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Why does this equation have 6 solutions in $\Bbb{C}$,

$$|z| = z^5$$

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    not a polynomial in $z$ – 2017-01-14
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    If I were asked to *find* the solutions, I would start by taking absolute values on both sides and discovering what value(s) $|z|$ can take. – 2017-01-14
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    @SoHCahToha Please do not vandalize your own question. – 2017-01-14
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    sorry, I am desperate for knowledge – 2017-01-14

3 Answers 3

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Take $r = |z|$ so that real $r \geq 0.$ With $r^5 - r = 0,$ we have real values $1,0.$ Back to $z,$ this means five fifth roots of $1$ and then $z=0$

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    I don't really understand how you get the roots.. – 2017-01-14
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    @SoHCahToha what are the solutions of $z^5 = 1?$ – 2017-01-14
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    how did you go from z^5 to r^5 if r= |z| – 2017-01-14
1

This isn't a polynomial! A polynomial is a linear combination of a finite number of powers of the variable. There is no way to calculate $\overline{z}$ or $|z|$ in that way, so any expression containing either function isn't a polynomial.

0

From $|z|=z^5$, write

$$r=r^5(\cos 5\theta+i\sin 5\theta)$$

$$r-r^5\cos 5\theta=0 \quad ; \quad \sin 5\theta=0$$

Then $5\theta=k\pi$ from this discover roots:

$$r=0 \quad , \quad 1-r^4\cos k\pi=0$$ then $r=0 \; $ or $\; r^4=\dfrac{1}{\cos k\pi}$

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    Your second equation should be $1-r^4\cos(k\pi)=0$ as the factors $5$ cancel. Since $\cos(k\pi)=(-1)^k$, you only get solutions for even $k$. – 2017-01-14