Complex Numbers

Let, complex number ${\left( {\overline z } \right)^2} + {1 \over {{z^2}}}$ is a new complex number $w$.

$\therefore$ $w = {\left( {\overline z } \right)^2} + {1 \over {{z^2}}}$

Now, let $z = r{e^{i\theta }}$ where $r = |z|$ and $\theta$ = argument

$\therefore$ $w = {\left( {r{e^{ - i\theta }}} \right)^2} + {1 \over {{r^2}}}({e^{ - 2i\theta }})$

$ = \left( {{r^2} + {1 \over {{r^2}}}} \right){e^{ - 2i\theta }}$

$ = \left( {{r^2} + {1 \over {{r^2}}}} \right)(\cos 2\theta - i\sin 2\theta )$

$\therefore$ Real part of $w = {\mathop{\rm Re}\nolimits} (w) = \left( {{r^2} + {1 \over {{r^2}}}} \right)\cos 2\theta $

Imaginary part of $w = {\mathop{\rm Im}\nolimits} (w) = \left( {{r^2} + {1 \over {{r^2}}}} \right)( - \sin 2\theta )$

Given both ${\mathop{\rm Re}\nolimits} (w)$ and ${\mathop{\rm Im}\nolimits} (w)$ are integer.

$\therefore$ Let $Re(w) = {I_1}$

and ${\mathop{\rm Im}\nolimits} (w) = {I_2}$

$\therefore$ $I_1^2 + I_2^2 = {\left( {{r^2} + {1 \over {{r^2}}}} \right)^2} = \alpha $ (where $\alpha$ is an integer)

[Note : As I₁ and I₂ are integer so $I_1^2 + I_2^2$ is also an integer.]

Now, ${\left( {{r^2} + {1 \over {{r^2}}}} \right)^2} = \alpha $

$ \Rightarrow {r^4} + 2 + {1 \over {{r^4}}} = \alpha $

$ \Rightarrow {r^8} + 2{r^4} + 1 = {r^4}\alpha $

$ \Rightarrow {r^8} - (\alpha - 2){r^4} + 1 = 0$

$\therefore$ ${r^4} = {{(\alpha - 2)\, \pm \,\sqrt {{{(\alpha - 2)}^2} - 4\,.\,1\,.\,1} } \over {2\,.\,1}}$

$ \Rightarrow r = {\left( {{{(\alpha - 2)\, \pm \,\sqrt {{\alpha ^2} - 4\alpha } } \over 2}} \right)^{{1 \over 4}}}$

$ \Rightarrow |z| = r = {\left( {{{(\alpha - 2)\, \pm \,\sqrt {{\alpha ^2} - 4\alpha } } \over 2}} \right)^{{1 \over 4}}}$

In option only positive sign is given so ignoring negative sign we get,

$|z| = r = {\left( {{{(\alpha - 2) + \sqrt {{\alpha ^2} - 4\alpha } } \over 2}} \right)^{{1 \over 4}}}$ ....... (1)

From option (A),

$|z| = {\left( {{{43 + 3\sqrt {205} } \over 2}} \right)^{{1 \over 4}}}$

Comparing with (1), we get

$\alpha - 2 = 43$

$ \Rightarrow \alpha = 45$

Putting $\alpha = 45$ in (1), we get

$|z| = {\left( {{{43 + \sqrt {45(41)} } \over 2}} \right)^{{1 \over 4}}}$

$ = {\left( {{{43 + \sqrt {9 \times 5 \times 41} } \over 2}} \right)^{{1 \over 4}}}$

$ = {\left( {{{43 + 3\sqrt {205} } \over 2}} \right)^{{1 \over 4}}}$

$\therefore$ Option (A) is correct.

We can re-write

$|z| = {\left( {{{(\alpha - 2) + \sqrt {{\alpha ^2} - 4\alpha } } \over 2}} \right)^{{1 \over 4}}}$

as $|z| = {\left( {{{2(\alpha - 2) + 2\sqrt {\alpha (\alpha - 4)} } \over 4}} \right)^{{1 \over 4}}}$

Comparing with option (B) we get,

$2(\alpha - 2) = 7$

$ \Rightarrow 2\alpha = 11$

$ \Rightarrow \alpha = {{11} \over 2}$ $\ne$ Integer

$\therefore$ Option (B) is incorrect.

Similarly option (C) and (D) also incorrect.

