Complex Numbers

Given, $\alpha, \beta$ and $\gamma$ are the roots of the equation $x^3+2 x^2-x-2=0$

On comparing the given equation with $a x^3+b x^2+c x+d=0$, we get

$ \begin{aligned} & a=1, b=2, c=-1 \text { and } d=-2 \\ & \therefore \alpha+\beta+\gamma=-\frac{b}{a}=-2 \\ & \alpha \beta \gamma=\frac{-d}{a}=2 \text { and } \alpha \beta+\beta \gamma+\alpha \gamma=\frac{c}{a}=-1 \end{aligned} $

Now, $(\alpha+\beta+\gamma)^2=\alpha^2+\beta^2+\gamma^2+2$

$ \begin{aligned} & \quad(\alpha \beta+\beta \gamma+\alpha \gamma) \\ \Rightarrow & \quad(-2)^2=\alpha^2+\beta^2+\gamma^2+2(-1) \\ \Rightarrow & \alpha^2+\beta^2+\gamma^2=4+2=6 \\ \therefore & \left(\alpha^2+\beta^2+\gamma^2\right)^3=(6)^3 \\ \Rightarrow & \left(\alpha^6+\beta^6+\gamma^6\right)+3\left[\left(\alpha^2+\beta^2+\gamma^2\right)\right. \\ & \left.\quad\left(\alpha^2 \beta^2+\beta^2 \gamma^2+\alpha^2 \gamma^2\right)-\alpha^2 \beta^2 \gamma^2\right]=216 \\ \Rightarrow & \quad \alpha^6+\beta^6+\gamma^6+3[(6)(9)-4]=216 \\ & \quad\left[\because \alpha^2 \beta^2+\beta^2 \gamma^2+\alpha^2 \gamma^2=9\right] \\ \Rightarrow & \alpha^6+\beta^6+\gamma^6=216-150 \\ \Rightarrow & \alpha^6+\beta^6+\gamma^6=66 \end{aligned} $

Given,

$ \begin{aligned} & \frac{3 x+2}{(x+1)\left(2 x^2+3\right)}=\frac{A}{x+1}+\frac{B x+C}{2 x^2+3} \\ & \Rightarrow 3 x+2=A\left(2 x^2+3\right) \\ & +(B x+C)(x+1) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

On putting $x=-1$ into the Eq. (i), we get

$ \begin{aligned} 3(-1)+2=A\left[2(-1)^2+3\right] & \\ & +(B(-1)+C)(-1+1) \\ -1=5 A \Rightarrow A= & -\frac{1}{5} \end{aligned} $

On putting $x=0$ and $A=-\frac{1}{5}$ into the

Eq. (i), we get

$ \begin{aligned} & 3(0)+2=-\frac{1}{5}\left(2(0)^2+3\right)+(B(0)+C)(0+1) \\ & \Rightarrow \quad 2=-\frac{3}{5}+C \Rightarrow C=\frac{13}{5} \end{aligned} $

On putting $x=1, A=-\frac{1}{5}$ and $C=\frac{13}{5}$ into the Eq. (i), we get

$ \begin{aligned} & 3(1)+2=-\frac{1}{5}\left(2(1)^2+3\right)+\left(B(1)+\frac{13}{5}\right)(1+1) \\ & \Rightarrow \quad 5=-1+2 B+\frac{26}{5} \Rightarrow 2 B=6-\frac{26}{5} \\ & \Rightarrow \quad 2 B=\frac{4}{5} \\ & \Rightarrow \quad B=\frac{2}{5} \\ & \therefore A-B+C=-\frac{1}{5}-\frac{2}{5}+\frac{13}{5}=\frac{10}{5}=2 \end{aligned} $

$ \begin{aligned} &\text { We know that, }\\ &\begin{aligned} & \sinh ^{-1} x=\log \left(x+\sqrt{x^2+1}\right) \\ & \text { So, } \sinh ^{-1} y=\log \left(y+\sqrt{y^2+1}\right) \\ & \Rightarrow \sinh ^{-1} y=x \\ & \quad \sinh x=y \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { Here, }\\ &\begin{aligned} 2^{2022}\left(\frac{1}{2}\right. & \left.+\frac{\sqrt{3}}{2} i\right)^{2022}-2^{2022}\left(\frac{\sqrt{3}}{2}-\frac{i}{2}\right)^{2022} \\ & =2^{2022} e^{i\left(\frac{\pi}{3}\right)(2022)}-2^{2022} e^{i\left(-\frac{\pi}{6}\right)(2022)} \\ & =2^{2022} e^{i(674) \pi}-2^{2022} e^{i(-337) \pi} \\ & =2^{2022}(1+0 i)-2^{2022}(-1+0 i) \\ & =2^{2022}+2^{2022}=2^{2023} \end{aligned} \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { Given, }\left(\frac{\sqrt{3}+i}{\sqrt{3}-i}\right)^4+\left(\frac{\sqrt{3}-i}{\sqrt{3}+i}\right)^4=r \text { cis } \theta \\ & \Rightarrow \quad \frac{(\sqrt{3}+i)^8+(\sqrt{3}-i)^8}{[(\sqrt{3}-i)(\sqrt{3}+i)]^4}=r \text { cis } \theta \end{aligned} \end{aligned} $

