {66\pi + {{2\pi } \over 3}} \right) - i\sin \left( {66\theta + {{2\pi } \over 3}} \right)} \right)$

$ = {2^{200}}\left( {\cos {{2\pi } \over 3} - i\sin {{2\pi } \over 3}} \right)$

$ = {2^{200}}\left( { - {1 \over 2} - {{\sqrt 3 } \over 2}i} \right)$

$ = {2^{199}}\left( { - 1 - \sqrt 3 i} \right)$

$ = {2^{199}}\left( {p + iq} \right)$

$\therefore$ p = $-$1 and q = $- \sqrt3$

Now, $p - q + {q^2} = - 1 + \sqrt 3 + 3 = 2 + \sqrt 3 = \alpha $

and $p + q + {q^2} = - 1 - \sqrt 3 + 3 = 2 - \sqrt 3 = \beta $

$\therefore$ $\alpha + \beta = 4$

$\alpha \beta = \left( {2 + \sqrt 3 } \right)\left( {2 - \sqrt 3 } \right) = 4 - 3 = 1$

$\therefore$ Quadratic equation is

${x^2} - (\alpha + \beta )x + \alpha \beta = 0$

$ \Rightarrow {x^2} - 4x + 1 = 0$

Let $\alpha=\omega$ and $\beta=\omega^2$

We know that

$ (a+b)\left(a \omega+b \omega^2\right)\left(a \omega^2+b \omega\right)=a^3+b^3 $

Now, check

$ \begin{aligned} & x+y+z=a(1+\alpha+\beta)+b(1+\alpha+\beta) \\ & \Rightarrow \quad x+y+z=0 \quad\left[\because 1+\omega+\omega^2=0\right] \\ & \Rightarrow \quad x^3+y^3+z^3=3 x y z=3\left(a^3+b^3\right) \end{aligned} $

$ \begin{aligned} & \text { Given, } z=\frac{3+2 i \cos \theta}{1-2 i \sin \theta} \times \frac{1+2 i \sin \theta}{1+2 i \sin \theta} \\ & =\frac{3-4 \sin \theta \cos \theta+6 i \sin \theta+2 i \cos \theta}{1+4 \sin ^2 \theta} \\ & \because \text { Purely imaginary } \Rightarrow \operatorname{Re}(z)=0 \\ & \frac{3-4 \sin \theta \cos \theta}{1+4 \sin ^2 \theta}=0 \Rightarrow 2 \sin 2 \theta=3 \\ & \Rightarrow \sin 2 \theta=\frac{3}{2}, \text { which is not possible. } \end{aligned} $

$ \begin{aligned} &\text { Given, } z=x+i y\\ &\begin{aligned} & \because \quad z \bar{z}^3+\bar{z} z^3=350 \\ & \Rightarrow \quad z \bar{z}\left(z^2+\bar{z}^2\right)=350 \\ & \Rightarrow\left(x^2+y^2\right)\left(x^2-y^2+2 i x y+x^2-y^2\right. \\ & -2 i x y)=350 \\ & \Rightarrow\left(x^2+y^2\right)\left(x^2-y^2\right)=175=25 \times 7 \\ & \Rightarrow\left(x^2+y^2\right)\left(x^2-y^2\right)=(16+9)(16-9) \\ & x^2=16, y^2=9 \\ & \text { Now, }|z|=\sqrt{x^2+y^2} \\ & =\sqrt{16+9}=5 \end{aligned} \end{aligned} $

