Complex Numbers

Given $\operatorname{Re}\left(\frac{z-2}{z-4 i}\right)=0$ and

$ \begin{aligned} & m\left(\frac{z-2}{z-4 i}\right)=1 \\ & \Rightarrow \quad \frac{z-2}{z-4 i}=i \Rightarrow z-2=i(z-4 i) \\ & \Rightarrow \quad z-2=i z+4 \Rightarrow z(1-i)=6 \\ & \Rightarrow \quad z=\frac{6}{1-i} \Rightarrow z=3(1+i) \end{aligned} $

∴ Only one $z$ in Argand plane.

Let $z=\sqrt{3}+i \Rightarrow|z|=\sqrt{3+1}=2$ and $\arg (z)=\tan ^{-1} \frac{1}{\sqrt{3}}=\frac{\pi}{6}$

∴ Polar form of $z=2\left(\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}\right)$

$ \begin{aligned} & \Rightarrow \quad z^{10}=2^{10}\left(\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}\right)^{10} \\ & \Rightarrow \quad z^{10}=2^{10}\left(\cos \frac{10 \pi}{6}+i \sin \frac{10 \pi}{6}\right) \\ & \Rightarrow \quad(\sqrt{3}+i)^{10}=2^{10}\left(\frac{1}{2}-i \frac{\sqrt{3}}{2}\right) \\ & \Rightarrow \quad(\sqrt{3}+i)^{10}=512-i 512 \sqrt{3}=a+i b \\ & \Rightarrow \quad a=512 \text { and } b=-512 \sqrt{3} \end{aligned} $

$ \begin{aligned} &\text { Given, } z^2+z+1=0\\ &\begin{aligned} & \therefore \quad z=\frac{-1 \pm \sqrt{1-4}}{2}=\frac{-1 \pm \sqrt{3} i}{2} \\ & \therefore z=\omega, \omega^2 \\ & \text { Now } z+\frac{1}{Z}=\omega+\frac{1}{\omega}=\omega+\omega^2=-1 \\ & \quad z^2+\frac{1}{z^2}=\omega^2+\frac{1}{\omega^2}=\omega^2+\omega=-1 \\ & \therefore\left(z+\frac{1}{z}\right)^3+\left(z^2+\frac{1}{z^2}\right)^3+ \\ & \quad \ldots+\left(z^{2020}+\frac{1}{z^{2020}}\right)^3 \\ & \Rightarrow \quad(-1)^3+(-1)^3+\ldots+(-1)^3=-2020 \end{aligned} \end{aligned} $

Given $n_1, n_2, n_3$ are roots of the equation

$ \begin{array}{r} E_1: x^3+x^2+l x+n=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ \therefore \text { Sum of roots } x_1+x_2+x_3=-1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{array} $

Now, $E_2: x^3+a x^2+b x+c=0$ is reciprocal equation of class one $\Rightarrow c=1$ and $a=b$

$ \therefore \quad E_2: x^3+a x^2+a x+1=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$

Given, Eq. (iii) have roots

$ \begin{aligned} & \frac{x_1-1}{2}, \frac{x_2-1}{2}, \frac{x_3-1}{2} \\ \Rightarrow & \frac{x_1-1}{2}+\frac{x_2-1}{2}+\frac{x_3-1}{2}=-a \\ \Rightarrow & \frac{x_1+x_2+x_3}{2}-\frac{3}{2}=-a \end{aligned} $

From Eq. (ii),

$ \begin{aligned} & \quad x_1+x_2+x_3=-1 \\ & \therefore \quad \frac{-1}{2}-\frac{3}{2}=-a \Rightarrow a=2 \\ & \therefore \text { Eq. }(\text { iii }), x^3+2 x^2+2 x+1=0 \\ & (x+1)\left(x^2+x+1\right)=0 \quad \therefore \quad x=-1, \\ & x^2+x+1=0 \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad x=\frac{-1 \pm \sqrt{3} i}{2} \\ & \therefore \text { Given, } \frac{x_1-1}{2}=-1 \Rightarrow x_1=-1 \\ & \quad \frac{x_2-1}{2}=\frac{-1+\sqrt{3} i}{2} \\ & \Rightarrow \quad x_2=\sqrt{3} i \\ & \quad \frac{x_3-1}{2}=\frac{-1-\sqrt{3} i}{2} \\ & \Rightarrow \quad x_3=-\sqrt{3} i \end{aligned} $

