Complex Numbers

$\begin{aligned} & z_1+z_2=5 \\ & z_1^3+z_2^3=20+15 i \\ & z_1^3+z_2^3=\left(z_1+z_2\right)^3-3 z_1 z_2\left(z_1+z_2\right) \\ & z_1^3+z_2^3=125-3 z_1 \cdot z_2(5) \\ & \Rightarrow 20+15 i=125-15 z_1 z_2 \\ & \Rightarrow 3 z_1 z_2=25-4-3 i \\ & \Rightarrow 3 z_1 z_2=21-3 i \\ & \Rightarrow z_1 \cdot z_2=7-i \\ & \Rightarrow\left(z_1+z_2\right)^2=25 \\ & \Rightarrow z_1^2+z_2^2=25-2(7-i) \\ & \Rightarrow 11+2 i \\ & \left(z_1^2+z_2^2\right)^2=121-4+44 i \\ & \Rightarrow z_1^4+z_2^4+2(7-i)^2=117+44 i \\ & \Rightarrow z_1^4+z_2^4=117+44 i-2(49-1-14 i) \\ & \Rightarrow\left|z_1^4+z_2^4\right|=75 \end{aligned}$

$\begin{array}{ll} \text { } & z^{1985}+z^{100}+1=0 \& z^3+2 z^2+2 z+1=0 \\ & (z+1)\left(z^2-z+1\right)+2 z(z+1)=0 \\ & (z+1)\left(z^2+z+1\right)=0 \\ \Rightarrow \quad & z=-1, \quad z=w, w^2 \\ & \text { Now putting } z=-1 \text { not satisfy } \\ & \text { Now put } z=w \\ \Rightarrow \quad & w^{1985}+w^{100}+1 \\ \Rightarrow \quad & w^2+w+1=0 \\ \Rightarrow \quad & \text { Also, } z=w^2 \\ \Rightarrow \quad & w^{3970}+w^{200}+1 \\ \Rightarrow & w+w^2+1=0 \end{array}$

Two common root

$\begin{aligned} & z^2=-i \bar{z} \\ & \left|z^2\right|=|i \bar{z}| \\ & \left|z^2\right|=|z| \\ & |z|^2-|z|=0 \\ & |z|(|z|-1)=0 \\ & |z|=0 \text { (not acceptable) } \\ & \therefore|z|=1 \\ & \therefore|z|^2=1 \end{aligned}$

$\begin{aligned} & z=2-i\left(2 \tan \frac{5 \pi}{8}\right)=x+i y(\text { let }) \\ & r=\sqrt{x^2+y^2} ~\& ~\theta=\tan ^{-1} \frac{y}{x} \\ & r=\sqrt{(2)^2+\left(2 \tan \frac{5 \pi}{8}\right)^2} \\ & =\left|2 \sec \frac{5 \pi}{8}\right|=\left|2 \sec \left(\pi-\frac{3 \pi}{8}\right)\right| \\ & =2 \sec \frac{3 \pi}{8} \\ & \& ~\theta = {\tan ^{ - 1}}\left( {{{ - 2\tan {{5\pi } \over 8}} \over 2}} \right) \\ & =\tan ^{-1}\left(\tan ^2\left(\pi-\frac{5 \pi}{8}\right)\right) \\ & =\frac{3 \pi}{8} \end{aligned}$

To begin with, let's analyze the given equation:

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

First, we compute the modulus of the left side:

Let's take the given value of z,

$z = \frac{1}{2} - 2i$

Now, we find $z + 1$:

$z + 1 = \left(\frac{1}{2} - 2i\right) + 1 = \frac{3}{2} - 2i$

Then, the modulus of $z + 1$ is calculated as follows:

$|z + 1| = \left|\frac{3}{2} - 2i\right| = \sqrt{\left(\frac{3}{2}\right)^2 + (-2)^2}$

$|z + 1| = \sqrt{\frac{9}{4} + 4}$

$|z + 1| = \sqrt{\frac{9}{4} + \frac{16}{4}}$

$|z + 1| = \sqrt{\frac{25}{4}}$

$|z + 1| = \frac{5}{2}$

Now we need to equate the modulus to the right-hand side of the equation and solve for $\alpha$ and $\beta$. Let's rewrite the equation:

$\frac{5}{2} = \alpha z + \beta(1 + i)$

Substitute $z$ with its value:

$\frac{5}{2} = \alpha \left(\frac{1}{2} - 2i\right) + \beta(1+i)$

Rewrite the equation separating real and imaginary parts:

$\frac{5}{2} = \alpha \left(\frac{1}{2}\right) - 2\alpha i + \beta + \beta i$

$\frac{5}{2} = \left(\alpha \frac{1}{2} + \beta\right) + \left(-2\alpha + \beta\right)i$

For the above equality to hold, both real and imaginary parts must be equal.

