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Find all complex numbers $z$ such that ($z^6 - i) \in \mathbb R$

My solution:

Let's set $x^6 = (z - i)^6$. Then $$x^6 = |x| e^{6\theta i} \\ x^6 \in \mathbb R \iff 6\theta = k\pi \land k\in \mathbb Z$$ $$\theta = \frac{k\pi}{6}$$ Therefore $z - i = |z - i|(\cos(\frac{k\pi}{6}) + i\sin(\frac{k\pi}{6})) \\$ $$z = |z-i|\left(\cos\left(\frac{k\pi}{6}\right)+i\sin\left(\frac{k\pi}{6}\right)\right)+i$$ Now, imagine that I have plotted the solution in terms of $x$. If I wanted to have a plot in terms of $z$, would it be enough to simply shift all of my solutions one imaginary unit upwards, to satisfy the $+i$ term?

Aemilius
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3 Answers3

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Known, $w\in \Bbb R$ iff $w=\bar{w}$

Then set $z= re^{it}$ $$(z^6 - i)\in\Bbb R\Longleftrightarrow (z^6 - i)= (\bar{z}^6 + i)\\\Longleftrightarrow (z^6 -\bar{z}^6 = 2 i) \Longleftrightarrow Im(z^6) = 1\\ \Longleftrightarrow \color{blue}{r^6\sin (6t) }=Im(r^6 e^{i6t}) = 1 $$

Conclusion $$(z^6 - i)\in\Bbb R \Longleftrightarrow \color{blue}{r^6\sin (6t) = 1}$$ with $z= re^{it}$

Guy Fsone
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I am going to us Doug M.'s nice diagram. Let $ a \ge 0$. Then we are looking for $$ z^6=a+i $$ Writing $a+i$ in polar form, we have $$ z^6=\sqrt{a^2+1}e^{\theta_a i} $$ where $\theta_a=arctan(1/a)$. Hence $$ z=(a^2+1)^{1/12}e^{\theta_a i/6} $$ We should also consider all the primitive roots of unity. Thus $$ z=(a^2+1)^{1/12}e^{\theta_a i/6}e^{k\pi i /3} $$ where $k \in \{0, 1,..., 5\}$ In particular, when $a=0$ $$ z=e^{\pi i/12}e^{k\pi i /3} $$ We also have $$ z^6 = -a+i $$ in which case $$ z=(a^2+1)^{1/12}e^{-\theta_a i/6}e^{k\pi i /3} $$ where I have used the fact that $arctan(1/a)=-arctan(-1/a)$.

So there is a whole continuum of such $z$, parameterized by $a \in \mathbb{R}$ and $k \in \{0, 1, ..., 5\}$

Eric Fisher
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enter image description here $z^6 = \cot \theta + i\\ z^6 = \csc \theta (\cos \theta + i\sin\theta)\\ z = (\csc \theta)^{\frac 16} e^{i(\frac {\theta}{6}+\frac {k\pi}{3})}\\$

Doug M
  • 57,877
  • This is a pretty diagram. It shows that there are two such complex numbers. If $z^6=a+i$ in Cartesian form, then the other one has $z^6 = - a +i$ – Eric Fisher Dec 05 '17 at 23:21
  • actually, are there not 6? I have updated in response to your post. – Doug M Dec 05 '17 at 23:24
  • I think there are actually twelve. Six of them have $z^6 = a + i$ and the other six have $z^6 = -a +i$. Thanks for a great picture. – Eric Fisher Dec 05 '17 at 23:29