Definition of $z^0, z=a+bi, a,b \in \mathbb{R}, z \neq 0$ Today I found in a True-False the question; Does the equality $$z^0=1, z=a+bi, a,b \in \mathbb{R}$$ hold $\forall z \in \mathbb{C^*}$?
The thing is, this was never clearly defined in the book, and polar/exponential forms of complex numbers aren't taught. My (EDIT; as it turns out partial)explanation would have been $$z=a+bi=\sqrt{a^2+b^2} e^{i\left (\tan^{-1}\left (\frac{b}{a}\right )\right )}=e^{i\left (\tan^{-1}\left (\frac{b}{a}\right )\right )+\ln(\sqrt{a^2+b^2})}$$
Then $$z^0=\left (e^{i \left (\tan^{-1}\left (\frac{b}{a}\right )\right )+\ln \left (\sqrt{a^2+b^2} \right )} \right )^0=e^{0\cdot (i\left (\tan^{-1}\left (\frac{b}{a}\right )\right )+\ln \left (\sqrt{a^2+b^2}\right ))}=e^0=1$$
It's a true-false questionaire and it doesn't ask for an explanation but I want to see one that uses more elementary methods.
EDIT#2; Another question came up. We have learnt that $x^0=1 \; , \; \forall x \in \mathbb{I^*}$ and $y^0=1 \; , \; \forall y \in \mathbb{R^*}$. Is there any statement that transfers properties of subsets to the equivalent sets?
 A: Consider $z=0 \in \mathbb{C}.$
Then $z^0$ is undefined.

Since $z$ can be written in the form $re^{i\theta}$, where $r \in \mathbb{R}$ and $\theta \in (-\pi,\pi]$, we have
$$z^0 =\left[re^{i\theta}\right]^0 =r^0e^{i\theta(0)}=r^0e^0=1 \cdot 1=1$$ since $\alpha^0=1 \quad$ for all $\alpha \in \mathbb{R} \setminus\{0\}
.$
Alternatively, you can use the fact that 
$$\underbrace{z^0= z^{1-1}=\frac{z^1}{z^1}=1}_{\substack{\text{provided} \quad z \neq 0,\\ \quad \text{in which case we've got } \\ \quad \\ \text{the indeterminate form} \quad 0/0}}$$

Also, be careful of how you write the principle argument, $\theta.$ You wrote it as $\arctan\left(\frac{b}{a}\right)$ which is not necessarily true (it's only true if $a,b>0$).
In general, $$\theta= \begin{cases} \arctan\left(\frac{b}{a}\right) \quad \text{if} \quad a,b>0 \\ \pi- \arctan\left(\frac{b}{a}\right) \quad \text{if} \quad a<0, b>0 \\-\arctan\left(\frac{b}{a}\right) \quad \text{if} \quad a>0, b<0 \\-\left[\pi-\arctan\left(\frac{b}{a}\right)\right] \quad \text{if} \quad a<0, b<0 \\ \quad \\ 0  \quad \text{if} \quad a>0, b=0 \\ \quad \\ \frac{\pi}{2} \quad \text{if} \quad a=0, b>0\ \\ \quad \\ \pi \quad \text{if} \quad a<0, b=0 \\ \quad \\ -\frac{\pi}{2}  \quad \text{if} \quad a=0, b<0 \\ \quad \\ \text{undefined} \quad \text{if} \quad a=b=0 \end{cases} \quad ,$$ which can be denoted as the much-more-succinct $$\operatorname{atan2} (b,a).$$ 