$ \begin{aligned} & \Rightarrow \frac{\left(2 \operatorname{cis} \frac{\pi}{6}\right)^8+\left(2 \operatorname{cis}\left(-\frac{\pi}{6}\right)\right)^8}{2^8}=r \operatorname{cis} \theta \\ & \begin{aligned} & \Rightarrow \operatorname{cis} \frac{8 \pi}{6}+\operatorname{cis}\left(-\frac{8 \pi}{6}\right)=r \operatorname{cis} \theta \\ & \Rightarrow \cos \frac{8 \pi}{6}+i \sin \frac{8 \pi}{6}+\cos \left(-\frac{8 \pi}{6}\right) +i \sin \left(-\frac{8 \pi}{6}\right)=r \operatorname{cis} \theta \end{aligned} \\ & \Rightarrow 2 \cos \left(\frac{8 \pi}{6}\right)=r \operatorname{cis} \theta \Rightarrow 2\left(-\frac{1}{2}\right)=r \operatorname{cis} \theta \\ & \Rightarrow-1=r \operatorname{cis} \theta \Rightarrow \pm \sqrt{-1}=\sqrt{r \operatorname{cis} \theta} \\ & \Rightarrow \sqrt{r \operatorname{cis} \theta}= \pm i \\ & \Rightarrow \sqrt{r \operatorname{cis} \theta}=\operatorname{cis}\left(\frac{\pi}{2}\right), \operatorname{cis}\left(\frac{3 \pi}{2}\right) \end{aligned} $

$ \begin{aligned} &\text { Given equation, }\\ &\begin{aligned} & \quad|z-2|+|z-2 i|=4 \\ & \Rightarrow \quad|(x+i y)-2|+|(x+i y)-2 i|=4 \\ & \Rightarrow \sqrt{(x-2)^2+y^2}+\sqrt{x^2+(y-2)^2}=4 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \\ & \text { Let, } l=(x-2)^2+y^2=x^2+y^2-4 x+4 \\ & \text { and } m=x^2+(y-2)^2=x^2+y^2-4 y+4 \end{aligned} \end{aligned} $

Eq. (i), can be written as

$ \sqrt{l}+\sqrt{m}=4 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Now, $l-m=4 y-4 x$

$ \begin{aligned} \Rightarrow & & (\sqrt{l}+\sqrt{m})(\sqrt{l}-\sqrt{m}) & =4(y-x) \\ \Rightarrow & & 4(\sqrt{l}-\sqrt{m}) & =4(y-x) \\ & \Rightarrow & (\sqrt{l}-\sqrt{m}) & =y-x \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{aligned} $

On adding Eqs. (ii) and (iii), we get

$ 2 \sqrt{1}=y-x+4 $

On squaring both sides, we get

$ \begin{gathered} 4 l=(y-x+4)^2 \\ \Rightarrow 4\left(x^2+y^2-4 x+4\right)=y^2+x^2+4^2 -2 x y-8 x+8 y \\ \Rightarrow 3 x^2+2 x y+3 y^2-8 x-8 y=0 \end{gathered} $

$ \begin{aligned} & \text { Given, }(\sqrt{3}-i)^{2 / 5}=\left[2\left(\frac{\sqrt{3}}{2}-\frac{i}{2}\right)\right]^{2 / 5} \\ & =2^{2 / 5}\left[\cos \left(-\frac{\pi}{6}\right)+i \sin \left(-\frac{\pi}{6}\right)\right]^{2 / 5} \\ & =2^{2 / 5}\left[\cos \left(-\frac{\pi}{6}+6 \pi\right)+i \sin \left(-\frac{\pi}{6}+6 \pi\right)\right]^{2 / 5} \\ & =2^{2 / 5}\left[\cos \left(\frac{35 \pi}{6}\right)+i \sin \left(\frac{35 \pi}{6}\right)\right]^{2 / 5} \\ & =2^{2 / 5}\left[\left(\cos \frac{35 \pi}{6} \times \frac{2}{5}\right)+i \sin \left(\frac{35 \pi}{6} \times \frac{2}{5}\right)\right] \\ & =2^{2 / 5}\left[\cos \left(\frac{7 \pi}{3}\right)+i \sin \left(\frac{7 \pi}{3}\right)\right] \\ & =2^{2 / 5}\left(\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}\right)=2^{2 / 5}\left[\frac{1}{2}+i\left(\frac{\sqrt{3}}{2}\right)\right] \\ & =\frac{2^{2 / 5}}{2}(1+\sqrt{3} i)=2^{-3 / 5}(1+\sqrt{3 i}) \end{aligned} $

Given, $\alpha, \beta, \gamma$ and $\delta$ are the roots of the equation $x^4+x^2+1=0$

Also, given that, $\alpha+\beta=-1, \gamma+\delta=1$, $\alpha^2=\beta$ and $\gamma^2=-\delta$

We know that, roots of the equation $x^4+x^2+1=0$ are $\pm \omega^{1 / 2}$ and $\pm \omega$

Now, it can be conclude that

$ \begin{aligned} & \alpha=-\omega^{1 / 2}, \beta=\omega, \gamma=\omega^{1 / 2} \text { and } \delta=-\omega \\ & \therefore \alpha^{2023}+\beta^{2023}+\gamma^{2022}+\delta^{2022} \\ & =\left(-\omega^{1 / 2}\right)^{2023}+(\omega)^{2023}+\left(\omega^{1 / 2}\right)^{2022}+(-\omega)^{2022} \\ & =-\omega^{1011} \cdot \omega^{1 / 2}+\omega+1+1 \\ & \begin{array}{l} =-(1) \cdot \omega^{1 / 2}+2+\omega=-\omega^{1 / 2}+2+\omega \\ =\omega^2+\omega+2 \quad\left[\because-\omega^{1 / 2}=\omega^2\right] \\ =1 \quad\left[\because 1+\omega+\omega^2=0\right]\\ \\ \end{array} \end{aligned} $

$ \begin{aligned} & (x-2)(x+3)(2 x-1)=0 \\ & \alpha=2, \beta=-3 \text { and } \gamma=\frac{1}{2} \\ & \therefore \alpha^3+\beta^3+\gamma^3=8-27+\frac{1}{8} \\ & \quad=\frac{64-216+1}{8}=-\frac{151}{8} \end{aligned} $

Given, $\alpha, \beta$ and $\gamma$ are the real roots of the equation $18 x^3-15 x^2-4 x+4=0$ such that $\alpha=\beta$ and $\alpha>\gamma$

So, $\alpha+\beta+\gamma=-\frac{(-19)}{18} \Rightarrow 2 \alpha+\gamma=\frac{5}{6}$

$ \begin{array}{ll} \text { and } & \alpha \beta+\beta \gamma+\gamma \alpha \\ \Rightarrow & \alpha^2+2 \alpha \gamma=-\frac{4}{18} \\ \Rightarrow & \alpha^2+2 \alpha\left(\frac{5}{6}-2 \alpha\right)=-\frac{2}{9} \\ \Rightarrow & 27 \alpha^2-15 \alpha-2=0 \end{array} $

$ \Rightarrow \quad \alpha=-\frac{1}{9}, \frac{2}{3} $

If $\alpha=-\frac{1}{9}$, then $\gamma=\frac{5}{6}-2\left(-\frac{1}{9}\right)=\frac{19}{18}$

But $\alpha>\gamma$.

