Complex Numbers

Given, arg $\left(\frac{z-3 i}{z+4}\right)=\frac{\pi}{2}$ and $z=x+i y$

$ \begin{aligned} & \text { Now, let } \omega=\frac{z-3 i}{z+4}=\frac{x+i y-3 i}{x+i y+4} \\ & =\frac{x+(y-3) i}{(x+4)+i y} \times \frac{(x+4)-i y}{(x+4)-i y} \\ & =\frac{x(x+4)-i x y+(y-3)(x+4) i+y(y-3)}{(x+4)^2+y^2} \\ & \Rightarrow \frac{x(x+4)+y(y-3)}{(x+4)^2+y^2}+i \frac{(y-3)(x+4)-xy}{(x+4)^2+y^2} \end{aligned} $

Now,

$ \begin{aligned} & \arg (\omega)=\tan ^{-1}\left(\frac{(y-3)(x+4)-xy)}{x(x+4)+y(y-3)}\right)= \frac{\pi}{2}\\ & \Rightarrow \frac{(y-3)(x+4)-x y}{x(x+4)+y(y-3)}=\tan \frac{\pi}{2} \\ & \Rightarrow \frac{4 y-3 x-12}{x^2+y^2+4 x-3 y}=\infty \end{aligned} $

So, clearly

$ x^2+y^2+4 x-3 y=0 $

and $3 x-4 y<0$

Given,

$\alpha_1, \alpha_2, \alpha_3, \alpha_4$ and $\alpha_5$ are roots of the equation

$ \begin{aligned} & x^5-5 x^4+9 x^3-9 x^2+5 x-1=0 \\ \Rightarrow & x^5-x^4-4 x^4+4 x^3+5 x^3-5 x^2 -4 x^2+4 x+x-1=0 \\ \Rightarrow & x^4(x-1)-4 x^3(x-1)+5 x^2(x-1) -4 x(x-1)+1(x-1)=0 \\ \Rightarrow & (x-1)\left[x^4-4 x^3+5 x^2-4 x+1\right]=0 \\ \Rightarrow & (x-1)\left(x^2-3 x+1\right)\left(x^2-x+1\right)=0 \\ \Rightarrow & x=1, \frac{3+\sqrt{5}}{2}, \frac{3-\sqrt{5}}{2}, \frac{1+\sqrt{3} i}{2} \end{aligned} $

$ \begin{aligned} & \text { and } \frac{1-\sqrt{3} i}{2} \\ & \alpha_1=1 \Rightarrow \alpha_1^2=1-\frac{1}{\alpha_1^2}=1 \\ & \alpha_2=\frac{3+\sqrt{5}}{2} \Rightarrow \alpha_2^2=\frac{(3+\sqrt{5})^2}{4} \\ & \Rightarrow \quad \frac{1}{\alpha_2^2}=\frac{4}{(3+\sqrt{5})^2} \\ & \Rightarrow \quad \begin{array}{l} \alpha_3=\frac{3-\sqrt{5}}{2} \Rightarrow \alpha_3^2=\frac{(3-\sqrt{5})^2}{4} \\ \Rightarrow \quad \frac{1}{\alpha_3^2}=\frac{4}{(3-\sqrt{5})^2} \end{array} \end{aligned} $

$ \begin{aligned} \alpha_4 & =\frac{1+\sqrt{3} i}{2} \\ & \Rightarrow \alpha_4^2=\frac{(1+\sqrt{3} i)^2}{4} \\ \Rightarrow & \frac{1}{\alpha_4^2}=\frac{4}{(1+\sqrt{3} i)^2} \\ \alpha_5 & =\frac{1-\sqrt{3} i}{2} \Rightarrow \alpha_5^2=\frac{(1-\sqrt{3} i)^2}{4} \\ & \Rightarrow \quad \frac{1}{\alpha_5^2}=\frac{4}{(1-\sqrt{3} i)^2} \end{aligned} $

$|Z| \leq 3 \Rightarrow$ Complex number Z lies within a circle centered at the origin with radius 3 .

$\frac{-\pi}{2} \leq \arg (z) \leq \frac{\pi}{2}$. This restricts z to the right half of the complex plane.

The intersection of these region is a semicircle with radius 3 . The area of semicircle is given by half the area of the full circle

The area of circle with radius $r$ is $=\pi r^2$

$ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\pi(3)^2=9 \pi $

Half of this area $=\frac{9 \pi}{2}$

So, area of region $=\frac{9 \pi}{2}$ sq units.