∴ Roots of these two equation excluding common roots are

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

Given $\alpha, \beta, \gamma, \delta$ are roots of

$ \begin{aligned} & x^4+x^2+1=0 \\ & \therefore \quad x^2=\frac{-1 \pm \sqrt{1-4}}{2}=\frac{-1 \pm \sqrt{3} i}{2} \\ & \therefore \quad x= \pm \sqrt{\frac{1+\sqrt{3} i}{2}}, x= \pm \sqrt{\frac{-1-\sqrt{3} i}{2}} \\ & \text { Let } \alpha=\sqrt{\frac{1+\sqrt{3} i}{2}}, \beta=-\sqrt{\frac{1+\sqrt{3} i}{2}}, \\ & \quad \gamma=\sqrt{\frac{-1-\sqrt{3} i}{2}} \\ & \quad \delta=-\sqrt{\frac{-1-\sqrt{3} i}{2}} \\ & \Rightarrow \quad \alpha^3+\beta^3+\gamma^3+\delta^3=0 \end{aligned} $

∴ Given expression,

$ \frac{\alpha^3+\beta^3+\gamma^3+\delta^3}{\alpha^6+\beta^6+\gamma^6+\delta^6}=0 $

We have, $|z|-z=2+i$

$ \begin{aligned} & \text { Let } \quad z=x+i y \\ & \therefore \sqrt{x^2+y^2}-x-i y=2+i \end{aligned} $

Equating real part and imaginary part, we get

$ \begin{aligned} & & \sqrt{x^2+y^2}-x & =2 \text { and } y=-1 \\ & \therefore & \sqrt{x^2+1} & =x+2 \end{aligned} $

$ \begin{array}{rlrl} \Rightarrow & x^2+1 =x^2+4 x+4 \\ \Rightarrow & x =-\frac{3}{4} \\ \Rightarrow & z =-\frac{3}{4}-i \\ \therefore & |z| =\sqrt{\frac{9}{16}+1} \\ & =\sqrt{\frac{25}{16}}=\frac{5}{4} \end{array} $

Given, $\arg (z-2-3 i)=\frac{\pi}{4}$

$ \begin{aligned} \Rightarrow \quad z & =x+i y \\ z-2-3 i & =x+i y-2-3 i \\ & =(x-2)+(y-3) i \\ \arg (z-2-3 i) & =\tan ^{-1}\left(\frac{y-3}{x-2}\right)=\frac{\pi}{4} \\ \Rightarrow \quad \frac{y-3}{x-2} & =\tan \frac{\pi}{4} \\\quad y-3 & =x-2 \\ \Rightarrow \quad x-y+1 & =0 \end{aligned} $

∴ Locus of $z$ is $x-y+1=0$.

Given, $z_1, z_2, z_3, \ldots, z_n$ are roots of equation

$ \begin{aligned} (z+1)^n & =z^n \Rightarrow\left(\frac{z+1}{z}\right)=(1)^{1 / n} \\ \frac{z+1}{z} & =\left(\cos \frac{2 k \pi}{n}+i \sin \frac{2 k \pi}{n}\right) \\ z+1 & =Z\left(\cos \frac{2 k \pi}{n}+i \sin \frac{2 k \pi}{n}\right) \\ z & =\frac{1}{\cos \frac{2 k \pi}{n}+i \sin \frac{2 k \pi}{n}-1} \\ z & =\frac{1}{2}\left(1-i \cot \frac{k \pi}{n}\right) \\ \operatorname{Re} z & =\frac{1}{2^{\prime}} \operatorname{Im}(z)=\left(\frac{-1}{2} \cot \frac{k \pi}{n}\right) \\ 2 \operatorname{Re}(z i) & =1, \cot ^{-1}[2|\operatorname{Im} z i|]=\frac{k \pi}{n} \end{aligned} $

$ \begin{aligned} & \sum_{i=1}^v \frac{\cot ^{-1}(2(\operatorname{Im} z i)-1}{2 \operatorname{Re}(z i)}=\sum_{i=1}^v\left(\frac{k \pi}{n}-1\right) \\ & \quad=\frac{\pi}{n}(1+2+3+\ldots . . n)-n \\ & \quad=\pi \frac{(n)(n+1)}{2(n)}-n \\ & \quad=\pi \frac{(n+1)}{2}-n \\ & \quad=\frac{1}{2}[n \pi+\pi-2 n] \\ & \quad=\frac{1}{2}[\pi-(\pi-2) n] \end{aligned} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} z_1 & =x_1+i y_1 \\ z_2 & =x_2+i y_2 \\ z_3 & =x_1+\frac{i x_2}{2} \\ z_4 & =2 y_1+i y_2 \\ \left|z_1\right| & =1,\left|z_2\right|=2 \text { and } \operatorname{Re} \\ \left(z_1 z_2\right)=0 & \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { Let } z_1=e^{i \alpha} \text { and } z_2=2 e^{i \beta}\\ &z_1 z_2=2 e^{i(\alpha+\beta)} \end{aligned} $

