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Why does $\vert z\vert = z^5 $ have 6 solutions in $\mathbb{C}$

I tried to solve it with the polar method and I only found 5 solutions.

$$ z_{(n+1)} = e^{\left(\frac{2\cdot\pi}{5} +\frac{2\cdot\pi\cdot n}{5} \right)} \quad with\quad n=0,1,2,3,4 $$

There must be something special with the $\vert z\vert$ , but I can't find what.

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    If someone could also explain me how to get the complex C in the title it would be very nice – 2017-01-14
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    Use `\mathbb`, not the double stroke font. – 2017-01-14
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    Are you user @SoHCahToha? – 2017-01-14
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    no he is not he is a friend and we genuinly have problem with this :( – 2017-01-14
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    The solutions are $\{0, 1, \exp(2\pi i/5), \exp(4\pi i/5), \exp(6\pi i/5), \exp(8\pi i/5)\}$ – 2017-01-14
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    @Thomas is 0 an answer because of the $\vert z \vert$ ? – 2017-01-14
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    @SoHCahToha So the two of you decided knowingly to post this duplicate... Not nice. – 2017-01-14

2 Answers 2

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you're just forgetting about the 6th solution $z=0$.

Oke here is a completely worked out solution:

We write $z=re^{i\alpha}$. Then we are looking for solutions $$r=|z|=z^5=r^5e^{5i\alpha}$$ It follows that we must have $r=0$, and so $z=0$, or otherwise we divide both sides by $r$ to find: $$r^4e^{5i\alpha}=1$$ And so by taking the absolute value of both sides: $$r^4=|r^4e^{5i\alpha}|=|1|=1$$ Since $r> 0$ we conclude that we must have $r=1$. Hence we want to find $\alpha$ satisfying $$e^{5i\alpha}=1$$ which gives the other $5$ solutions that you already found.

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    but what is |z| is not equal to 1 say |z| = 2? – 2017-01-14
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    @SoHCahToha I dont understand that sentence – 2017-01-14
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    We can find 5 solutions if we set |z| = 1, where we have $$ z^5=1$$, then using polar form we find 5 solutions but what happens if we set |z| to something other than 1? – 2017-01-14
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    @SoHCahToha see my edit – 2017-01-14
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    I love you so much thank you I fully understood – 2017-01-14
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$z=0$ is clearly a solution.

Now suppose $z$ is a non-zero solution. Taking the modulus of each side in the given equation, we get $\vert z\vert=\vert z\vert^5$, and then $\vert z\vert=1$. From this, we deduce that $z^5=1$ and so $z\in\{1,\omega,\omega^2,\omega^3,\omega^4\}$ where $\omega=\exp(\frac{2i\pi}{5})$.

Finally the set of solutions of the given equation is included in $\{0, 1,\omega,\omega^2,\omega^3,\omega^4\}$

Conversely, it is readily seen that all those six complex numbers are effectively solutions.

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    thank you, this is the answer I was looking for. – 2017-01-14
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    but how can we say |z|=1 ? why not |z| = 200? – 2017-01-14
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    @SoHCahToha Because the relation $\vert z\vert=\vert z\vert^5$ says that the positive real number $\vert z\vert$ is a solution of the equation $x=x^5$ which is equivalent to $x(1-x)(1+x)(1+x^2)=0$. Therefore $\vert z\vert$ must be equal to $1$. – 2017-01-15