how to show that $\lim_{z \to 0}z^z$ does not exist?

Question

What makes $0^0$ indeterminate? Here is a video by numberphile that claims that $z^z$ does not exist as $z \to 0$ where $z \in \mathbb C $. I tried$$\lim_{x \to 0}(x+ix)^{(x+ix)}$$ and replaced $x$ by $-x$ and couple of tricks Limit[(x + I x^2)^(x + I x^2), x -> 0] in Mathematica, but I got $1$. Did I misunderstand what is tried to be shown on that video?

Answer 1

Answering, instead, your first question: What makes $0^0$ indeterminate?

This is not just a matter of $$ \lim_{x \to 0} x^x = ? \tag{1} $$ but more generally of $$ \lim_{t \to a} f(t)^{g(t)} = ? \qquad\text{where } \lim_{t \to a} f(t) = 0\text{ and } \lim_{t\to a}g(t) = 0. \tag{2}$$ We say "$0^0$ is indeterminate" to mean that the answer in (2) can vary, depending on $f$ and $g$.

For example, if $$ f(t) := e^{-(\log 2)/|t|},\qquad g(t) := |t|, $$ then $$ \lim_{t \to 0} f(t) = 0,\qquad \lim_{t\to 0}g(t) = 0,\qquad \lim_{t\to 0} f(t)^{g(t)} = \frac{1}{2} . $$ You can probably make minor changes in this to get other results instead of $1/2$.

Answer 2

I disagree.

The subtlety is $z^{z}$ is a multi-valued complex function. If it is $x^{x}$ then it is easy to find the limit by $e^{x\log[x]}$, and we know $\lim_{x\rightarrow 0}x\log[x]=\frac{1/x}{-1/x^{2}}=-x=0$(repeat use of L'Hospital rule at here). So you can conclude $x^{x}\rightarrow 1$.

In the complex case the same strategy would give you complications, because $\log[z]=\log[|z|]+irad(z)+2\pi ni$. This is because if we let $z=re^{i\theta}$, then we have $\log[z]=\log[r]+i\theta+2\pi n i$. So we have extra complication aring from $z* rad(z)$, or $re^{i\theta}*i\theta$. Clearly when $r\rightarrow 0$ this goes to $0$ as well. So we can conclude $e^{z\log[z]}$ goes to 1 as we did earlier.