Equating real parts:

$\alpha \frac{1}{2} + \beta = \frac{5}{2}$

Equating imaginary parts:

$-2\alpha + \beta = 0$

We now have a system of two linear equations:

1) $\frac{\alpha}{2} + \beta = \frac{5}{2}$

2) $-2\alpha + \beta = 0$

Let's solve the system by isolating $\beta$ from the second equation and then substituting it into the first one:

$\beta = 2\alpha$

Now substitute $\beta$ in the first equation:

$\frac{\alpha}{2} + 2\alpha = \frac{5}{2}$

$\alpha \left(\frac{1}{2} + 2\right) = \frac{5}{2}$

$\alpha \left(\frac{1}{2} + \frac{4}{2}\right) = \frac{5}{2}$

$\alpha \left(\frac{5}{2}\right) = \frac{5}{2}$

To find the value of $\alpha$, we divide both sides by $\frac{5}{2}$:

$\alpha = 1$

Now, we use the value of $\alpha$ to find $\beta$:

$\beta = 2\alpha$

$\beta = 2 \cdot 1$

$\beta = 2$

Finally, we add both $\alpha$ and $\beta$ to find $\alpha + \beta$:

$\alpha + \beta = 1 + 2 = 3$

The value of $\alpha + \beta$ is 3.

So, the correct answer is Option C) 3.

$|z-i|=|z+i|=|z-1|$

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

So $n(S)=1$

$\begin{aligned} \text{(A)} \quad & (-1+\sqrt{2})^{\mathrm{n}}=\mathrm{m}+\sqrt{2} \mathrm{n}, \mathrm{m}, \mathrm{n} \in \mathbb{Z} \\ & (1+\sqrt{2})^{\mathrm{n}}=\mathrm{m}_1+\sqrt{2} \mathrm{n}_1, \mathrm{~m}_1, \mathrm{n}_1 \in \mathbb{Z} \\ & \Rightarrow \mathbb{Z} \cup \mathrm{T}_1 \cup \mathrm{T}_2 \subseteq \mathrm{S} \end{aligned}$

but $b \sqrt{2} \in S$ for negative $b \in \mathbb{Z}$.

So $\quad \mathbb{Z} \cup T_1 \cup T_2 \subset \mathrm{S}$

$\begin{aligned} \text{(B)}\quad& (\sqrt{2}-1)^{\mathrm{n}}=\frac{1}{(\sqrt{2}+1)^{\mathrm{n}}}<\frac{1}{2024} \\ & \Rightarrow 2024<(\sqrt{2}+1)^{\mathrm{n}}, \exists \mathrm{n} \in \mathbb{N} \\ & \Rightarrow \mathrm{T}_1 \cap\left(0, \frac{1}{2024}\right) \neq \phi \end{aligned}$

$\begin{aligned} \text{(C)}\quad& (1+\sqrt{2})^{\mathrm{n}}>2024, \exists \mathrm{n} \in \mathbb{N} \\ & \Rightarrow \mathrm{T}_2 \cap(2024, \infty) \neq \phi \end{aligned}$

$\mathrm{(D)}$ $\quad \sin (\pi(a+b \sqrt{2})=0) \Rightarrow b=0, a \in \mathbb{Z}$.

$\Rightarrow$ Options (A), (C), (D) are Correct.

$z=\frac{(2-i)(1+i)^3}{(1-i)^2}$

$ \begin{aligned} z & =\frac{(2-i)\left(1+i^3+3 i^2+3 i\right)}{1+i^2-2 i} \\\\ & =\frac{(2-i)(1-i-3+3 i)}{-2 i} \\\\ & =\frac{(2-i)(-2+2 i)}{-2 i}=\frac{(2-i)(1-i)}{i} \end{aligned} $

$ \begin{aligned} & =\frac{2-i-2 i+i^2}{i} \\\\ & =\frac{1-3 i}{i}=-i-3 \\\\ z & =-3-i \\\\ \arg z= & \tan ^{-1}\left(\frac{1}{3}\right)-\pi \end{aligned} $