So, $\alpha \neq-\frac{1}{9}$.

If $\alpha=\frac{2}{3}$, then $\gamma=\frac{5}{6}-2\left(\frac{2}{3}\right)=-\frac{1}{2}$

$ \begin{aligned} \therefore \alpha+\beta^2 & +\gamma^3 \\ & =\frac{2}{3}+\left(\frac{2}{3}\right)^2+\left(-\frac{1}{2}\right)^3 \\ & =\frac{2}{3}+\frac{4}{9}-\frac{1}{8}=\frac{71}{72} \end{aligned} $

Given, $\alpha$ is a multiple root of the equation

$ x^5-6 x^4+11 x^3-2 x^2-12 x+8=0 $

The possible rational roots of this equation be $\pm 1, \pm 2, \pm 4$ and $\pm 8$

Let $f(x)=x^5-6 x^4+11 x^3-2 x^2-12 x+8$

$ \begin{aligned} &\text { Now, }\\ &\begin{array}{|c|c|c|c|} \hline \boldsymbol{x} & \boldsymbol{f}(\boldsymbol{x}) & \boldsymbol{x} & \boldsymbol{f}(\boldsymbol{x}) \\ \hline-1 & 0 & -4 & -3240 \\ \hline+1 & 0 & 4 & 120 \\ \hline-2 & -192 & -8 & -63000 \\ \hline+2 & 0 & 8 & 13608 \\ \hline \end{array} \end{aligned} $

Thus, $-1,1$ and 2 be three roots of

$ x^5-6 x^4+11 x^3-2 x^2-12 x+8=0 $

So, $(x+1)(x-1)(x-2)$ or $x^3-2 x^2-x+2$

be a factor of

$ \begin{aligned} & x^5-6 x^4+11 x^3-2 x^2-12 x+8 \\ & \text { Now, } x^5-6 x^4+11 x^3-2 x^2-12 x+8 \\ & \div x^3-2 x^2-x+2=x^2-4 x+4 \end{aligned} $

The roots of equation $x^2-4 x+4=0$ are 2, 2.

Thus, the roots of the equation

$ x^5-6 x^4+11 x^3-2 x^2-12 x+8=0 $

be, $-1,1,2,2$ and 2 .

So, $\alpha=2$

$ \begin{aligned} \therefore 3 \alpha^2-2 \alpha+1 & =3(2)^2-2(2)+1 \\ & =12-4+1=9 \end{aligned} $

$ \text { } \begin{aligned} & 3^{2023}=3^{4 \times 505+3}=3^3\left(3^4\right)^{505} \\ = & 27\left[(1+80)^{505}\right]=27(1+80 K) \\ = & 27+27 \times 80 K=11+16+16 \lambda \\ = & 11+16(\lambda+1) \end{aligned} $

∴ Required remainder $=11$

$ \sqrt{-5-12 i}= \pm\binom{\sqrt{\frac{\sqrt{(-5)^2+(-12)^2}+(-5)}{2}}-i}{\sqrt{\frac{\sqrt{(-5)^2+(-12)^2}-(-5)}{2}}} $

$ = \pm(\sqrt{4}-i \sqrt{9})= \pm(2-3 i) $

$ \begin{aligned} \sqrt{7+24 i}= \pm\left(\sqrt{\frac{\sqrt{7^2+24^2}+7}{2}}\right. & \left.+i \sqrt{\frac{\sqrt{7^2+24^2}-7}{2}}\right) \end{aligned} $

$ \begin{aligned} &\begin{array}{ll} & = \pm(\sqrt{16}+i \sqrt{9})= \pm(4+3 i) \\ \because & \sqrt{-5-12 i}+\sqrt{7+24 i}=k \end{array}\\ &\text { (a negative real number) }\\ &\begin{array}{lrl} \Rightarrow & -(2-3 i)-(4+3 i) & =k \\ \Rightarrow & k & =-6 \end{array} \end{aligned} $

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

$ \begin{aligned} & \Rightarrow \mathrm{amp}\left(\frac{(x-3)+i y}{x+i(y+2)} \times \frac{x-i(y+2)}{x-i(y+2)}\right)=\frac{\pi}{2} \\ & \Rightarrow \mathrm{amp} \\ & \begin{array}{r} \left(\frac{x(x-3)+y(y+2)+i(x y-(x-3)(y+2)}{x^2+(y+2)^2}\right) =\frac{\pi}{2} \end{array} \end{aligned} $

Here, $\operatorname{amp}(z)=\frac{\pi}{2}$ means $z$ is purely imaginary

$ \begin{array}{ll} \Rightarrow & \text { Real part }=0 \\ \therefore & x(x-3)+y(y+2)=0 \\ & x^2-3 x+y^2+2 y=0 \\ \Rightarrow & x^2+y^2-3 x+2 y=0 \end{array} $

which represents a circle containing the origin.

