The number of complex numbers z satisfying ov | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given equation is $\overline{z} = i z^2 \dots (i)$Taking modulus on both sides,$|\overline{z}| = |i z^2|$ $\Rightarrow |z| = |i| |z|^2$ $\Rightarr

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) Given equation is $\overline{z} = i z^2 \dots (i)$Taking modulus on both sides,$|\overline{z}| = |i z^2|$ $\Rightarrow |z| = |i| |z|^2$ $\Rightarrow |z| = |z|^2$.This implies $|z|(|z| - 1) = 0$,so $|z| = 0$ or $|z| = 1$.Case $1$: If $|z| = 0$,then $z = 0$. This is one solution.Case $2$: If $|z| = 1$,then $\overline{z} = 1/z$. Substituting into $(i)$,$1/z = i z^2 \Rightarrow z^3 = 1/i = -i$.We know $-i = e^{i(3\pi/2 + 2k\pi)}$ for $k = 0, 1, 2$.Thus,$z = e^{i(\pi/2 + 2k\pi/3)}$.For $k = 0$,$z = e^{i\pi/2} = i$.For $k = 1$,$z = e^{i(7\pi/6)} = -\frac{\sqrt{3}}{2} - \frac{i}{2}$.For $k = 2$,$z = e^{i(11\pi/6)} = \frac{\sqrt{3}}{2} - \frac{i}{2}$.Including $z = 0$,there are $1 + 3 = 4$ solutions.

The number of complex numbers $z$ satisfying $\overline{z} = i z^2$ is

Similar Questions

The $A + iB$ form of $\frac{(\cos x + i\sin x)(\cos y + i\sin y)}{(\cot u + i)(1 + i\tan v)}$ is

Let $\alpha, \beta$ be the roots of the equation $x^2-x+2=0$ with $\operatorname{Im}(\alpha)>\operatorname{Im}(\beta)$. Then $\alpha^6+\alpha^4+\beta^4-5 \alpha^2$ is equal to

$(r, \theta)$ denotes $r(\cos \theta + i \sin \theta)$. If $x = (1, \alpha)$,$y = (1, \beta)$,$z = (1, \gamma)$ and $x + y + z = 0$,then $\sum \cos (2\alpha - \beta - \gamma) = $

If $\frac{1+i \cos \theta}{1-2 i \cos \theta}$ is purely real,then $\cos ^3 \theta+\sin ^2 \theta+\cos \theta+1=$

The values of $x$ for which $\sin x + i \cos 2x$ and $\cos x - i \sin 2x$ are conjugate to each other are

Vedclass Test Series

Exam Paper Generator

Online Exam Module