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

If $x+iy = (1+i)^6 - (1-i)^6$,then which one of the following is true?

If $3+i$ and $2-\sqrt{3}$ are the roots of the equation $f(x)=a_0+a_1 x+a_2 x^2+\ldots+a_{n} x^{n}$ where $a_0, a_1, \ldots, a_{n} \in \mathbb{Z}$,then the least value of $n$ and the value of $a_0$ are respectively:

The value of $\left(\frac{1+\sin \frac{2 \pi}{9}+i \cos \frac{2 \pi}{9}}{1+\sin \frac{2 \pi}{9}-i \cos \frac{2 \pi}{9}}\right)^3$ is

If the real part of the complex number $z = \frac{3 + 2i \cos \theta}{1 - 3i \cos \theta}$, where $\theta \in (0, \frac{\pi}{2})$, is zero, then the value of $\sin^2 3\theta + \cos^2 \theta$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module