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Why $|z^5|-2^5=0$ has $infty$ solutions in $mathbb{C}$?

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0 $begingroup$ Given $z=x+iy$ a complex number, I can't understand why $|z^5|-2^5=0$ has infinite solutions in $mathbb{C}$ . complex-numbers share | cite | improve this question asked Jan 16 at 18:15 Kevin Kevin 162 11 $endgroup$ $begingroup$ For any $thetain[0, 2pi/5),$ the complex number $z=2(costheta+isintheta)$ satisfies your equation. $endgroup$ ...