Given, $\alpha$ and $\beta$ are the roots of equation $x^2+x+1=0$

$ \because x=\frac{-1 \pm i \sqrt{3}}{2}=\omega, \omega^2 $

$ \begin{aligned} &\text { Now, }\\ &\begin{aligned} & (\alpha+\beta)^2+\left(\alpha^2+\beta^2\right)^2+\left(\alpha^3+\beta^3\right)^2 \\ & +\ldots .+\left(\alpha^{12}+\beta^{12}\right)^2 \\ & =\left(\omega+\omega^2\right)^2+\left(\omega^2+\omega^4\right)^2+\left(\omega^3+\omega^6\right)^2 \\ & \ldots\left(\omega^{12}+\omega^{24}\right)^2 \\ & =\left(\omega+\omega^2\right)^2+\left(\omega^2+\omega\right)^2 \\ & +(1+1)^2+\left(\omega+\omega^2\right) \ldots(1+1)^2 \\ & =8(-1)^2+4(1+1)^2 \\ & {\left[\because \omega+\omega^2=-1 \text { and } \omega^3=1\right]} \\ & =24 \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { }\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)^n=1 \\ & \Rightarrow\left(\frac{1+\cos \frac{5 \pi}{18}+i \sin \frac{5 \pi}{18}}{1+\cos \frac{5 \pi}{18}-i \sin \frac{5 \pi}{18}}\right)^n=1 \\ & {\left[\because \frac{2 \pi}{9}=\frac{\pi}{2}-\frac{5 \pi}{18} \text { and } \sin \left(\frac{\pi}{2}-\theta\right)=\cos \theta\right.} \\ & \left.\cos \left(\frac{\pi}{2}-\theta\right)=\sin \theta\right] \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \Rightarrow\left(\frac{2 \cos ^2 \frac{5 \pi}{36}+i 2 \sin \frac{5 \pi}{36} \cos \frac{5 \pi}{36}}{2 \cos ^2 \frac{5 \pi}{36}-i 2 \sin \frac{5 \pi}{36} \cos \frac{5 \pi}{36}}\right)^n=1 \\ & {\left[\because 1+\cos \theta=2 \cos ^2 \frac{\theta}{2}\right.} \text { and sin } \left.\theta=2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}\right] \\ & \Rightarrow\left(\frac{\cos \frac{5 \pi}{36}+i \sin \frac{5 \pi}{36}}{\cos \frac{5 \pi}{36}-i \sin \frac{5 \pi}{36}}\right)^n=1 \\ & \{\text { multiplying with conjugate }\} \\ & \Rightarrow\left\{\frac{\left(\cos \frac{5 \pi}{36}+i \sin \frac{5 \pi}{36}\right)^2}{\cos ^2 \frac{5 \pi}{36}+\sin \frac{5 \pi}{36}}\right\}^n=1 \\ & \Rightarrow\left[\cos \frac{5 n \pi}{18}+i \sin \frac{5 n \pi}{18}\right]^n=1 \end{aligned}\\ &\text { So, least positive integral value of } n=18 \end{aligned} $

$ \begin{aligned} & P(x)=2 x^4+a x^3+b x^2+c x+d \\ & P(1)=1^2+3=4 \\ & P(2)=2^2+3=7 \\ & P(3)=3^2+3=12 \\ & P(4)=4^2+3=19 \end{aligned} $

$ \begin{aligned} & \text { Let } \\ & P(x)=2(x-1)(x-2)(x-3)(x-4)+x^2+3 \\ & \Rightarrow P(5)=2 \times 4 \times 3 \times 2 \times 1+5^2+3=76 \end{aligned} $

$ \begin{aligned} & \text { Given, } x^3+x^2+x+1=0 \\ & \Rightarrow \quad x^2(x+1)+1(x+1)=0 \\ & \Rightarrow \quad\left(x^2+1\right)(x+1)=0 \\ & x=-1, i,-i \end{aligned} $

$ \begin{aligned}(i) \frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma} & =\frac{1}{-1}+\frac{1}{i}+\frac{1}{-i} \\ & =-1-i+i=-1(\mathrm{a}) \end{aligned} $

$ \begin{array}{r}(ii) \alpha^3+\beta^3+\gamma^3=(-1)^3+(i)^3+(-i)^3 \\ =-1+(-i)+i=-1(\mathrm{a}) \end{array} $

$ \begin{gathered}(iii) \alpha^4+\beta^4+\gamma^4=(-1)^4+(i)^4+(-i)^4 \\ =1+1+1=3(\mathrm{~d}) \end{gathered} $

$ \text { } \begin{aligned} (iv) & (\alpha-\beta)^2+(\beta-\gamma)^2+(\gamma-\alpha)^2 \\ = & (-1-i)^2+(2 i)^2+(-i+1)^2 \\ = & 2 i-4-2 i=-4 \text { (b) } \end{aligned} $