\begin{array}{l} \operatorname{amp}\left(\frac{2 z-i}{z+2 i}\right)=\frac{\pi}{4}\quad \text { (given) } \\\\ \Rightarrow \operatorname{amp}(2 z-i)-\operatorname{amp}(z+2 i)=\frac{\pi}{4} \\\\ \Rightarrow \operatorname{amp}[2 x+i(2 y-1)]-\operatorname{amp} {[x+i(y+2)]=\frac{\pi}{4}} \\\\ \Rightarrow \tan ^{-1}\left(\frac{2 y-1}{2 x}\right)-\tan ^{-1}\left(\frac{y+2}{x}\right)=\tan ^{-1} 1 \\\\ \Rightarrow \tan ^{-1}\left(\frac{2 y-1}{2 x}\right)=\tan ^{-1} 1+\tan ^{-1}\left(\frac{y+2}{x}\right) \\\\ \Rightarrow \tan ^{-1}\left(\frac{2 y-1}{2 x}\right)=\tan ^{-1} \frac{1+\frac{y+2}{x}}{1-\frac{y+2}{x}} \\\\ \Rightarrow \quad \frac{2 y-1}{2 x}=\frac{x+y+2}{x-y-2} \\\\ \Rightarrow(2 y-1)(x-y-2)=2 x(x+y+2) \\\\ \Rightarrow 2 x y-2 y^2-4 y-x+y+2 = 2 x^2+2 x y+4 x \\\\ \Rightarrow 2 x^2+2 y^2+5 x+3 y-2=0, (x, y) \neq(0,-2) . \end{array}

$x^2+2 x+4=0$

$ \begin{aligned} & x=\frac{-2 \pm \sqrt{4-16}}{2}=\frac{-2 \pm 2 i \sqrt{3}}{2} \\\\ & \alpha=2 e^{i \frac{2 \pi}{3}} \\\\ & x=-1 \pm i \sqrt{3}=2\left(\frac{\cos 2 \pi}{3}+i \sin \frac{2 \pi}{3}\right) \\\\ & \text { and } \quad \beta=2 e^{-i \frac{2 \pi}{3}} \\\\ & \alpha^{2024}-\beta^{2024}=\left[\frac{\alpha^{2025}}{\alpha}-\frac{\beta^{2025}}{\beta}\right] \\\\ & =\frac{2^{2025} e^{i \frac{2 \pi}{3} \times 2025}}{\alpha}-\frac{2^{2025} e^{-i \frac{2 \pi}{3} \times 2025}}{\beta} \\\\ & =2^{2025}\left[\frac{e^{i 1350 \pi}}{\alpha}-\frac{e^{-i 1350 \pi}}{\beta}\right] \\\\ & =2^{2025}\left[\frac{1}{\alpha}-\frac{1}{\beta}\right]=2^{2025}\left(\frac{1}{2} e^{-i \frac{2 \pi}{3}}-\frac{1}{2} e^{-i \frac{i \pi}{3}}\right) \end{aligned} $

$ \begin{aligned} & =-2^{2024}\left(e^{i \frac{2 \pi}{3}}-e^{-i \frac{2 \pi}{3}}\right) \\\\ & =-2^{2024}\left(i 2 \sin \frac{2 \pi}{3}\right) \\\\ & =-2^{2024} \cdot 2 \times \frac{\sqrt{3}}{2} i=i K \\\\ & \therefore \quad K=-2^{2024} \sqrt{3} \end{aligned} $

To solve the given equation $ z^{2} + az + a^{2} = 0 $, where $ z = x + iy $ and $ a \in \mathbb{R} $, follow these steps:

First, multiply both sides of the equation by $ (z-a) $:

$ (z-a)(z^{2} + az + a^{2}) = 0 $

Apply the identity for the difference of cubes:

$ z^{3} - a^{3} = 0 $

This implies:

$ z^{3} = a^{3} $

Taking the modulus (or absolute value) of both sides:

$ |z|^{3} = |a|^{3} $

Thus, we conclude:

$ |z| = |a| $

Given the complex numbers $ z_1, z_2, z_3 $ with unit modulus, we know:

$ z_1 \bar{z}_1 = z_2 \bar{z}_2 = z_3 \bar{z}_3 = 1 $

This implies that each complex number has a magnitude of 1. We are also given the condition:

$ |z_1 - z_2|^2 + |z_1 - z_3|^2 = 4 $

We can expand the squared magnitudes as follows:

$ \begin{aligned} |z_1 - z_2|^2 &= (z_1 - z_2)(\bar{z}_1 - \bar{z}_2) = z_1 \bar{z}_1 - z_1 \bar{z}_2 - \bar{z}_1 z_2 + z_2 \bar{z}_2, \\ |z_1 - z_3|^2 &= (z_1 - z_3)(\bar{z}_1 - \bar{z}_3) = z_1 \bar{z}_1 - z_1 \bar{z}_3 - \bar{z}_1 z_3 + z_3 \bar{z}_3. \end{aligned} $

Substitute the expansions back into the equation:

$ \begin{aligned} & (z_1 \bar{z}_1 - z_1 \bar{z}_2 - \bar{z}_1 z_2 + z_2 \bar{z}_2) \\ & + (z_1 \bar{z}_1 - z_1 \bar{z}_3 - \bar{z}_1 z_3 + z_3 \bar{z}_3) = 4. \end{aligned} $

Since $ z_1 \bar{z}_1 = z_2 \bar{z}_2 = z_3 \bar{z}_3 = 1 $, we simplify:

$ (1 - z_1 \bar{z}_2 - \bar{z}_1 z_2 + 1) + (1 - z_1 \bar{z}_3 - \bar{z}_1 z_3 + 1) = 4 $

Simplifying further gives:

$ 4 - (z_1 \bar{z}_2 + \bar{z}_1 z_2 + z_1 \bar{z}_3 + \bar{z}_1 z_3) = 4 $

Thus:

$ z_1 \bar{z}_2 + \bar{z}_1 z_2 + z_1 \bar{z}_3 + \bar{z}_1 z_3 = 0 $

This is the required result.