          $ x^2-2 x+2=0 $

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

$ \begin{aligned} &\text { }\\ &\begin{aligned}Taking\,\,\, \alpha & =1+i, \beta=1-i \\ \alpha^2 & =(1+i)^2=1+(i)^2+2 i \\ & =1-1+2 i=2 i \\ \beta^2 & =(1-i)^2=1+(i)^2-2 i \\ & =1-1-2 i=-2 i \end{aligned} \end{aligned} $

$ \text { } \begin{aligned} Now, \alpha^{2020} & +\beta^{2020} \\ & =\left(\alpha^2\right)^{1010}+\left(\beta^2\right)^{1010} \\ & =(2 i)^{1010}+(-2 i)^{1010} \\ & =2^{1010}\left[i^{1010}+(-i)^{1010}\right] \\ & =2^{1010}\left[2\left(i^{1010}\right)\right]=2^{1011}(i)^2 \\ & =2^{1011}(-1)=-2^{1011} \end{aligned} $

$ \begin{aligned} z & =\frac{-1-i \sqrt{3}}{2} \\ z & =\omega^2 \cdot\left(z^k+\frac{1}{z^k}\right)^2=\left(\omega^{2 k}+\frac{1}{\omega^{2 k}}\right)^2 \\ & =\omega^{4 k}+\frac{1}{\omega^{4 k}}+2 \\ & =\omega^{4 k}+\frac{\omega^{2 k}}{\omega^{4 k} \cdot \omega^{2 k}}+2 \\ & =\omega^{4 k}+\frac{\omega^{2 k}}{\omega^{6 k}}+2 \\ & =\omega^{4 k}+\omega^{2 k}+2 \quad\left[\because \omega^{6 k}=\left(\omega^3\right)^{2 k}=(1)^{2 k}=1\right]\\ & =\omega^k+\omega^{2 k}+2\,\,\,\,\,\, \left[\because \omega^{3 k}=1\right]\end{aligned} $

$ \begin{aligned} & \text { } \begin{aligned}Now, \sum_{k=1}^{2022} & \omega^k+\omega^{2 k}+2 \\ = & {\left[\omega+\omega^2+\omega^3+\ldots .+\omega^{2022}\right] } \\ & +\left[\omega^2+\omega^4+\omega^6+\ldots+\omega^{4044}\right] \end{aligned} \\ & \begin{aligned} \frac{\omega \cdot\left(\omega^{2022}-1\right)}{(\omega-1)}+\frac{\omega^2\left(\left(\omega^2\right)^{2022}-1\right)}{\omega^2-1}+\sum_{k=1}^{2022} 2 \end{aligned} \\ & =\frac{\omega(1-1)}{(\omega-1)}+\frac{\omega^2(1-1)}{\omega^2-1}+2 \times(1+1+\ldots .+1) \\ & =0+0+2 \times 2022 \\ & =4044 \end{aligned} $

For $x \in[0,2 \pi]$

$\sin x+i \cos 2 x$ and $\cos x-i \sin 2 x$ are conjugate to each other.

$ \begin{aligned} \Rightarrow \quad \sin x+i \cos 2 x & =\overline{\cos x-i \sin 2 x} \\ \sin x+i \cos 2 x & =\cos x+i \sin 2 x \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

On comparing both side of Eq. (i), we get

$ \begin{array}{rlrlrl} & & \sin x & =\cos x ; & \sin 2 x=\cos 2 x \\ \Rightarrow & & x & =\frac{\pi}{4}, \frac{5 \pi}{4} ; & \tan 2 x=1 \\ \Rightarrow & & 2 x & =\frac{\pi}{4}, \frac{5 \pi}{4}, \frac{9 \pi}{4}, \frac{13 \pi}{4} \\ \Rightarrow & & x & =\frac{\pi}{8}, \frac{5 \pi}{8}, \frac{9 \pi}{8}, \frac{13 \pi}{8} \end{array} $

But there is no common value of $x$ which satisfy both the condition. So, $x=\phi$

Given, $|x+i y|=\sqrt{x^2+y^2}$

$ \begin{aligned}Since,\,\ & (1-\sqrt{3} i)^9+(\sqrt{3}+i)^9 \\ & \quad=i^{-9}\left(\frac{1}{i}-\sqrt{3}\right)^9+(\sqrt{3}+i)^9 \\ & {\left[\because i^{-9}=\frac{1}{i^9}=\frac{1}{i^8 \cdot i}=\frac{1}{\left(i^2\right)^4 i}\right.} \left.=\frac{1}{(-1)^4 i}=\frac{1}{i}=-i\right] \end{aligned} $