$ \begin{aligned} & z_1 z_2=2(\cos (\alpha+\beta)+i \sin (\alpha+\beta)) \\ & \operatorname{Re}\left(z_1+z_2\right)=2 \cos (\alpha+\beta)=0 \\ & \therefore \quad \alpha+\beta=\frac{\pi}{2} \\ & \quad \beta=\frac{\pi}{2}-\alpha \end{aligned} $

$ \begin{aligned} \therefore \quad z_1 & =\cos \alpha+i \sin \alpha \\ z_2 & =2(\cos \beta+i \sin \beta) \\ z_2 & =2(\sin \alpha+i \cos \alpha) \\ \therefore \quad z_3 & =\cos \alpha+i\left(\frac{2 \sin \alpha}{2}\right) \\ z_3 & =\cos \alpha+i \sin \alpha \\ \left|z_3\right| & =1 \\ z_4 & =2(\sin \alpha+i \cos \alpha) \\ \left|z_4\right| & =2 \\ z_3 z_4 & =2(\cos \alpha+i \sin \alpha)(\sin \alpha+i \cos \alpha) \\ z_3 z_4 & =2[(\cos \alpha \sin \alpha-\cos \alpha \sin \alpha) \\ & \left.\quad+i\left(\sin ^2 \alpha+\cos ^2 \alpha\right)\right] \end{aligned} $

$ \begin{gathered} z_3 z_4=2 i \\ \operatorname{Re}\left(z_3 z_4\right)=0 \end{gathered} $

We have, $|z| \geq 3$

Then, $\left|z+\frac{3}{z}\right| \leq|z|+\left|\frac{3}{z}\right|$

$ =\left|z+\frac{3}{z}\right| \leq 3+\left|\frac{3}{z}\right| $

∴ Least value of $\left|z+\frac{3}{z}\right|$ is not 1

And $\left|z_1-z_2\right| \leq\left|z_1\right|+\left|z_2\right|$ which is true.

$ \begin{aligned} & \text { At } \theta=\frac{\pi}{2} \\ & \left(\frac{\cos \theta+i \sin \theta}{\sin \theta+i \cos \theta}\right)^{2020}+\left(\frac{1+\cos \theta+i \sin \theta}{1-\cos \theta+i \sin \theta}\right)^{2021} \end{aligned} $

$ \begin{aligned} & =\left(\frac{0+i \cdot 1}{1+i \cdot 0}\right)^{2020}+\left(\frac{1+0+i}{1-0+i}\right)^{2021} \\ & =(i)^{2020}+\left(\frac{1+i}{1+i}\right)^{2021} \\ & =1+(1)^{2021}=1+1 \\ & =2=x+i y \Rightarrow x=2, y=0 \\ \therefore \quad x & +y=2+0=2 \end{aligned} $

As, $\omega$ is complex root of unity, then

$ \begin{gathered} 1+\omega+\omega^2=0, \omega^3=1 \\ \therefore \quad(\omega x+2)\left(\omega^2 x+2\right)-3 \\ =\omega^3 x^2+2 \omega x+2 \omega^2 x+4-3 \\ =x^2+2 x\left(\omega+\omega^2\right)+1 \quad\left(\because \omega^3=1\right) \\ =x^2+2 x(-1)+1=x^2-2 x+1 \\ \therefore \quad \sum_{x=1}^{10}\left((\omega x+2)\left(\omega^2 x+2\right)-3\right) \\ =\sum_{x=1}^{10}\left(x^2-2 x+1\right)=\sum_{x=1}^{10} x^2-2 \sum_{x=1}^{10} x+\sum_{x=1}^{10} 1 \\ =\frac{10(10+1)(2 \times 10+1)}{6}-\frac{2 \times 10(10+1)}{2}+10 \\ =\frac{10 \times 11 \times 21}{6}-10 \times 11+10 \\ =385-100=285 \end{gathered} $

We have, $|z| \leq 2$

$ \Rightarrow \sqrt{x^2+y^2} \leq 2 \Rightarrow x^2+y^2 \leq 4 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Again, $(1-i) z+(1+i) \bar{z} \geq 4$