Given the equation:

$ \left(\frac{a+b \omega+c \omega^{2}}{c+a \omega+b \omega^{2}}\right)^{k}+\left(\frac{a+b \omega+c \omega^{2}}{b+a \omega^{2}+c \omega}\right)^{l}=2 $

where $\omega$ is the complex cube root of unity, meaning $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$.

To solve this, we first manipulate the terms by using properties of complex numbers. Consider multiplying and dividing:

Multiply and divide the first term by $\omega^2$:

$ \left(\frac{\omega^2(a+b \omega+c \omega^2)}{\omega^2(c+a \omega+b \omega^2)}\right)^{k} $

Multiply and divide the second term by $\omega$:

$ \left(\frac{\omega(a+b \omega+c \omega^2)}{\omega(b+a \omega^2+c \omega)}\right)^{l} $

This simplifies to:

$ \left(\omega^{2k}\right) + \left(\omega^l\right) = 2 $

For these powers to sum to 2, and knowing the properties of cube roots of unity, we have specific implications for the exponents:

Since $\omega^n$ can only take values 1, $\omega$, or $\omega^2$, for the sum $\omega^{2k} + \omega^l = 2$ to be a real number (i.e., 2), both $\omega^{2k}$ and $\omega^l$ must equal 1.

Thus, this requires both 2k and l to be multiples of 3 (since $\omega^3 = 1$).

Hence, the expression $2k + l$ is also a multiple of 3, indicating that $2k + l$ is always divisible by 3.

Given, $ z_{1}=\sqrt{3}+i \sqrt{3} $

$ \begin{aligned} z_{2} & =\sqrt{3}+i \\ \frac{z_{1}}{z_{2}} & =\frac{\sqrt{3}(1+i)}{\sqrt{3}+i}=\frac{\sqrt{3}(1+i)}{\sqrt{3}+i} \frac{(\sqrt{3}-i)}{(\sqrt{3}-i)} \\ & =\frac{\sqrt{3}\left(\sqrt{3}-i+\sqrt{3} i-i^{2}\right)}{3-i^{2}} \\ & =\frac{\sqrt{3}}{4}[\sqrt{3}+1+i(\sqrt{3}-1)] \end{aligned} $

multiply and divide by $ 2 \sqrt{2} $.

$ \begin{aligned} \Rightarrow \quad \frac{z_{1}}{z_{2}} & =\frac{2 \sqrt{2} \sqrt{3}}{4} \quad\left[\cos 15^{\circ}+i \sin 15^{\circ}\right] \\ \frac{z_{1}}{z_{2}} & =\frac{\sqrt{6}}{2}\left[e^{i 15^{\circ}}\right] \\ \Rightarrow\left(\frac{z_{1}}{z_{2}}\right)^{50} & =\left(\frac{\sqrt{6}}{2}\right)^{50} e^{i 750^{\circ}} \\ \theta & =750^{\circ}-720^{\circ}=30^{\circ} \end{aligned} $

$ \because $ First quadrant

$ x^{3}-3 x^{2}+3 x+7=0 $

$ \begin{array}{l} \Rightarrow \quad(x+1)\left(x^{2}-4 x+7\right)=0 \\ \Rightarrow \quad x=-1 \text { or } \frac{4 \pm \sqrt{16-28}}{2(1)} \\ \Rightarrow \quad x=-1 \text { or } 2 \pm \sqrt{3} i \\ \therefore \quad \alpha=-1, \beta=2+\sqrt{3} i \text { and } \\ \gamma=2-\sqrt{3} i \end{array} $

New root are $ -1-h_{,}(2-h)+\sqrt{3} i $ and $ (2-h)-\sqrt{3} i $

Equation which have these roots is $ (x+1+h)(x+h-2-\sqrt{3} i) $

$ \begin{array}{l} \quad(x+h-2+\sqrt{3} i)=0 \\ \Rightarrow(x+1+h)\left[(x+h-2)^{2}-(\sqrt{3} i)^{2}\right]=0 \\ \Rightarrow(x+1+h)\left[x^{2}+(h-2)^{2}\right. \\ +2 x(h-2)+3]=0 \end{array} $

In this equation terms containing $ x^{2} $ and $ x $ should be missing.