$ \begin{aligned} Now,\,\,&\left|(1-\sqrt{3} i)^9+(\sqrt{3}+i)^9\right| \\ &=\left|(i+\sqrt{3})^9(i+1)\right| \\ &=|i+1|\left|(i+\sqrt{3})^9\right| \\ &=|i+1||i+\sqrt{3}|^9 \\ &=\left(\sqrt{1^2+1^2}\right)^{\left(\sqrt{1^2+(\sqrt{3})^2}\right)^9} \\ &=\sqrt{2}(\sqrt{4})^9 \\ &=2^{\frac{1}{2}} \cdot 2^9=2^{\frac{19}{2}} \end{aligned} $

If $x^3=1$

Then, $\quad x=1, \omega, \omega^2$ and $x^4=1$

$ \begin{array}{lr} \Rightarrow & x^4-1=0 \\ \Rightarrow & x \pm 1, \pm i \end{array} $

So, roots are $1, i,-1,-i$

i.e. $\quad \alpha=i$

$\left[\because 1, \alpha, \alpha^2, \alpha^3\right.$ are fourth roots of unity]

Now, $\alpha+\alpha \omega-\alpha^3 \omega^2=\alpha+\alpha \omega-\frac{\omega^2}{\alpha}$

$ \begin{aligned} {\left[\because \alpha=i \Rightarrow \alpha^3=i^3\right.} & \Rightarrow \alpha^3=i^2 \cdot i \\ & \left.\Rightarrow \alpha^3=-i \Rightarrow \alpha^3=-\alpha\right] \\ = & \alpha(1+\omega)-\frac{\omega^2}{\alpha}=-\alpha \omega^2-\frac{\omega^2}{\alpha} \\ = & -\omega^2\left(\alpha+\frac{1}{\alpha}\right)=-\omega^2\left(\frac{\alpha^2+1}{\alpha}\right) \\ = & -\omega^2 \frac{(-1+1)}{i}=0 \end{aligned} $

$ \begin{aligned} &\text { Given, } z=\alpha+i \beta \text {, }\\ &\begin{array}{ll} |z|-z=1+2 i ;|z|=\sqrt{\alpha^2+\beta^2} \\ \sqrt{\alpha^2+\beta^2}-\alpha-i \beta=1+2 i \\ \Rightarrow \quad \beta=-2 \\ \text { and } \sqrt{\alpha^2+4}-\alpha=1 \end{array} \end{aligned} $

$ \begin{array}{lc} \Rightarrow & \alpha^2+4=(1+\alpha)^2 \\ & \quad[\because \text { on squaring both sides }] \\ \Rightarrow & \alpha^2+4=\alpha^2+1+2 \alpha \\ \Rightarrow & 2 \alpha=3 \\ \Rightarrow & \alpha=\frac{3}{2} \end{array} $

$ \quad \begin{aligned}\therefore\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, z \bar{z} & =|z|^2=\alpha^2+\beta^2 \\ & =\left(\frac{3}{2}\right)^2+(-2)^2 \\ & =\frac{9}{4}+4=\frac{25}{4} \end{aligned} $

Given, $i z^2-2(i+1) z+(2-i)=0$

$\therefore-i$ and $\alpha$ are the roots of above equation.

$ \begin{aligned} & & -i \times \alpha & =\frac{2-i}{i} \\ \Rightarrow & & \alpha & =\frac{2-i}{-i^2} \\ \Rightarrow & & \alpha & =2-i \end{aligned} $

Clearly, $\tan \theta=-\frac{1}{2}, \gamma=\sqrt{5}$

$ \sin \theta=\frac{-1}{\sqrt{5}}, \cos \theta=\frac{2}{\sqrt{5}} $

[ ∵ from the polar form]

$ \begin{aligned} & \cos 2 \theta=\cos ^2 \theta-\sin ^2 \theta=\frac{4}{5}-\frac{1}{5}=\frac{3}{5} \\ & \cos 6 \theta=4 \cos ^3 2 \theta-3 \cos 2 \theta \\ &=4\left(\frac{3}{5}\right)^3-3 \times \frac{3}{5}=\frac{108}{125}-\frac{9}{5} \\ &=\frac{108-225}{125}=-\frac{117}{125} \\ & \therefore \quad 5^3 \cos 6 \theta=5^3 \times-\left(\frac{117}{125}\right)=-117 \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { Let } z^n=1 \\ & z^n-1=(z-1)\left(z-\alpha_1\right)\left(z-\alpha_2\right)\left(z-\alpha_3\right) \\ & \ldots\left(z-\alpha_{n-1}\right) \end{aligned} \end{aligned} $