$ \text { Given, } i=\sqrt{-1} $

$ \begin{aligned} \text { Consider, } z= & \frac{(1+i)^{2025}}{(1-i)^{2022}}=\frac{(1+i)^{2022}}{(1-i)^{2022}}(1+i)^3 \\ & =\frac{\left[(1+i)^2\right]^{1011}(1+i)^3}{\left[(1-i)^2\right]^{1011}} \\ & =\frac{[1-1+2 i]^{1011}}{[1-1-2 i]^{1011}}(1+i)^3 \\ & =\left(\frac{2 i}{-2 i}\right)^{1011}(1+i)^2(1+i) \\ & =(-1)^{1011}[1-1+2 i](1+i) \\ & =-1(2 i)(1+i)=-\left[2 i+2(i)^2\right] \\ & =-2 i-(-2)=2-2 i \end{aligned} $

We have,

$ \begin{aligned} \frac{|z-i|}{|z+i|} & =2, \text { where } z=x+ \\ |z-i| & =2|z+i| \\ |x+(y-1) i| & =2|x+(y+1) i| \end{aligned} $

Squaring on both sides, we get

$ \begin{gathered} |x+(y-1) i|^2=(2|x+(y+1) i|)^2 \\ x^2+(y-1)^2=4\left(x^2+(y+1)^2\right) \\ x^2+y^2+1-2 y=4 x^2+4 y^2+4+8 y \\ 3 x^2+3 y^2+8 y+2 y+4-1=0 \\ 3 x^2+3 y^2+10 y+3=0 \end{gathered} $

Hence, the locus of $z$ is

$ 3 x^2+3 y^2+10 y+3=0 $

$ \begin{aligned} &\text { Given, } x_n=\cos \frac{\pi}{2^n}+i \sin \frac{\pi}{2^n}\\ &\prod_{n=1}^{\infty} x_n=x_1 \cdot x_2 \cdot x_3 \ldots \end{aligned} $

$ \begin{aligned} & =\left(\cos \frac{\pi}{2}+i \sin \frac{\pi}{2}\right)\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right) \left(\cos \frac{\pi}{8}+i \sin \frac{\pi}{8}\right) \\ & =\cos \left(\frac{\pi}{2}+\frac{\pi}{4}+\frac{\pi}{8}+\ldots\right)+i \sin \left(\frac{\pi}{2}+\frac{\pi}{4}+\frac{\pi}{8}+\ldots\right) \end{aligned} $

$ \begin{aligned} & =\cos \left(\frac{\pi}{2}\left(1+\frac{1}{2}+\frac{1}{4}+\ldots\right)\right) +i \sin \left(\frac{\pi}{2}\left(1+\frac{1}{2}+\frac{1}{4}+\ldots\right)\right) \\ & =\cos \left(\frac{\pi}{2} \times \frac{1}{1-\frac{1}{2}}\right)+i \sin \left(\frac{\pi}{2} \times \frac{1}{1-\frac{1}{2}}\right) \\ & =\cos \pi+i \sin \pi \\ & =-1+i(0)=-1+0=-1 \end{aligned} $

Given, the roots of the equation $z^2-i=0$ are $\alpha$ and $\beta$.

Now, $z^2-i=0$

$ \begin{array}{ll} \Rightarrow & z^2=i \\ \Rightarrow & z= \pm \sqrt{i} \\ \Rightarrow & z= \pm \sqrt{\cos \frac{\pi}{2}+i \sin \frac{\pi}{2}} \\ \Rightarrow & z= \pm\left(\cos \frac{\pi}{2}+i \sin \frac{\pi}{2}\right)^{1 / 2} \\ \Rightarrow & z= \pm\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right) \end{array} $

$ \begin{aligned} &\text { Thus, } \alpha=\cos \frac{\pi}{4}+i \sin \frac{\pi}{4} \text { and }\\ &\begin{aligned} & \beta=-\cos \frac{\pi}{4}-i \sin \frac{\pi}{4} \\ & \alpha=\cos \frac{\pi}{4}+i \sin \frac{\pi}{4} \text { and } \\ & \beta=\cos \frac{5 \pi}{4}+i \sin \frac{5 \pi}{4} \\ & \therefore|(\arg \beta-\arg \alpha)|=\left|\frac{5 \pi}{4}-\frac{\pi}{4}\right|=\left|\frac{4 \pi}{4}\right|=\pi \end{aligned} \end{aligned} $