$ \therefore(1+h)+2(h-2)=0 $ and

$ 2(1+h)(h-2)+3+(h-2)^{2}=0 $

Now, $ \frac{\alpha-h}{\beta-h}+\frac{\beta-h}{\gamma-h}+\frac{\gamma-h}{\alpha-h} $

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

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

$ \begin{array}{l} =\frac{-2(1-\sqrt{3} i)}{4}+\frac{(1+\sqrt{3} i)^{2}}{4}+\frac{(-2)(1-\sqrt{3} i)}{4} \\ =\frac{1}{4}[-2+2 \sqrt{3} i+1-3+2 \sqrt{3} i-2+2 \sqrt{3} i] \\ =\frac{1}{4}[-6+6 \sqrt{3} i] \\ =3\left[-\frac{1}{2}+\frac{\sqrt{3}}{2} i\right] \end{array} $

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

$ \begin{aligned} & =2+3 i \\\\ \therefore x+i y & =\frac{13(2+3 i)}{(2-3 i)(3+2 i)} \\\\ & =\frac{13(2+3 i)}{6-9 i+4 i-6 i^{2}}=\frac{13(2+3 i)}{12-5 i} \\\\ & =\frac{13(2+3 i)(12+5 i)}{(12-5 i)(12+5 i)} \\\\ & =\frac{13\left(24+36 i+10 i+15 i^{2}\right)}{169} \\\\ & =\frac{13(9+46 i)}{169}=\frac{9}{13}+\frac{46 i}{13} \\\\ \Rightarrow \quad x & =\frac{9}{13} \text { and } y=\frac{46}{13} \\\\ 13 y-26 \dot{x} & =46-18=28 \end{aligned} $

Given, $ z=x+i y $

$ \begin{array}{l} |x+i y-1|+|x+i y+i|=2 \\\\ \sqrt{(x-1)^{2}+y^{2}}+\sqrt{x^{2}+(y+1)^{2}}=2 \\\\ \sqrt{(x-1)^{2}+y^{2}}=2-\sqrt{x^{2}+(y+1)^{2}} \\\\ x^{2}+y^{2}-2 x+1=4+x^{2}+y^{2}+2 y+1 -4 \sqrt{x^{2}+(y+1)^{2}} \\\\ x+y+2=2 \sqrt{x^{2}+y^{2}+2 y+1} \end{array} $

$ \begin{array}{r} x^{2}+y^{2}+4+2 x y+4 x+4 y \\\\ =4 x^{2}+4 y^{2}+8 y+4 \\\\ 3 x^{2}+3 y^{2}-2 x y-4 x+4 y=0 \end{array} $

To find one of the values of $(-64i)^{5/6}$, let's proceed with the computation:

Let $ Z = (-64i)^{5/6} $.

We can express $ Z $ as:

$ Z = \left((-i)^5\right)^{1/6} \times (64)^{5/6} $

Calculate $(64)^{5/6}$:

$ 64^{5/6} = 32 $

Handle the imaginary unit:

Since $(-i)^5 = -i$, we represent this in polar coordinates:

$ -i = \cos\left(\frac{3\pi}{2}\right) + i\sin\left(\frac{3\pi}{2}\right) $

Find $(-i)^{1/6}$:

Using De Moivre’s theorem, $ (-i)^{1/6} $ can be expressed as:

$ (-i)^{1/6} = \left[\cos\left(\frac{3\pi}{2}\right) + i\sin\left(\frac{3\pi}{2}\right)\right]^{1/6} $

$ = \cos\left(\frac{3\pi}{12}\right) + i\sin\left(\frac{3\pi}{12}\right) $

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

Combine the results:

Therefore, $ Z $ becomes:

$ Z = 32 \times \frac{1}{\sqrt{2}}(1+i) = 16\sqrt{2}(1+i) $

The calculation yields one of the values of $(-64i)^{5/6}$ as $ 16\sqrt{2}(1+i) $.

We have,

$ \begin{aligned} & \frac{(2-i) x+(1+i)}{2+i}+\frac{(1-2 i) y+(1-i)}{1+2 i}=1-2i \\\\ & \Rightarrow \frac{(2 x+1)+(1-x) i}{2+i}+\frac{(y+1)+(-2 y-1) i}{1+2 i}=1-2i \\\\ & \Rightarrow \frac{[(2 x+1)+(1-x) i](2-i)}{4-i^2}+\frac{[(y+1)+(-2 y-1) i](1-2 i)}{1-4 i^2}=1-2i \\\\ & \Rightarrow {[2(2 x+1)+(1-x)+(y+1)+2(-2 y-1)] } \\\\ &+i[-2 x-1+2(1-x)-2(y+1)+(-2 y-1)] \\\\ &=5(1-2 i) \\\\ & \Rightarrow(3 x-3 y+2)+i(-4 x-4 y-2)=5-10 i \end{aligned} $

