The number of complex numbers z satisfying ov | vedclass

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(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

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