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?

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    $\begingroup$ Now that there is one damned engaging question! Endorsed! $\endgroup$ Jul 16, 2015 at 1:08
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    $\begingroup$ Often, these types of problems are solved in terms of the Lambert W function. $\endgroup$ Jul 16, 2015 at 1:24
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    $\begingroup$ Problem is that under the condition that $a$ is not positive real and $b$ is not integral, $a^b$ has no unambiguous definition. In particular, when you take $\log$ of both sides to get $z\log z=\log z$, it's not even clear to me that you want the same branch of the logarithm on both sides of the equation. $\endgroup$
    – Lubin
    Jul 16, 2015 at 1:35
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    $\begingroup$ @johannesvalks They solve the original equation. I wanted to avoid the issue with logarithms. $\endgroup$ Jul 16, 2015 at 15:55
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    $\begingroup$ @johannesvalks Wolfram Alpha is really bad at evaluating complex powers. WA and Mathematica do weird things with branch cuts. This is a very well-known issue with the software. You have to write things in polar form to get a better formed problem (for their purposes). $\endgroup$ Jul 16, 2015 at 16:06

2 Answers 2


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 $k=-2$.

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    $\begingroup$ It is not just the logarithm... the expression $z^z$ itself is multivalued by definition. $\endgroup$
    – Masacroso
    Jul 16, 2015 at 18:57
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    $\begingroup$ Yes, @Masacroso, I’m starting to agree strongly with your position, and thinking that my analysis did not take sufficient account of the multivaluedness of the logarithm. $\endgroup$
    – Lubin
    Jul 17, 2015 at 3:38

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

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

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 $r \ne 1$.

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

as the only solutions.

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    $\begingroup$ Not sure what you mean by "the right part can be positive or negative". It sounds like you might be trying to solve "all values of $z^z$ are equal to $z$ for every branch cut of $\ln$", which I would call a very strong restriction as it essentially forces $z$ to be real. My interpretation of the question was that the OP was interested in just one of the multiple values of $z^z$ being equal to $z$. $\endgroup$
    – Erick Wong
    Jul 16, 2015 at 15:40
  • $\begingroup$ Just jumping in in medias res, your sixth display seems to be of the form $a+bi=0$, where $a$ and $b$ are real. Doesn’t that mean that both $a$ and $b$ are zero? $\endgroup$
    – Lubin
    Jul 16, 2015 at 15:45
  • $\begingroup$ Yes, and when you have these two equation $a=0$ and $b=0$, you can work it out to remove the $\sin()$ and $\cos()$. $\endgroup$ Jul 16, 2015 at 15:51
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    $\begingroup$ I think @ErickWong's suggestion is spot on here. In the usual branch cut of the complex plane, the solutions you presented are the only ones. However if you do not place restrictions on $\theta$, there are many other solutions. $\endgroup$ Jul 16, 2015 at 15:58

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