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So I was revising complex numbers and I came across this question:

Find the number of solution(s) of the equation

$$\ z^3+\frac{3(\bar z)^2}{|z|}=0$$ where z is a complex number.

The answer given in my booklet is 5 solutions, but I am unable to understand how. So far I am able to find only one solution i.e. z=0. I tried to solve it by assuming$\ z=re^{i\theta} $ but I got stuck at this equation: $$\cos 5\theta=\frac {-3}{r^2} $$
Please help me understand this question.

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It's almost correct, except $z=0$ isn't a solution and your last line should be: $$e^{5i\theta}=-3/r^2$$

Taking modulus of both sides shows $r=\sqrt3$, and then the LHS is just the five roots of $-1$.

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Notice that $z=0$ is NOT a solution (see the denominator $|z|$). So $z\not=0$ and by multiplying the equation by $z^2$ we obtain $$z^5+\frac{3z^2(\bar z)^2}{|z|}=z^5+3|z|^3=0\Rightarrow |z|^5=3|z|^3\Rightarrow r=|z|=\sqrt{3}.$$ Hence, by letting $z=\sqrt{3}e^{i\theta}$, it remains to solve $$z^5+3(\sqrt{3})^3=0\Leftrightarrow \left(\frac{z}{\sqrt{3}}\right)^5=-1\Leftrightarrow e^{i5\theta}=\cos(5\theta)+i\sin(5\theta)=-1.$$ Can you take it from here?

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Hint: Obviously, $0$ is not a solution. On the other hand, if $z$ is a solution, then$$z^3+3\frac{\overline z^2}{|z|}=0\iff z^3=-3\frac{\overline z^2}{|z|}\implies|z|^3=3|z|\iff|z|=\sqrt3,$$since $0$ is not a solution. So, you can assume that $z=\sqrt3e^{i\theta}$ for some real $\theta$.

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  • $\begingroup$ Oh yes, 0 is definitely not a solution. My mind is definitely not on track. $\endgroup$ – user527679 Feb 3 '18 at 9:27

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