I know that $z = a + ib$ and that $\overline{z} = a - ib$, but when I try and calculate the solutions I get an unsolvable equation.
Would appreciate any help.
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Forget about real and imaginary parts and note that every solution $z$ is such that $|z|^3=|z^3|=|\bar z|=|z|$ hence $|z|=0$ or $|z|=1$. Furthermore, $z^4=z^3\cdot z=\bar z\cdot z=|z|^2$.
Can you finish in both cases?
Did
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Shouldn't I use the real and imaginary numbers as the solutions? For example: |z|=1 then: |z|= |a + ib| ---> |z|^2 = a^2 + b^2 ---> |z| = sqare of (a^2+b^2) = 1 a^2+b^2 = 1. Not sure what to do from this point. – David Dec 14 '14 at 16:26
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The method using modulus and argument is swift and painless. But if you prefer the path of suffering, please use real and imaginary parts (only, I prefer not to walk this path with you). – Did Dec 14 '14 at 16:28
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Sorry, I'm not familiar with the modulus. Can you show me how to finish the two cases with this method? – David Dec 14 '14 at 16:31
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Just to be clear, I meant if you can you show me how to solve with the modulus method you suggested. – David Dec 14 '14 at 16:55
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First note that, if $|z|=r$, then $|z^3|=|z|^3=|\overline{z}|=|z|$, giving $r\in\{0,1\}$, so $z=0$ works and $z=1 \text{cis } \theta$ works, for some $\theta$. Plugging in gives $z^3=\text{cis } 3\theta$ and $\overline{z}=\text{cis } -\theta$, so we have $ 3 \theta \equiv -\theta \pmod {2\pi} $.
That means that $4\theta$ is a multiple of $2\pi$, which is true iff $\theta$ is a multiple of $\frac{\pi}{2}$.
Can you finish?
Ahaan S. Rungta
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