Solving $z^z=z$ in Complex Numbers

Solution 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 $k=-2$.

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

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