$ \begin{array}{r} \frac{z-(2+i)}{z+(1-2 i)}=\frac{(x-2)+i(y-1)}{(x+1)+i(y-2)} \times \frac{(x+1)-i(y-2)}{(x+1)-i(y-2)} \end{array} $

$ =\frac{\begin{array}{l} (x-2)(x+1)+(y-1)(y-2) \\ +i[(x+1)(y-1)-(x-2)(y-2)] \end{array}}{(x+1)^2+(y-2)^2} $

$ \begin{aligned} & \because \text { This is purely real. } \\ & \Rightarrow \text { Imaginary point }=0 \\ & \Rightarrow(x+1)(y-1)-(x-2)(y-2)=0 \\ & \qquad[x \neq-1, y \neq 2] \end{aligned} $

$ \begin{aligned} & \Rightarrow x y-x+y-1=x y-2 x-2 y+4 \\ & \Rightarrow-x+2 x+y+2 y-1-4=0 \\ & \Rightarrow x+3 y-5=0, \quad[x \neq-1, y \neq 2] \end{aligned} $

We know that $1, \omega$ and $\omega^2$ are cube roots of unit, where

$ \begin{aligned} & \omega=\frac{-1+i \sqrt{3}}{2} \\ & \omega^2=\frac{-1-i \sqrt{3}}{2} \end{aligned} $

and $\quad \omega^3=1$

$ \begin{array}{lrl} \therefore -2 \omega & =1-i \sqrt{3} \\ \text { and }-2 \omega^2 & =1+i \sqrt{3} \end{array} $

Now, $(1+i \sqrt{3})^{2023}+(1-\sqrt{3} i)^{2023}$

$ \begin{aligned} & =\left(-2 \omega^2\right)^{2023}+(-2 \omega)^{2023} \\ & =(-2)^{2023}\left(\omega^{4046}+\omega^{2023}\right) \\ & =(-2)^{2023}\left(\omega^2+\omega\right) \quad\left[\because \omega^{3 n}=1\right] \\ & =(-2)^{2023}(-1) \quad\left[\omega+\omega^2=-1\right] \\ & =2^{2023} \end{aligned} $

Let $z=\sqrt{3}-i$

$ \begin{aligned} r & =|z|=\sqrt{3+1}=2 \\ \theta & =\arg (z)=-\tan ^{-1} \frac{1}{\sqrt{3}}=-\frac{\pi}{6} \\ \therefore \sqrt{3}-i & =2 \operatorname{cis}\left(-\frac{\pi}{6}\right) \\ \sqrt{3}-i & =2 \operatorname{cis}\left(2 n \pi-\frac{\pi}{6}\right) \\ (\sqrt{3}-i)^{1 / 6} & =2^{1 / 6} \operatorname{cis}\left(\frac{2 n \pi}{6}-\frac{\pi}{36}\right) \\ n & =0,1,2,3,4,5 \\ & =2^{1 / 6} \operatorname{cis} \frac{(12 n-1) \pi}{36} \end{aligned} $

For $n=5$ one of the value is $2^{1 / 6} \mathrm{cis}\left(\frac{59 \pi}{36}\right)$.

On comparing,

$ a x^2-x y-3 y^2-5 x+20 y+c=0 $

with $A x^2+2 H x y+B y^2+2 G x+2 F y+C=0$

we have,

$ \begin{aligned} & A=a, H=-\frac{1}{2}, B=-3 \\ & G=\frac{-5}{2}, F=10, C=c \end{aligned} $

Given, equation represents a pair of straight line, if

$ \begin{gathered} \Delta=\left|\begin{array}{rrr} A & H & G \\ H & B & F \\ G & F & C \end{array}\right|=0 \\ \Rightarrow \quad\left|\begin{array}{rrr} a & -\frac{1}{2} & -\frac{5}{2} \\ -\frac{1}{2} & -3 & 10 \\ -\frac{5}{2} & 10 & c \end{array}\right|=0 \\ \Rightarrow \quad-12 a c-400 a-c+175=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{gathered} $

Given, pair of line passing through $(2,3)$, then

$ 4 a+c+17=0 \Rightarrow c=-4 a-17 $

On substituting value of $c$ in Eq. (i), we get

$ \begin{aligned} & \quad a^2-4 a+4=0 \Rightarrow a=2 \\ & \therefore \quad c=-25 \\ & \text { Now, } a-c=27 \end{aligned} $

Given that

$ \left(z=\sin \left(\frac{6 \pi}{5}\right)+i\left(1+\cos \frac{6 \pi}{5}\right)\right) $

Thus, clearly $\left(\sin \frac{6 \pi}{5}, 1+\left(\cos \frac{6 \pi}{5}\right)\right)$ lies in the 2 nd quadrant of complex plane. Hence, its argument is

$ \begin{aligned} & \arg (z)=\pi-\tan ^{-1}\left|\left(\frac{y}{z}\right)\right| \\ & \quad\left(\because z=x+i y, \forall x<0 \text { and } y^{}\right. \geqslant 0)\end{aligned} $

$ \begin{aligned} &\begin{aligned} & =\pi-\tan ^{-1}\left|\frac{1+\cos \frac{6 \pi}{5}}{\sin \frac{6 \pi}{5}}\right| \\ & =\pi-\tan ^{-1}\left|\frac{1-\cos \frac{\pi}{5}}{\sin \frac{\pi}{5}}\right| \\ & =\pi-\tan ^{-1}\left|\frac{2 \sin ^2 \frac{\pi}{10}}{2 \sin \frac{\pi}{10} \cos \frac{\pi}{10}}\right| \\ & =\pi-\tan ^{-1}\left|\frac{\sin \frac{\pi}{10}}{\cos \frac{\pi}{10}}\right| \\ & =\pi-\tan ^{-1}\left(\tan \frac{\pi}{10}\right) \\ & =\pi-\frac{\pi}{10}\left(\because-\frac{\pi}{2} \leq \tan ^{-1} x \leq \frac{1}{2}\right) \\ & =\frac{10 \pi-\pi}{10}=\frac{9 \pi}{10} \end{aligned}\\ &\text { Thus, } \arg (z)=\frac{9 \pi}{10} \end{aligned} $

