how to show that $\lim_{z \to 0}z^z$ does not exist? 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 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 am getting $1$. Did I misunderstand what is tried to be shown on that video?
 A: 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 $1/2$.
A: 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. 