We have,

$ z=1-\sqrt{3} i=2 e^{-i \frac{\pi}{3}} $

Now, $z^3-3 z^2+3 z$

$ \begin{aligned} & \begin{array}{r} =\left(2 e^{-\frac{\pi}{3} i}\right)^3-3\left(2 e^{-\frac{\pi}{3} i}\right)^2+3\left(2 e^{-\frac{\pi}{3} i}\right) \\\\ \begin{aligned} &= 8 e^{-\pi i}-12 e^{-\frac{2 \pi}{3} i}+6 e^{-\frac{\pi}{3} i} \\\\ &=8(\cos (-\pi)+i \sin (-\pi)) \end{aligned} \\\\ \quad-12\left(\cos \left(-\frac{2 \pi}{3}\right)+i \sin \left(-\frac{2 \pi}{3}\right)\right) \\\\ \quad+6\left(\cos \left(-\frac{\pi}{3}\right)+i \sin \left(-\frac{\pi}{3}\right)\right) \\\\ =8(-1+i(0))-12\left(-\frac{1}{2}+i\left(-\frac{\sqrt{3}}{2}\right)\right) \\\\ +6\left(\frac{1}{2}+i\left(-\frac{\sqrt{3}}{2}\right)\right)\end{array} \\\\ & =(-8+6+3)+i(0+6 \sqrt{3}-3 \sqrt{3}) \\\\ & =1+3 \sqrt{3 i} \end{aligned} $

Let, $(\sqrt{3}-i)^{2 / 5}=x$

$ \begin{array}{ll} \Rightarrow & (\sqrt{3}-i)^2=x^5 \\\\ \Rightarrow & x^5=(\sqrt{3})^2+(i)^2-2 \sqrt{3} i \\\\ \Rightarrow & x^5=3+(-1)-2 \sqrt{3} i \\\\ \Rightarrow & x^5=2-2 \sqrt{3} i=2(1-\sqrt{3} i) \end{array} $

Clearly, the product of all the values of $(\sqrt{3}-i)^{2 / 5}$ is $2(1-\sqrt{3} i)$

12 th roots of unity are $l, e^{i \frac{\pi}{6}}, e^{i \frac{2 \pi}{6}}$, $e^{i \frac{3 \pi}{6}}, e^{i \frac{4 \pi}{6}}, \ldots, e^{i \frac{11 \pi}{6}}$.

30th roots of unity are $1, e^{i \frac{\pi}{15}}, e^{i \frac{2 \pi}{15}}$, $e^{i \frac{3 \pi}{15}}, e^{i \frac{4 \pi}{15}}, \ldots, e^{i \frac{29 \pi}{15}}$.

Clearly, common roots are $1, e^{i \frac{\pi}{3}}, e^{i \frac{2 \pi}{3}}$, $e^{i \pi}, e^{i \frac{4 \pi}{3}}$ and $e^{i \frac{5 \pi}{3}}$

Hence, the number of common roots is 6.

We have,

$ \begin{aligned} & \sqrt{5}-i \sqrt{15}=r(\cos \theta+i \sin \theta) \\\\ & \begin{aligned} \Rightarrow \quad r & =\sqrt{(\sqrt{5})^2+(-\sqrt{15})^2} \\\\ & =\sqrt{5+15}=2 \sqrt{5} \end{aligned} \\\\ & \text { and } \theta=\tan ^{-1}\left(\frac{\text { Imaginary part }}{\text { Real part }}\right) \\\\ & =\tan ^{-1}\left(\frac{-\sqrt{15}}{\sqrt{5}}\right)=\tan ^{-1}(-\sqrt{3}) \end{aligned} $

$\begin{aligned} & \Rightarrow \quad \theta=-\frac{\pi}{3} \\\\ & \text { So, } r^2\left(\sec \theta+3 \operatorname{cosec}^2 \theta\right) \\\\ & =(2 \sqrt{5})^2\left[\sec \left(-\frac{\pi}{3}\right)+3 \operatorname{cosec}^2\left(-\frac{\pi}{3}\right)\right] \\\\ & =20\left[2+3 \times \frac{4}{3}\right]=20 \times 6=120\end{aligned}$

