Solving $z^z=z$ in Complex Numbers

Question

I wanted to find all complex numbers $z\neq0$ such that $z^z=z$. I observed that $z=\pm1$ satisfies the equation. But I had problems when tried to find all the possible solutions since $z^z$ may take more than one value. I attempted this: $$z=r(\cos\theta+\mathrm i\sin\theta)$$ $$\therefore\log z=\log r+\mathrm i\theta$$ $$z^{z-1}=1$$ $$\therefore (z-1)\log z=2\pi n\mathrm i\qquad n\in\mathbb Z$$ $$\therefore\cases{(r\cos\theta-1)\log r-r\theta\sin\theta=0\\\theta(r\cos\theta-1)+r\log r\sin\theta=2\pi n}$$ But I couldn't go further. I'm also aware that adding an integer multiple of $2\pi$ to $\theta$ may give another possible value for $\log z$. Can anyone help please?

Answer 1

The more I think about this question, the more I like it. The key to it is to have a precise idea of what we’re talking about.

We need an unambiguous definition of the natural logarithm, $\log$. It can be defined as a single-valued function only on a simply-connected domain in $\Bbb C$ that omits the origin. Since we know all the positive real solutions of our equation, namely $z=1$, we might as well omit from the plane the whole nonnegative real axis. Then we may specify that $0<\Im(\log z)<2\pi$, so that this logarithm maps onto the open strip between the real axis and a line parallel to it and $2\pi$ units above.

Now, to the equation $z^z=z$ we apply log and get $z\log z=\log z+2k\pi i$, and so $(z-1)\log z=2k\pi i$. This is really infinitely many equations, one for each integer $k$. The value $k=0$ gives us our known value $z=1$, and if you try it for $k=-1$, you can check that since $\log(-1)=\pi i$, there’s your other known solution. It would be fun to see whether there are other solutions for this value of $k$, but I’m going to bet that each other value of $k$ leads to at least one solution.

I’m posting this incomplete answer, and will look for a value with $k=-2$.

Answer 2

We can write $$ z = z(r,\theta), $$

so we actually have $$ z(r,\theta)^{ z(r,\theta) } = z(r,\theta). $$

Note that $$ z(r,\theta) = e^{\ln(r) + \mathbf{i} \theta} = r \cos(\theta) + \mathbf{i} r \sin(\theta), $$

so we obtain $$ e^{ \big[ \ln(r) + \mathbf{i} \theta \big] \big[ r \cos(\theta) + \mathbf{i} r \sin(\theta) \big] } = e^{\ln(r) + \mathbf{i} \theta}, $$

therefore $$ \big[ \ln(r) + \mathbf{i} \theta \big] \big[ r \cos(\theta) + \mathbf{i} r \sin(\theta) \big] = \ln(r) + \mathbf{i} \big\{ \theta + 2 n \pi \big\}. $$

So we obtain $$ \Big[ r \ln(r) \cos(\theta) - r \theta \sin(\theta) - \ln(r) \Big] + \mathbf{i} \Big[ r \ln(r) \sin(\theta) + r \theta \cos(\theta) - \theta - 2 n \pi \Big] = 0. $$

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(added to show step)

We get $$ \left[ \begin{array}{rcl} r \ln(r) \cos(\theta) - r \theta \sin(\theta) &=& \ln(r)\\ r \ln(r) \sin(\theta) + r \theta \cos(\theta) &=& \theta + 2 n \pi \end{array} \right. $$

So $$ \Big[ r \ln(r) \cos(\theta) - r \theta \sin(\theta) \Big]^2 + \Big[ r \ln(r) \sin(\theta) + r \theta \cos(\theta) \Big]^2 = \ln^2(r) + \big( \theta + 2 n \pi \big)^2 $$

Thus $$ r^2 \ln^2(r) + r^2 \theta^2 = \ln^2(r) + \big( \theta + 2 n \pi \big)^2 $$

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So we obtain

$$ r^2 \ln^2(r) + r^2 \theta^2 = \ln^2(r) + \big( \theta + 2 n \pi \big)^2, $$

or $$ \big[ r^2 - 1 \big] \big[ \ln^2(r) + \theta^2 \big] = \big( \theta + 2 n \pi \big)^2 - \theta^2. $$

The case $r=1$

We obtain $$ \big( \theta + 2 n \pi \big)^2 - \theta^2 = 0, $$

whence $$ \theta = - n \pi, $$

Thus $$ z = \pm 1 $$

The case $r \ne 1$

We obtain $$ \big[ r^2 - 1 \big] \big[ \ln^2(r) + \theta^2 \big] = 4 n \pi \Big( \theta + n \pi \Big). $$

But as $$ z(r,\theta) = z(r,\theta + 2 k \pi), $$

so the right part can be positive or negative, while the left part does not change sign. There are no solutions for the case $r \ne 1$.

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So $$ z^z=z \Rightarrow z = \pm 1, $$

as the only solutions.