To determine the locus of the complex number $ Z $, where $ Z = x + i y $ (with $ x $ and $ y $ as real numbers), we analyze the given condition:

$ \arg \left(\frac{Z-1}{Z+1}\right) = \frac{\pi}{4} $

This equation implies that the complex number $ A = \frac{Z-1}{Z+1} $ forms an angle of $ \frac{\pi}{4} $ with the positive real axis. Therefore, $ A $ lies on a line making an angle of $ \frac{\pi}{4} $.

To simplify, we express:

$ \frac{Z-1}{Z+1} = e^{i \frac{\pi}{4}} = \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}} $

Thus, we have:

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

This calculation indicates that the complex number $ \frac{Z-1}{Z+1} $ is on the unit circle. Therefore, we equate:

$ \left|\frac{Z-1}{Z+1}\right| = 1 \Rightarrow |Z-1| = |Z+1| $

This describes the locus as the perpendicular bisector of the line segment joining 1 and -1, which corresponds to the imaginary axis $ x = 0 $. Thus, the locus is a straight line on the imaginary axis.

To find the values of $(8i)^{\frac{1}{3}}$, we utilize De Moivre's theorem. Express $8i$ in polar form as $8e^{i(\frac{\pi}{2} + 2k\pi)}$, then take the cube root:

$ (8i)^{\frac{1}{3}} = 8^{\frac{1}{3}} e^{i\left(\frac{\theta + 2k\pi}{3}\right)} = 2e^{i\left(\frac{\pi}{6} + \frac{2k\pi}{3}\right)} $

Next, calculate this for different values of $k$:

For $k = 0$:

$ 2e^{i\left(\frac{\pi}{6}\right)} = 2\left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right) = 2\left(\frac{\sqrt{3}}{2} + i\frac{1}{2}\right) = \sqrt{3} + i $

For $k = 1$:

$ 2e^{i\left(\frac{\pi}{6} + \frac{2\pi}{3}\right)} = 2e^{i\left(\frac{5\pi}{6}\right)} = 2\left(\cos\frac{5\pi}{6} + i\sin\frac{5\pi}{6}\right) = 2\left(-\frac{\sqrt{3}}{2} + i\frac{1}{2}\right) = -\sqrt{3} + i $

For $k = 2$:

$ 2e^{i\left(\frac{\pi}{6} + \frac{4\pi}{3}\right)} = 2e^{i\left(\frac{3\pi}{2}\right)} = 2(0 - i) = -2i $

Therefore, all the values of $(8i)^{\frac{1}{3}}$ are $\sqrt{3} + i$, $-\sqrt{3} + i$, and $-2i$, or in other words, $\pm \sqrt{3} + i$, $-2i$.

$ x^9-x^5+x^4-1=0 $

$ \begin{array}{lrl} \Rightarrow & x^5\left(x^4-1\right)+1\left(x^4-1\right) & =0 \\ \Rightarrow & \left(x^5+1\right)\left(x^4-1\right) & =0 \\ \Rightarrow & x^5+1=0 \text { or } x^4-1=0 \\ \Rightarrow & x^5=-1 \text { or } x^4=1 \\ \Rightarrow & x^5=-1 \text { or } x=1,-1, i,-i \end{array} $

$\therefore$ Number of real roots $(n)=3$

Number of complex roots $(m)=2$

Number of complex roots having argument in 2nd quadrant $(k)=1$

$ \therefore m \cdot n \cdot k=3(2)(1)=6 $

We have,

$ \begin{aligned} & \frac{(1-i)^3}{(2-i)(3-2 i)}=\frac{1^3-i^3-3.1 . i(1-i)}{\left(6-4 i-3 i+2 i^2\right)} \\ & \quad\left[\because(a-b)^3=a^3-b^3-3 a b(a-b)\right] \\ &= \frac{1-(-i)-3 i+3 i^2}{(6-7 i-2)}=\frac{1+i-3 i-3}{4-7 i} \\ &= \frac{-2-2 i}{4-7 i} \\ &= \frac{-2-2 i}{4-7 i} \times \frac{4+7 i}{4+7 i} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,[rationalising]\\ &= \frac{-(2+2 i)(4+7 i)}{4^2-(7 i)^2} \\ &= \frac{-(8+14 i+8 i-14)}{16+49}=\frac{6-22 i}{65} \end{aligned} $

On comparing with $Z=x+i y$, we get

Imaginary part is $\frac{-22}{65}$.

To find the square root of $7 + 24i$, set it equal to $x + yi$, where $x, y \in \mathbb{R}$.