Given,

$ \begin{aligned} & x+i y =\sqrt{\frac{3+i}{1+3 i}} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ \Rightarrow & \quad \overline{x+i y} =\sqrt{\frac{\overline{3+i}}{1+3 i}} \\ \Rightarrow & x-i y =\sqrt{\frac{3-i}{1-3 i}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \end{aligned} $

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

$ \begin{aligned} & x^2+y^2=\sqrt{\frac{(3+i)(3-i)}{(1+3 i)(1-3 i)}}=\sqrt{\frac{9+1}{1+9}}=1 \\ \Rightarrow & \left(x^2+y^2\right)^2=(1)^2=1 \end{aligned} $

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

Now, let $z=x+i y$

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

$ =\frac{\begin{array}{r} (2 x+1)(1-y)+i\left(2 y-2 y^2\right) -i\left(2 x^2+x\right)+2 x y \end{array}}{(1-y)^2+x^2} $

From Eq. (i), we get

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

which is a straight line.

If $ i = \sqrt{-1} $, then consider evaluating $ (1+i)^{10} + (1-i)^{10} $.

Solution

We know that $ i^2 = -1 $.

First, let's compute $ (1+i)^2 $ and $ (1-i)^2 $:

Calculate $ (1+i)^2 $:

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

Calculate $ (1-i)^2 $:

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

Now, raise each result to the 5th power:

$ \{(1+i)^2\}^5 = (2i)^5 $

$ (2i)^5 = 2^5 \cdot i^5 = 32 \cdot i^5 $

Since $ i^5 = i^4 \cdot i = 1 \cdot i = i $, it follows that:

$ (2i)^5 = 32i $

$ \{(1-i)^2\}^5 = (-2i)^5 $

$ (-2i)^5 = (-2)^5 \cdot i^5 = -32 \cdot i^5 $

Similarly, $ i^5 = i $, thus:

$ (-2i)^5 = -32i $

Finally, sum these results:

$ (1+i)^{10} + (1-i)^{10} = 32i - 32i = 0 $

Thus, the final answer is $ 0 $.

To solve the problem, we have the condition that the magnitude of the sum of two complex numbers, $ z_1 $ and $ z_2 $, equals the sum of their magnitudes:

$ \left| z_1 + z_2 \right| = \left| z_1 \right| + \left| z_2 \right| $

We start by squaring both sides of this equation:

$ \begin{aligned} & \left( \left| z_1 + z_2 \right| \right)^2 = \left( \left| z_1 \right| + \left| z_2 \right| \right)^2 \\ & \left| z_1 \right|^2 + \left| z_2 \right|^2 + 2 \left| z_1 \right| \left| z_2 \right| \cos(\theta_1 - \theta_2) = \left| z_1 \right|^2 + \left| z_2 \right|^2 + 2 \left| z_1 \right| \left| z_2 \right| \end{aligned} $

Upon simplifying, we have:

$ 2 \left| z_1 \right| \left| z_2 \right| \cos(\theta_1 - \theta_2) = 2 \left| z_1 \right| \left| z_2 \right| $

This implies:

$ \cos(\theta_1 - \theta_2) = 1 $

The condition $\cos(\theta_1 - \theta_2) = 1$ indicates that:

$ \theta_1 - \theta_2 = 0 $

Therefore, the difference in amplitude (or argument) between $ z_1 $ and $ z_2 $ is:

$ \arg(z_1) - \arg(z_2) = 0 $

Thus, the two complex numbers have the same direction or angle in the complex plane.

Given, $1+i^2+i^4+i^6+\ldots+i^{2024}$

$ \begin{aligned} & =1+\left(i^2+i^4+i^6+\ldots+1012 \text { terms }\right) \\ & =1+(-1+1-1+1-\ldots+1-1+1) \end{aligned} $

i.e. 506 terms will be +1 and 506 terms will be $-1=1+0=1$

To solve the problem, let's first express the given complex number:

$ z = \frac{1 + i \cos \theta}{1 - 2 i \cos \theta} $

To simplify this expression and ensure it is purely real, we'll multiply the numerator and denominator by the conjugate of the denominator:

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

Recall that multiplying by the conjugate will result in a real denominator:

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

Simplifying the denominator, since $(2i \cos \theta)^2 = -4 \cos^2 \theta$, we get:

$ 1 - (2i \cos \theta)^2 = 1 + 4 \cos^2 \theta $

Thus, the expression for $ z $ becomes:

$ z = \frac{1 + 3i \cos \theta - 2 \cos^2 \theta}{1 + 4 \cos^2 \theta} $

For $ z $ to be purely real, the imaginary part must be zero. The imaginary part is:

$ \operatorname{Im}(z) = \frac{3 \cos \theta}{1 + 4 \cos^2 \theta} = 0 $

This implies that $ \cos \theta = 0 $.

Substituting $ \theta = \frac{\pi}{2} $, we note:

$\cos \theta = 0$

$\sin \theta = 1$

Now, substitute these into the expression $ \cos^3 \theta + \sin^2 \theta + \cos \theta + 1 $:

$ \cos^3 \theta + \sin^2 \theta + \cos \theta + 1 = 0^3 + 1^2 + 0 + 1 = 2 $

Therefore, the final result is 2.