$ \begin{aligned} & \Rightarrow \quad(1-i)(x+i y)+(1+i)(x-i y) \geq 4 \\ & \Rightarrow \quad x+i y-i x+y+x-i y+i x+y \geq 4 \\ & \Rightarrow \quad 2 x+2 y \geq 4 \Rightarrow x+y \geq 2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

So, $A \cap B=\left\{x^2+y^2 \leq 4\right\} \cap\{x+y \geq 2\}$

$ \therefore \quad z=\sqrt{3}+\frac{1}{2} i $

Given equation, $z^2\left(1-z^2\right)=16, z \in \mathbf{C}$

Now, let $z^2=w=r(\cos \theta+i \sin \theta)$

where $r>0$.

$ \begin{aligned} & \therefore \quad 1-z^2=\frac{16}{z^2} \Rightarrow z^2+\frac{16}{z^2}=1 \\ & \Rightarrow \quad\left(r+\frac{16}{r}\right) \cos \theta+i\left(r-\frac{16}{r}\right) \sin \theta=1 \end{aligned} $

On comparing real and imaginary parts, we get

$ \left(r+\frac{16}{r}\right) \cos \theta=1 \text { and }\left(r-\frac{16}{r}\right) \sin \theta=0 $

∴ Either $\sin \theta=0$

$ \Rightarrow \quad \cos \theta=1 \Rightarrow r+\frac{16}{r}=1 $

(not possible)

or $\quad r-\frac{16}{r}=0 \Rightarrow r^2=16$

$ \begin{array}{lc} \Rightarrow & r=4 \\ \Rightarrow & |z|^2=4 \Rightarrow|z|=2 \end{array} $

Let $z=x+i y$

Then, vertices of rectangle are $(x, y)$, $(x,-y),(-x,-y),(-x, y)$.

Now, area of rectangle $=(2 x)(2 y)=4 x y$

It is given that,

$ \begin{array}{rlrl} & & 4 x y & =2 \sqrt{3} \Rightarrow 2 x y=\sqrt{3} \\ & \therefore & & x =\frac{1}{2}, y=\sqrt{3} \\ & \therefore & & z=\frac{1}{2}+\sqrt{3} i \end{array} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} & \left(\frac{\cos \theta+i \sin \theta}{\sin \theta+i \cos \theta}\right)^8+\left(\frac{1+\cos \theta-i \sin \theta}{1+\cos \theta+i \sin \theta}\right)^{16} \\ & =\left(\frac{\cos \theta+i \sin \theta}{i(\cos \theta-i \sin \theta)}\right)^8 +\left(\frac{2 \cos ^2 \frac{\theta}{2}-i 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \cos ^2 \frac{\theta}{2}+i 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}\right)^{16} \end{aligned} \end{aligned} $

$ \begin{aligned} = & \frac{1}{i^8}\left(\frac{\cos \theta+i \sin \theta}{\cos \theta-i \sin \theta}\right)^8+\left(\frac{\cos \frac{\theta}{2}-i \sin \frac{\theta}{2}}{\cos \frac{\theta}{2}+i \sin \frac{\theta}{2}}\right)^{16} \\ = & \frac{\cos 8 \theta+i \sin 8 \theta}{\cos 8 \theta-i \sin 8 \theta}+\frac{\cos 8 \theta-i \sin 8 \theta}{\cos 8 \theta+i \sin 8 \theta} \\ = & (\cos 8 \theta+i \sin 8 \theta)(\cos 8 \theta-i \sin 8 \theta)^{-1} \\ & \quad+\left(\cos 8 \theta-i \sin 8 \theta(\cos 8 \theta+i \sin 8 \theta)^{-1}\right. \\ = & (\cos 8 \theta+i \sin 8 \theta)(\cos 8 \theta+i(\sin 8 \theta)+ \\ & \quad(\cos 8 \theta-i \sin 8 \theta)(\cos 8 \theta-i \sin 8 \theta) \\ = & (\cos 8 \theta+i \sin 8 \theta)^2+(\cos 8 \theta-i \sin 8 \theta)^2 \\ = & \cos ^2 8 \theta+i^2 \sin ^2 8 \theta+2 i \cos 8 \theta \sin 8 \theta \\ & \quad+\cos ^2 8 \theta+i^2 \sin ^2 8 \theta-2 i \cos 8 \theta \sin 8 \theta \\ = & 2\left(\cos ^2 8 \theta-\sin ^2 8 \theta\right)=2 \cos 16 \theta \end{aligned} $