Complex numbers also give up to 0?

Any number (be it complex, natural or any other) raised to the power of 0 is 1. Proof:

$x^{0}=x^{y-y}=x^{y}:x^{y}=\frac{x^{y}}{x^{y}} = \frac{1}{1}$

This proof shows us how raising to $0^{th}$ power is done, down below is an explanation of why it works with imaginary numbers .

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Answer to why $i$ is treated as a real number:

$i^{0}$ is the same as: $-1^{\frac{1}{2} \times 0}$

$\frac{1}{2} \times 0 = 0$

That leaves us with a real number $-1^{0}$ (without any roots).

Thanks to Itamar Keren for clearing up confusion that has occurred.

We can write (√-1)^0 as (i)^0

i^0=[cos(π/2)+isin(π/2)]^0(since arg(i)=π/2) in polar form. =cos0+isin0(De Moivre theorem) =1.

Or i^0 in Euler's form is (e^(iπ/2))^0 =e^0=1.

But Di=[0,+00)

Nice question

Here

$i^0= e^{0 \cdot \ln i} = e^0 =1$

Note : $\sqrt{-1}$ still makes no sense. i is the number defined by $i^2=-1$ , but square root remains undefined for -1.