Squaring both sides gives:

$ (x + yi)^2 = x^2 - y^2 + 2xyi $

This leads to:

$ x^2 - y^2 + 2xyi = 7 + 24i $

By equating the real and imaginary parts, we obtain:

$ x^2 - y^2 = 7 \quad \text{(equation 1)} $

$ 2xy = 24 \quad \Rightarrow \quad xy = 12 \quad \text{(equation 2)} $

From equation 2, solve for $y$:

$ y = \frac{12}{x} $

Substitute $y = \frac{12}{x}$ into equation 1:

$ x^2 - \left(\frac{12}{x}\right)^2 = 7 $

Simplifying the equation, we get:

$ x^2 - \frac{144}{x^2} = 7 $

$ x^4 - 144 = 7x^2 $

$ x^4 - 7x^2 - 144 = 0 $

This can be factored as:

$ (x^2 - 16)(x^2 + 9) = 0 $

Since $x$ is real, we discard $x^2 + 9 = 0$ as it does not lead to real solutions for $x$. Therefore:

$ x^2 = 16 \quad \Rightarrow \quad x = \pm 4 $

For $x = 4$, $y = 3$.

For $x = -4$, $y = -3$.

Thus, the square roots of $7 + 24i$ are $\pm(4 + 3i)$.

Therefore, the correct solution is $4 + 3i$.

We have,

$ Z=\cos \theta+i \sin \theta, \theta \neq(2 n+1) \frac{\pi}{2} $

Now, $\frac{1+Z^{2 n}}{1-Z^{2 n}}=\frac{1+(\cos \theta+i \sin \theta)^{2 n}}{1-(\cos \theta+i \sin \theta)^{2 n}}$

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

[By De Moivre's theorem]

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

When calculating the complex conjugate of the expression $(4-3i)(2+3i)(1+4i)$, follow these steps:

First, compute the product:

$ (4-3i)(2+3i) = 8 + 12i - 6i - 9i^2 = 8 + 6i + 9 = 17 + 6i $

Here, we used the fact that $i^2 = -1$.

Next, multiply this result by $(1+4i)$:

$ (17+6i)(1+4i) = 17 + 68i + 6i + 24i^2 = 17 + 74i - 24 $

Simplifying this, we get:

$ -7 + 74i $

To find the complex conjugate of $-7 + 74i$, change the sign of the imaginary part:

$ -7 - 74i $

So, the complex conjugate of the expression is $-7 - 74i$.

To determine the locus of $ z $, given that the amplitude (also known as the argument) of $ (z-2) $ is $\frac{\pi}{2}$, we proceed as follows:

Let $ z = x + i y $, where $ x $ and $ y $ are real numbers. Given the condition:

$ \arg(z-2) = \frac{\pi}{2} $

Substituting $ z = x + iy $ into the expression, we have:

$ \arg((x-2) + iy) = \frac{\pi}{2} $

This implies:

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

From the property of arctangent, we know:

$ \frac{y}{x-2} \to \infty $

This scenario occurs when $ x-2 = 0 $. Hence,

$ x = 2 $

Thus, the locus of $ z $ is a vertical line at $ x = 2 $, which is parallel to the Y-axis, only considered for $ y > 0 $.

Therefore, the locus of $ z $ is a vertical line $ x = 2 $ for $ y > 0 $.

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} $.

According to question,

Let find slope of $ A B, B C $ and $ C A $

Slope of $ A B=\frac{1-3}{1-(-2)}=\frac{-2}{3} $

Slope of $ B C=\frac{3-\left(\frac{5}{3}\right)}{-2-0}=\frac{-2}{3} $

Slope of $ C A=\frac{\frac{5}{3}-1}{0-1}=\frac{-2}{3} $

Since, slope of all lines are same, therefore, points are collinear.

Let $ z=x+i y $

Given,

$ |z+3-i|=1 $

$ \begin{array}{lr} \Rightarrow & |(x+3)+(y-1) i|=1 \\ \Rightarrow & \sqrt{(x+3)^{2}+(y-1)^{2}}=1 \\ \Rightarrow & (x+3)^{2}+(y-1)^{2}=1 \end{array} \quad \ldots \text { (i) } $

Above is the equation of circle with centre $ \equiv(-3,1) $ and

radius $ =1 $.

Also, $ \arg (z)=\pi \Rightarrow \tan ^{-1}\left|\frac{y}{x}\right|=\pi $

$ \Rightarrow \quad\left|\frac{y}{x}\right|=\tan \pi=0 \Rightarrow y=0 $

By Eq. (i),

$ \begin{array}{lrl} \Rightarrow & x & =-3 \\ \therefore & z & =-3+0 i \\ \Rightarrow & \text { Modulus of } z & =3 \end{array} $

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$