$T_{10}=(\cos \theta+i \sin \theta)^9$

Using De-Moivre theorem,

$ \begin{aligned} & \cos (9 \theta)+i \sin (9 \theta) \\ = & \cos \left(\frac{9 \pi}{6}\right)+i \sin \left(\frac{9 \pi}{6}\right) \\ = & \cos \left(\frac{3 \pi}{2}\right)+i \sin \left(\frac{3 \pi}{2}\right) \\ = & 0+i(-1) \\ = & -i \end{aligned} $

To find the minimum value of $|z|^2$ where $z=(\alpha+i\beta)(2+7i)$ is purely imaginary, follow these steps:

First, express $z$ in terms of real and imaginary components:

$ z = (\alpha + i \beta)(2 + 7i) = 2\alpha + 7\alpha i + 2\beta i - 7\beta = (2\alpha - 7\beta) + (7\alpha + 2\beta)i $

Since $z$ is purely imaginary, its real part must be zero:

$ 2\alpha - 7\beta = 0 \implies \alpha = \frac{7}{2} \beta $

Given that $\alpha$ and $\beta$ are integers, $\beta$ must be such that $\alpha$ is also an integer. Therefore, $\beta$ should be chosen so that $\frac{7}{2}\beta$ is an integer. This implies $\beta$ must be even, so assume $\beta = 2k$ for some integer $k$.

Substituting $\beta = 2k$ into $\alpha = \frac{7}{2}\beta$, we get $\alpha = 7k$.

Next, calculate $|z|^2$:

$ |z|^2 = (7\alpha + 2\beta)^2 = (7(7k) + 2(2k))^2 = (49k + 4k)^2 = (53k)^2 $

To minimize $|z|^2$, we minimize $k$. The smallest positive integer value for $k$ is 1. Therefore, when $k = 1$, $|z|^2 = 53^2 = 2809$.

Thus, the minimum value of $|z|^2$ is:

$ \boxed{2809} $

We know,

$\left| {|{z_1}| - |{z_2}|} \right| \le \left| {{z_1} + {z_2}} \right| \le |{z_1}| + |{z_2}|$

$\therefore$ $\left| {|z| - {1 \over {|z|}}} \right| \le \left| {z - {1 \over z}} \right|$

$ \Rightarrow \left| {|z| - {1 \over {|z|}}} \right| \le 2$ [Given $\left| {z - {1 \over z}} \right| = 2$]

$ \Rightarrow \left| {{{|z{|^2} - 1} \over {|z|}}} \right| \le 2$

$ \Rightarrow - 2 \le {{|z{|^2} - 1} \over {|z|}} \le 2$

$\therefore$ ${{|z{|^2} - 1} \over {|z|}} \le 2$

$ \Rightarrow |z{|^2} - 1 \le 2|z|$

$ \Rightarrow |z{|^2} - 2|z| - 1 \le 0$

$ \Rightarrow |z{|^2} - 2|z| + 1 - 2 \le 0$

$ \Rightarrow {(|z| - 1)^2} - 2 \le 0$

$ \Rightarrow - \sqrt 2 \le |z| - 1 \le \sqrt 2 $

$ \Rightarrow 1 - \sqrt 2 \le |z| \le 1 + \sqrt 2 $ ..... (1)

or

$ - 2 \le {{|z{|^2} - 1} \over {|z|}}$

$ \Rightarrow |z{|^2} - 1 \le - 2|z|$

$ \Rightarrow |z{|^2} + 2|z| - 1 \le 0$

$ \Rightarrow |z{|^2} + 2|z| + 1 - 2 \le 0$

$ \Rightarrow {(|z| + 1)^2} - 2 \le 0$

$ \Rightarrow - \sqrt 2 \le |z| + 1 \le + \,\sqrt 2 $

$ \Rightarrow - \sqrt 2 - 1 \le |z| \le \sqrt 2 - 1$ ...... (2)

From (1) and (2) we get,

Maximum value of $|z| = \sqrt 2 + 1$ and minimum value of $|z| = - \sqrt 2 - 1$

$z = (2 + 3i)$

$ \Rightarrow {z^5} = (2 + 3i){\left( {{{(2 + 3i)}^2}} \right)^2}$

$ = (2 + 3i){( - 5 + 12i)^2}$

$ = (2 + 3i)( - 119 - 120i)$

$ = - 238 - 240i - 357i + 360$

$ = 122 - 597i$

${\overline z ^5} = 122 + 597i$

${z^5} + {\overline z ^5} = 244$

$\because$ ${\left| {{Z_2} + |{Z_2} - 1|} \right|^2} = {\left| {{Z_2} - |{Z_2} + 1|} \right|^2}$

$ \Rightarrow \left( {{Z_2} + |{Z_2} - 1|} \right)\left( {{{\overline Z }_2} + |{Z_2} - 1|} \right) = \left( {{Z_2} - |{Z_2} + 1|} \right)\left( {{{\overline Z }_2} - |{Z_2} + 1|} \right)$

$ \Rightarrow {Z_2}\left( {|{Z_2} - 1| + |{Z_2} + 1|} \right) + {\overline Z _2}\left( {|{Z_2} - 1| + |{Z_2} + 1|} \right) = |{Z_2} + 1{|^2} - |{Z_2} - 1{|^2}$

$ \Rightarrow \left( {{Z_2} + {{\overline Z }_2}} \right)\left( {|{Z_2} + 1| + |{Z_2} - 1|} \right) = 2\left( {{Z_2} + {{\overline Z }_2}} \right)$

$\Rightarrow$ Either ${Z_2} + {\overline Z _2} = 0$ or $|{Z_2} + 1| + |{Z_2} - 1| = 2$

So, Z₂ lies on imaginary axis or on real axis within $[ - 1,1]$

Also $|{Z_1} - 3| = {1 \over 2} \Rightarrow {Z_1}$ lies on the circle having center 3 and radius ${1 \over 2}$.