Sum of $n$th roots of unity $=0$

$ \begin{aligned} 1+\alpha_1+\alpha_2+\alpha_3+\ldots+\alpha_{n-1} & =0 \\ \Rightarrow \quad \alpha_1+\alpha_2+\alpha_3+\ldots+\alpha_{n-1} & =-1 \end{aligned} $

Also, the coefficient of $z^{n-1}$ is zero, so sum of product of zeroes taken two at a time $=0$

$ \begin{aligned} \Rightarrow \alpha_1+\alpha_2+\ldots & +\alpha_{n-1} \\ & +\sum_{1 \leq i < j \leq n-1} \alpha_i \alpha_j=0 \end{aligned} $

$ \Rightarrow \,\,\,\,\, - 1 + \sum\limits_{1 \le i < j \le n - 1}^{} {} {a_i}{a_j} = 0$

$ \Rightarrow \,\,\,\,\,\sum\limits_{1 \le i < j \le n - 1}^{} {} {a_i}{a_j} = 1$

Given, 2- $i$ is one of the roots of the equation

$ x^4-9 x^3+31 x^2-49 x+30=0 $

As coefficient of $x^4, x^3, x^2, x$ and constant are real numbers and $2-i$ is one of the root, then $2+i$ is the another root of the equation.

$ \begin{aligned} & & (x-2-i)(x-2+i) & =0 \\ \Rightarrow & & (x-2)^2-i^2 & =0 \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \Rightarrow \quad x^2-4 x+4+1=0 \\ & \Rightarrow \quad x^2-4 x+5=0 \end{aligned}\\ &\text { Now, } \end{aligned} $

$ \begin{array}{ll} \therefore & x^2-5 x+6=0 \\ \Rightarrow & x^2-3 x-2 x+6=0 \\ \Rightarrow & x(x-3)-2(x-3)=0 \\ \Rightarrow & (x-2)(x-3)=0 \\ \Rightarrow & x=2,3 \\ \therefore & \alpha=2, \beta=3 \\ & 2 \alpha-\beta=2 \times 2-3=1 \end{array} $

Given, $e^{i t}=\cos t+i \sin t$ and $e^{-i t}=\cos t-i \sin t$

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

$ \begin{aligned} = & \frac{1}{2}\left[e^x[\cos y+i \sin y-\cos y+i \sin y]\right. \left.+e^{-x}[\cos y-i \sin y-\cos y-i \sin y]\right] \\ = & \frac{1}{2}\left[e^x(2 i \sin y)+e^{-x}[-2 i \sin y]\right. \\ = & 2 i \sin y\left[\frac{e^x-e^{-x}}{2}\right] \\ = & 2 i \sin y \sinh x \end{aligned} $

$ \begin{aligned} &\text { Given, }\\ &(2 x-y+1)+i(x-2 y-1)=2-3 i \end{aligned} $

$ \begin{array}{llrl} \Rightarrow & & 2 x-y+1 & =2 \\ \text { and } & & x-2 y-1 & =-3 \\ \Rightarrow & & 2 x-y & =1 \\ \text { and } & & x & =2 y-2 \end{array} $

Now, substitute value of $x=2 y-2$ in equation $2 x-y=1$, we get

$ \begin{aligned} & & 2(2 y-2)-y & =1 \\ \Rightarrow & & 4 y-4-y & =1 \end{aligned} $

$ \begin{aligned} &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3 y=5 \Rightarrow y=\frac{5}{3}\\ &\text { Now, substitute } y=\frac{5}{3} \text { in } x=2 y-2\\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x=\frac{10}{3}-2=\frac{4}{3} \end{aligned} $

$ \begin{aligned} &\text { Multiplicative inverse of }(x-i y)=\frac{1}{x-i y}\\ &\begin{aligned} & =\frac{1}{\frac{4}{3}-i \frac{5}{3}}=\frac{3}{4-5 i} \times \frac{(4+5 i)}{(4+5 i)} \\ & =\frac{12}{4^2+5^2}+\frac{15}{4^2+5^2} i=\frac{12}{41}+\frac{15}{41} i \end{aligned} \end{aligned} $