We have, $z=x+i y$

$ \begin{aligned} \therefore \frac{2 z-i}{z-2}= & \frac{2(x+i y)-i}{x+i y-2} \\\\ & =\frac{2 x+(2 y-1) i}{x-2+i y} \\\\ & =\frac{2 x+(2 y-1) i}{(x-2)+y i} \times\left[\frac{(x-2)-y i}{(x-2)-y i}\right] \\\\ & 2 x(x-2)+y(2 y-1) \\\\ & =\frac{+[(x-2)(2 y-1)-2 x y] i}{(x-2)^2+y^2} \\\\ = & \frac{2 x^2-4 x+2 y^2-y}{(x-2)^2+y^2} \\\\ & \quad+\frac{[(x-2)(2 y-1)-2 x y] i}{(x-2)^2+y^2} \end{aligned} $

$\because \frac{2 z-i}{z-2}$ is purely real number

$ \begin{aligned} & \Rightarrow \quad \frac{(x-2)(2 y-1)-2 x y}{(x-z)^2+y^2}=0 \\\\ & \Rightarrow \quad 2 x y-x-4 y-2 x y+2=0 \\\\ & \Rightarrow \quad x+4 y=2 \end{aligned} $

Given $x$ and $y$ are two complex number with $|x|=|y|=1$ and $x=e^{i 2 \alpha}$

$ \begin{aligned} y & =e^{i .3 \beta}, \text { where } \alpha+\beta=\frac{\pi}{36} \\\\ \Rightarrow \quad x^6 & =\left(e^{i 2 \alpha}\right)^6=e^{i .12 \alpha} \\\\ y^4 & =\left(e^{i .3 \beta}\right)^4=e^{i .12 \beta} \end{aligned} $

So, $x^6 \cdot y^4=e^{i .12 \alpha} \cdot e^{i, 12 \beta}=e^{i .12(\alpha+\beta)}$

Since, $\alpha+\beta=\frac{\pi}{36}$

$ \Rightarrow \quad x^6 \cdot y^4=e^{i .12 \frac{\pi}{36}=e^{i \frac{\pi}{3}}} $

and $\frac{1}{x^6 \cdot y^4}=e^{-i \cdot \frac{\pi}{3}}$

$ \begin{aligned} \Rightarrow x^6 \cdot y^4+\frac{1}{x^6 \cdot y^4} & =e^{i \cdot \frac{\pi}{3}}+e^{-i \cdot \frac{\pi}{3}} \\\\ & =2 \cos \frac{\pi}{3}=1 \end{aligned} $

We have,

$ \begin{aligned} & x^{14}+x^9-x^5-1=0 \\\\ & \Rightarrow \quad\left(x^9-1\right)\left(x^5+1\right)=0 \\\\ & x^9=1 \\\\ & x=(1)^{1 / 9} \\\\ & x=\left(\cos 0^{\circ}\right. \\\\ & \left.+i \sin 0^{\circ}\right)^{1 / 9} \\\\ & =(\cos 2 K \pi \\\\ & +i \sin 2 K \pi)^{1 / 9} \\\\ & =\cos \frac{2 K \pi}{9}+i \sin \frac{2 K \pi}{9} \\\\ & \text { using Demoivre's theorm here, } K=0 \text {, } 1, \ldots ., 8 \\\\ & \begin{array}{l} x^5=-1 \\\\ x=(\cos \pi+i \sin \pi)^{1 / 5} \\\\ =[\cos (2 K \pi+\pi) \\\\ +i \sin (2 K \pi+\pi)]^{1 / 5} \\\\ =[\cos (2 K+1) \pi \\\\ +i \sin (2 K+1) \pi]^{1 / 5} \\\\ =\cos \frac{(2 K+1) \pi}{5} \\\\ +i \sin \frac{(2 K+1) \pi}{5} \\\\ \text { where } K=0,1, \ldots ., 5 \end{array} \end{aligned} $

So, one of root of the given equation is

$ \begin{aligned} & \text { So, one of root } \frac{2 \pi}{9}+i \sin \frac{2 \pi}{9} \text { for } K=1 \\\\ & =\frac{\sqrt{5}+1}{4}+i \frac{\sqrt{10-2 \sqrt{5}}}{4} \end{aligned} $

We have, $|z-1| \leq 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

$ \begin{aligned} & \text { and }\left|\omega^2 z-1-\omega\right|=a \\ & \qquad \begin{array}{l} \omega^2 z+\omega^2 \mid=a \quad\left[\because \omega^2=-1-\omega\right] \\ \left|\omega^2\right||z+1|=a \\ |z-1+2|=a \\ |z-1|+2 \geq a \end{array} \end{aligned} $

From Eq. (i), we get

$ \begin{aligned} & |z-1|+2=4 \\ \Rightarrow \quad & 0 \leq a \leq 4 \end{aligned} $

We have, $z^3+i z^2+2 i=0$

One root of $z^3+i z^2+2 i=0$ is $i$.