Clearly $|{Z_1} - {Z_2}{|_{\min }} = {3 \over 2}$

$\because$ ${{1 - i\sin \alpha } \over {1 + 2i\sin \alpha }}$ is purely imaginary

$\therefore$ ${{1 - i\sin \alpha } \over {1 + 2i\sin \alpha }} + {{1 + i\sin \alpha } \over {1 - 2i\sin \alpha }} = 0$

$ \Rightarrow 1 - 2{\sin ^2}\alpha = 0$

$\therefore$ $\alpha = {{5\pi } \over 4},\,{{7\pi } \over 4}$

and ${{1 + i\cos \beta } \over {1 - 2i\cos \beta }}$ is purely real

${{1 + i\cos \beta } \over {1 - 2i\cos \beta }} - {{1 - i\cos \beta } \over {1 + 2i\cos \beta }} = 0$

$ \Rightarrow \cos \beta = 0$

$\therefore$ $\beta = {{3\pi } \over 2}$

$\therefore$ $S = \left\{ {\left( {{{5\pi } \over 2},{{3\pi } \over 2}} \right),\left( {{{7\pi } \over 4},{{3\pi } \over 2}} \right)} \right\}$

${Z_{\alpha \beta }} = 1 - i$ and ${Z_{\alpha \beta }} = - 1 - i$

$\therefore$ $\sum\limits_{(\alpha ,\beta ) \in S} {\left( {i{Z_{\alpha \beta }} + {1 \over {i{{\overline Z }_{\alpha \beta }}}}} \right) = i( - 2i) + {1 \over i}\left[ {{1 \over {1 + i}} + {1 \over { - 1 + i}}} \right]} $

$ = 2 + {1 \over i}{{2i} \over { - 2}} = 1$

Let $z = x + iy$

$v = {x^2} + {y^2} + {(x - 3)^2} + {y^2} + {x^2} + {(y - 6)^2}$

$ = (3{x^2} - 6x + 9) + (3{y^2} - 12y + 36)$

$ = 3({x^2} + {y^2} - 2x - 4y + 15)$

$ = 3[{(x - 1)^2} + {(y - 2)^2} + 10]$

${v_{\min }}$ at $z = 1 + 2i = {z_0}$ and ${v_0} = 30$

so $|2{(1 + 2i)^2} - {(1 - 2i)^3} + 3{|^2} + 900$

$ = |2( - 3 + 4i) - (1 - 8{i^3} - 6i(1 - 2i) + 3{|^2} + 900$

$ = | - 6 + 8i - (1 + 8i - 6i - 12) + 3{|^2} + 900$

$ = |8 + 6i{|^2} + 900$

$ = 1000$

$|z - i| = |z + 5i|$

So, $\mathrm{z}$ lies on ${ \bot ^r}$ bisector of $(0,1)$ and $(0, - 5)$

i.e., line $y = - 2$

as $|z| = 2$

$ \Rightarrow z = - 2i$

$x = 0$ and $y = - 2$

so, $x + 2y + 4 = 0$

${{{z_2} - 0} \over {(1 + 2i) - 0}} = {{|OB|} \over {|OA|}}{e^{{{i\pi } \over 4}}}$

$ \Rightarrow {{{z_2}} \over {1 + 2i}} = \sqrt 2 {e^{{{i\pi } \over 4}}}$

OR ${z_2} = (1 + 2i)(1 + i)$

$ = - 1 + 3i$

$\arg {z_2} = \pi - {\tan ^{ - 1}}3$

$|{z_2}| = \sqrt {10} $

${z_1} - 2{z_2} = (1 + 2i) + 2 - 6i = 3 - 4i$

$\arg ({z_1} - 2{z_2}) = - {\tan ^{ - 1}}{4 \over 3}$

$|2{z_1} - {z_2}| = |2 + 4i + 1 - 3i| = |3 + i| = \sqrt {10} $

It is sum of distance of z from $\left( {3\sqrt 2 ,0} \right)$ and $\left( {0,p\sqrt 2 } \right)$

For minimising, z should lie on AB and $AB = 5\sqrt 2 $

${(AB)^2} = 18 + 2{p^2}$

$p = \, \pm \,4$

Given,

${{{{(1 + 2i)}^8}\,.\,{{(1 - 2i)}^2}} \over {(3 + 2i)\,.\,\overline {(4 - 6i)} }}$

$ = {{{{(1 + 2i)}^2}{{(1 - 2i)}^2}{{(1 + 2i)}^6}} \over {(3 + 2i)(4 + 6i)}}$

$ = {{{{(1 - 4{i^2})}^2}{{(1 + 2i)}^6}} \over {12 + 18i + 8i + 12{i^2}}}$

$ = {{{{(1 + 5)}^2}{{\left[ {{{(1 + 2i)}^2}} \right]}^3}} \over {12 + 26i - 12}}$

$ = {{25{{(1 + 4{i^2} + 4i)}^3}} \over {26i}}$

$ = {{25{{(1 - 4 + 4i)}^3}} \over {26i}}$

$ = {{25{{( - 3 + 4i)}^3}} \over {26i}}$

$ = {{25} \over {26i}}\left[ {{{( - 3)}^3} + {{(4i)}^3} + 3\,.\,{{( - 3)}^2}\,.\,4i + 3( - 3)\,.\,{{(4i)}^2}} \right]$

$ = {{25} \over {26i}}( - 27 - 64i + 108i + 144)$

$ = {{25} \over {26i}}(117 + 44i)$

$ = {{25i} \over {26{i^2}}}(117 + 44i)$

$ = {{25i} \over { - 26}}(117 + 44i)$

$ = {{25 \times 117i} \over { - 26}} - {{25 \times 44{i^2}} \over {26}}$

$ = {{25 \times 117i} \over { - 26}} + {{22 \times 25} \over {13}}$

$ = {{25 \times 117i} \over { - 26}} + {{550} \over {13}}$

$\therefore$ Real part $ = {{550} \over {13}}$

Let $z = x + iy$

$\therefore$ $|z| = \sqrt {{x^2} + {y^2}} $

Given, $|z| = 3$

$\therefore$ $\sqrt {{x^2} + {y^2}} = 3$

$ \Rightarrow {x^2} + {y^2} = 9 = {3^2}$

This represent a circle with center at (0, 0) and radius = 3

Now, given

$\arg (z - 1) - \arg (z + 1) = {\pi \over 4}$

$ \Rightarrow \arg (x + iy - 1) - \arg (x + iy + 1) = {\pi \over 4}$

$ \Rightarrow \arg (x - 1 + iy) - \arg (x + 1 + iy) = {\pi \over 4}$

$ \Rightarrow {\tan ^{ - 1}}\left( {{y \over {x - 1}}} \right) - {\tan ^{ - 1}}\left( {{y \over {x + 1}}} \right) = {\pi \over 4}$