$ \begin{aligned} & (-1)=\cos \pi+i \sin \pi=\operatorname{cis} \pi \\ & \begin{aligned} (-1)^{1 / 4} & =(\cos \pi+i \sin \pi)^{1 / 4} \\ & =\cos \frac{3 \pi}{4}+i \sin \frac{3 \pi}{4} \\ & {\left[\because \cos \frac{\pi}{4} \text { is not }-(\mathrm{ve})\right] } \end{aligned} \end{aligned} $

$ \begin{aligned} & =\operatorname{cis} \frac{3 \pi}{4}(-i)^{1 / 2}=\left(\cos \frac{3 \pi}{2}+i \sin \frac{3 \pi}{2}\right)^{1 / 2} \\ & =\left(\cos \frac{3 \pi}{4}+i \sin \frac{3 \pi}{4}\right)=\operatorname{cis} \frac{3 \pi}{4} \\ \therefore & \alpha=\frac{3 \pi}{4} \Rightarrow \tan \alpha=\tan \frac{3 \pi}{4}=-1 \end{aligned} $

The roots of the given equations are $\sqrt{3}+\sqrt{2} i$ and $\sqrt{3}-\sqrt{2}$.

$ \begin{aligned} \therefore & & x & =\sqrt{3}+\sqrt{2} i \\ \Rightarrow & & x-\sqrt{3} & =\sqrt{2} i \\ \Rightarrow & & (x-\sqrt{3})^2 & =(\sqrt{2} i)^2 \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad x^2+3-2 \sqrt{3} x=-2 \Rightarrow x^2+5=2 \sqrt{3} x \\ & \Rightarrow \quad\left(x^2+5\right)^2=(2 \sqrt{3} x)^2 \\ & \Rightarrow \quad x^4+25+10 x^2=12 x^2 \\ & \Rightarrow \quad x^4-2 x^2+25=0 \end{aligned} $

$ \begin{array}{ll} \text { Also, } & x=\sqrt{3}-\sqrt{2} \\ \Rightarrow & x^2=(\sqrt{3}-\sqrt{2})^2 \\ \Rightarrow & x^2=3+2-2 \sqrt{6} \end{array} $

$ \begin{aligned} &\begin{array}{lr} \Rightarrow & x^2-5=-2 \sqrt{6} \\ \Rightarrow & \left(x^2-5\right)^2=(-2 \sqrt{6})^2 \\ \Rightarrow & x^4+25-10 x^2=24 \\ \Rightarrow & x^4-10 x^2+1=0 \end{array}\\ &\text { ∴ The required equation is }\\ &\left(x^4-2 x^2+25\right)\left(x^4-10 x^2+1\right)=0 \end{aligned} $

$ \begin{aligned} & \text {}\text { }\frac{x+i(x-2)}{3+i}-i=\frac{2 y+i(1-3 y)}{i-3} \\ & \Rightarrow \frac{x+(x-2) i}{3+i}+\frac{2 y+(1-3 y) i}{3-i}=i \end{aligned} $

$ \Rightarrow \frac{\{(4 x-2)+i(2 x-6)\}+\{(9 y-1)+i(3-7 y)\}}{(3+i)(3-i)}=i $

$ \Rightarrow(4 x+9 y-3)+i(2 x-7 y-3) $

$ =0+10 i $

On comparing the real and imaginary parts

$ \begin{aligned} & 4 x+9 y-3=0 \text { and } 2 x-7 y-3=10 \\ & \Rightarrow \quad 4 x+9 y=3 \\ & 2 x-7 y=13 \end{aligned} $

On subtracting Eq. (ii) $\times 2$ from Eq. (i),

From Eq. (i), $x=3$

$ \therefore \quad x+y=2 $

$ \begin{aligned} & z_1=e^{i \alpha}, z_2=e^{i \beta}, z_3=e^{i \gamma} \\ & \frac{1}{z_1}=e^{-i \alpha}, \frac{1}{z_2}=e^{-i \beta}, \frac{1}{z_3}=e^{-i \gamma} \end{aligned} $

According to the questions,

$ z_1+z_2+z_3=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Now, $\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}$

$ \begin{aligned} = & (\cos \alpha-i \sin \alpha)+(\cos \beta-i \sin \beta) +(\cos \gamma-i \sin \gamma) \\ = & \Sigma \cos \alpha-i \Sigma \sin \alpha=0-i \cdot 0=0 \\ \Rightarrow & z_1 z_2+z_2 z_3+z_3 z_1=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ & z_1^2+z_2^2+z_3^2 \\ = & \left(z_1+z_2+z_3\right)^2-2\left(z_1 z_2+z_2 z_3+z_3 z_1\right) \\ \Rightarrow & z_1^2+z_2^2+z_3^2=0-0 \end{aligned} $