$ \begin{array}{ll} & {\left[\because i^3+i\left(i^2\right)+2 i=-i-i+2 i=0\right]} \\ & z^3+i z^2+2 i=0 \\ \Rightarrow \quad & (z-i)\left(z^2+2 i z-2\right)=0 \\ \Rightarrow \quad & (z-i)\left(z^2+2 i z+i^2+1-2\right)=0 \\ \Rightarrow \quad & (z-i)\left((z+i)^2-1\right)=0 \\ \Rightarrow \quad & (z-i)(z+(i+1))(z+(i-1))=0 \\ \Rightarrow \quad & z=i,-i-1,1-i \end{array} $

$\therefore$ The triangle formed by the vertices $A(0,1), B(-1,-1)$ and $C(1,-1)$.

$ A B=\sqrt{1+4}=\sqrt{5} \text { units } $

$B C=2$ units

$A C=\sqrt{1+4}=\sqrt{5}$ units

$\Rightarrow \triangle A B C$ is an isosceles triangle.

$ [\because A B=A C] $

Since, the non-equal side is less than the equal side, so $A B C$ is not a right-angled triangle.

In the complex plane, any point can be represented in polar form as $(r, \theta) = r(\cos \theta + i \sin \theta)$. Here, $x = (1, \alpha)$, $y = (1, \beta)$, and $z = (1, \gamma)$. Given the condition $x + y + z = 0$, we can express

$ x = \cos \alpha + i \sin \alpha $

Similarly,

$ y = \cos \beta + i \sin \beta $

$ z = \cos \gamma + i \sin \gamma $

These imply:

$ x + y + z = 0 \Rightarrow \cos \alpha + \cos \beta + \cos \gamma = 0 $

$ \sin \alpha + \sin \beta + \sin \gamma = 0 $

From the above, we isolate:

$ \cos \alpha + \cos \beta = -\cos \gamma \quad \text{...(i)} $

$ \sin \alpha + \sin \beta = -\sin \gamma \quad \text{...(ii)} $

Squaring and adding equations (i) and (ii), we get:

$ \begin{align*} (\cos \alpha + \cos \beta)^2 + (\sin \alpha + \sin \beta)^2 &= \cos^2 \gamma + \sin^2 \gamma \\ \cos^2 \alpha + \cos^2 \beta + 2 \cos \alpha \cos \beta + \sin^2 \alpha + \sin^2 \beta + 2 \sin \alpha \sin \beta &= 1 \end{align*} $

This simplifies to:

$ 2 \cos(\alpha - \beta) = -1 \quad \Rightarrow \quad \cos(\alpha - \beta) = -\frac{1}{2} $

By symmetry,

$ \cos(\beta - \gamma) = -\frac{1}{2} $

and

$ \cos(\gamma - \alpha) = -\frac{1}{2} $

The corresponding sine values are:

$ \sin (\alpha - \beta) = \sin (\beta - \gamma) = \sin (\gamma - \alpha) = \frac{\sqrt{3}}{2} $

Consider $\cos (2\alpha - \beta - \gamma)$:

$ \begin{align*} \cos(2\alpha - \beta - \gamma) &= \cos((\alpha - \beta) - (\gamma - \alpha)) \\ &= \cos(\alpha - \beta) \cos(\gamma - \alpha) + \sin(\alpha - \beta) \sin(\gamma - \alpha) \\ &= \left(-\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 \\ &= \frac{1}{4} + \frac{3}{4} \\ &= 1 \end{align*} $

Similarly,

$ \cos(2\beta - \gamma - \alpha) = 1 $

$ \cos(2\gamma - \alpha - \beta) = 1 $

Thus, we find:

$ \Sigma \cos (2\alpha - \beta - \gamma) = 1 + 1 + 1 = 3 $

We have, $\arg \left[\frac{(1+i \sqrt{3})(-\sqrt{3}-i)}{(1-i)(-i)}\right]$

$ \begin{aligned}Let\,\,\,\, z & =\frac{(1+i \sqrt{3})(-\sqrt{3}-i)}{(1-i)(-i)} \\ & =\frac{\left(i+i^2 \sqrt{3}\right)(-\sqrt{3}-i)}{(1-i)\left(-i^2\right)} \\ & =\frac{(-\sqrt{3}+i)(-\sqrt{3}-i)}{(1-i)} \\ & =\frac{(-\sqrt{3})^2-i^2}{1-i}=\frac{3+1}{1-i}=\left(\frac{4}{1-i}\right) \\ & \frac{4(1+i)}{(1-i)(1+i)}=4 \times \frac{1+i}{2} \end{aligned} $

$ \begin{aligned} \Rightarrow \quad z & =2(1+i) \\ \therefore \arg (z)=\arg (2+2 i) & =\tan ^{-1}(1) \\ & =\tan ^{-1}\left(\tan \frac{\pi}{4}\right) \\ \Rightarrow \quad \arg (z) & =\frac{\pi}{4} \end{aligned} $

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 $.