$ \Rightarrow {\tan ^{ - 1}}\left( {{{{y \over {x - 1}} - {y \over {x + 1}}} \over {1 + {y \over {x - 1}} \times {y \over {x + 1}}}}} \right) = {\pi \over 4}$

$ \Rightarrow {\tan ^{ - 1}}\left( {{{{{xy + y - xy + y} \over {{x^2} - 1}}} \over {{{{x^2} - 1 + {y^2}} \over {{x^2} - 1}}}}} \right) = {\pi \over 4}$

$ \Rightarrow {\tan ^{ - 1}}\left( {{{xy + y - xy + y} \over {{x^2} - 1 + {y^2}}}} \right) = {\pi \over 4}$

$ \Rightarrow {{2y} \over {{x^2} - 1 + {y^2}}} = \tan \left( {{\pi \over 4}} \right)$

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

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

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

This represent a circle with center at (0, 1) and radius $\sqrt 2 $.

From diagram you can see both the circles do not cut anywhere.

Given equation,

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

$ \Rightarrow {x^2} = 1 - 2i$

Let $\alpha$ and $\beta$ are the two roots of the equation.

As, we know roots of a equation satisfy the equation so

${\alpha ^2} = 1 - 2i$

and ${\beta ^2} = 1 - 2i$

$\therefore$ ${\alpha ^2} = {\beta ^2} = 1 - 2i$

$\therefore$ $|{\alpha ^2}| = \sqrt {{1^2} + {{( - 2)}^2}} = \sqrt {15} $

Now, $|{\alpha ^8} + {\beta ^8}|$

$|{\alpha ^8} + {\alpha ^8}|$

$ = 2|{\alpha ^8}|$

$ = 2|{\alpha ^2}{|^4}$

$ = 2{\left( {\sqrt 5 } \right)^4}$

$ = 2 \times 25$

$ = 50$

${C_1}:|z - (4 + 3i)| = 2$ and ${C_2}:|z| + |z - 4| = 6$, $z \in C$

C₁ represents a circle with centre (4, 3) and radius 2 and C₂ represents a ellipse with focii at (0, 0) and (4, 0) and length of major axis = 6, and length of semi-major axis 2$\sqrt5$ and (4, 2) lies inside the both C₁ and C₂ and (4, 3) lies outside the C₂

$\therefore$ number of intersection points = 2

$\overline z = i{z^2}$

Let $z = x + iy$

$x - iy = i({x^2} - {y^2} + 2xiy)$

$x - iy = i({x^2} - {y^2}) - 2xy$

$\therefore$ $x = - 2yx$ or ${x^2} - {y^2} = - y$

$x = 0$ or $y = - {1 \over 2}$

Case - I

$x = 0$

$ - {y^2} = - y$

$y = 0,\,\,1$

Case - II

$y = - {1 \over 2}$

$ \Rightarrow {x^2} - {1 \over 4} = {1 \over 2} \Rightarrow x = \pm {{\sqrt 3 } \over 2}$

$x = \left\{ {0,i,{{\sqrt 3 } \over 2} - {i \over 2},{{ - \sqrt 3 } \over 2} - {i \over 2}} \right\}$

Area of polygon $ = {1 \over 2}\left| {\matrix{ 0 & 1 & 1 \cr {{{\sqrt 3 } \over 2}} & {{{ - 1} \over 2}} & 1 \cr {{{ - \sqrt 3 } \over 2}} & {{{ - 1} \over 2}} & 1 \cr } } \right|$

$ = {1 \over 2}\left| { - \sqrt 3 \,\,\, - {{\sqrt 3 } \over 2}} \right| = {{3\sqrt 3 } \over 4}$

$\because$ ${{{z_1}} \over {{z_2}}} = - i \Rightarrow {z_1} = - i{z_2}$

$ \Rightarrow \arg ({z_1}) = - {\pi \over 2} + \arg ({z_2})$ ..... (i)

Also $\arg ({z_1}) - \arg ({\overline z _2}) = \pi $

$ \Rightarrow \arg ({z_1}) + \arg ({z_2}) = \pi $ ..... (ii)

From (i) and (ii), we get

$\arg ({z_1}) = {\pi \over 4}$ and $\arg ({z_2}) = {{3\pi } \over 4}$

${z_1} = 3 + 4i$, ${z_2} = 4 + 3i$ and ${z_3} = 5i$

Clearly, $C \equiv {x^2} + {y^2} = 25$

Let $z(x,y)$

$ \Rightarrow \left( {{{y - 4} \over {x - 3}}} \right)\left( {{2 \over { - 4}}} \right) = - 1$

$ \Rightarrow y = 2x - 2 \equiv L$

$\therefore$ z is intersection of C & L

$ \Rightarrow z \equiv \left( {{{ - 7} \over 5},{{ - 24} \over 5}} \right)$

$\therefore$ $Arg(z) = - \pi + {\tan ^{ - 1}}\left( {{{24} \over 7}} \right)$

Let, $z = x + iy$

Given, $1 \le \left| {z - (1 + i)} \right| \le 2$

$ \Rightarrow 1 \le \left| {x + iy - 1 - i} \right| \le 2$

$ \Rightarrow 1 \le \left| {(x - 1) + i(y - 1)} \right| \le 2$

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

It represent two concentric circle both have center at (1, 1) and radius 1 and 2.

Also given,

$\left| {z - (1 - i)} \right| = 1$

$ \Rightarrow \left| {x + iy - 1 + i} \right| = 1$

$ \Rightarrow \left| {(x - 1) + i(y + 1)} \right| = 1$

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

This represent a circle with center at (1, $-$1) and radius = 1.

In the common region infinite values of B possible.