[From Eqs. (i) and (ii)]

$ \begin{aligned} & \Rightarrow e^{i 2 \alpha}+e^{i 2 \beta}+e^{i 2 \gamma}=0 \\ & \Rightarrow \cos 2 \alpha+\cos 2 \beta+\cos 2 \gamma=0 \\ & \text { and } \sin 2 \alpha+\sin 2 \beta+\sin 2 \gamma=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii) \end{aligned} $

Now, $z_1 z_2+z_2 z_3+z_3 z_1=0$

$ \begin{aligned} & \Rightarrow \quad e^{i(\alpha+\beta)}+e^{i(\beta+\gamma)}+e^{i(\gamma+\alpha)}=0 \\ & \Rightarrow \quad \Sigma \cos (\alpha+\beta)=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)\end{aligned} $

and $\Sigma \sin (\alpha+\beta)=0$

From Eqs. (iii) and (iv),

$ \begin{aligned} & \cos 2 \alpha+\cos 2 \beta+\cos 2 \gamma \\ & =\cos (\alpha+\beta)+\cos (\beta+\gamma)+\cos (\gamma+\alpha) \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \text {Let } x=(-32 i)^{\frac{2}{5}}=(-1024)^{\frac{1}{5}} \\ & \therefore \quad x^5=-1024 \Rightarrow\left(\frac{-x}{4}\right)^5=1 \\ & \therefore \quad \frac{-x}{4}=\operatorname{cis}\left(\frac{2 n \pi}{5}\right), n=0,1,2,3,4 \end{aligned}\\ &\text { For } n=4 \text {, }\\ &x=-4 \operatorname{cis} \frac{8 \pi}{5}=4 \operatorname{cis} \frac{3 \pi}{5} \end{aligned} $

Given, $\sqrt{(-3+4 i)(8+6 i)}$

$ =\sqrt{-48+14 i}=\sqrt{2} \sqrt{-24+7 i} $

We know that,

$ \begin{aligned} & \sqrt{z}= \pm\left(\sqrt{\frac{|z|+\operatorname{Re}(z)}{2}}+i \sqrt{\frac{|z|-\operatorname{Re}(z)}{2}}\right) \text { for } \\ & \operatorname{Im}(z)>0 \\ & \therefore \sqrt{-24+7 i} \\ & = \pm\left(\sqrt{\frac{25-24}{2}}+i \sqrt{\frac{25+24}{2}}\right) \\ & = \pm \frac{1}{\sqrt{2}}(1+7 i) \\ & \therefore \sqrt{(-3+4 i)(8+6 i)}= \pm(1+7 i) \end{aligned} $

$ \begin{aligned} &\text { We know that, }\\ &\begin{aligned} & \omega=\frac{-1}{2}+i \frac{\sqrt{3}}{2} \text { and } \omega^2=\frac{-1}{2}-i \frac{\sqrt{3}}{2} \\ \Rightarrow & -2 i \omega=\sqrt{3}+i \text { and } 2 i \omega^2=\sqrt{3}-i \\ & \left(\frac{\sqrt{3}+i}{\sqrt{3}-i}\right)^m=1 \end{aligned} \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad\left(\frac{-2 i \omega}{2 i \omega^2}\right)^m=1 \Rightarrow(-\omega)^m=1 \\ & \Rightarrow \quad(-1)^m \omega^m=1 \end{aligned} $

$\Rightarrow m$ is even number and multiple of 3 . Clearly, 2028 is even and multiple of 3.

Here, 1, $\omega$ and $\omega^2$ are the cube root of unity, $n \in N, n>2$

$ \begin{array}{ll} \therefore \quad & 1+\omega+\omega^2=0, \omega^3=1 \\ & x^n-x=0 \rightarrow 1+\omega \text { is a root. } \end{array} $

$ \begin{aligned} & \text { So, }(1+\omega)^n-(1+\omega)=0 \\ & \Rightarrow(1+\omega)^n=(1+\omega) \\ & \quad(1+\omega)^{n-1}=1 \Rightarrow\left(-\omega^2\right)^{n-1}=1 \end{aligned} $

Least value of $n$ must be 7 .

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

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.