Arbitrary Power of Complex Number

Theorem
Let $z = a + i b$ be a complex number.

Let $n \in \Z_{>0}$ be a (strictly) positive integer.

Then:

Lemma
The proof proceeds by induction.

For all $n \in \Z_{>0}$, let $\map P n$ be the proposition:
 * $\ds z^n = \paren {\sum_{\substack {0 \mathop \le j \mathop \le n \\ \text {$j$ even} } } \paren {-1}^{j / 2} \dbinom n j a^{n - j} b^j} + i \paren {\sum_{\substack {0 \mathop \le j \mathop \le n \\ \text {$j$ odd} } } \paren {-1}^{j / 2 - 1} \dbinom n j a^{n - j} b^j}$

$\map P 1$ is the case:

Thus $\map P 1$ is seen to hold.

Basis for the Induction
$\map P 2$ is the case:

Thus $\map P 2$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.

So this is the induction hypothesis:
 * $\ds z^k = \paren {\sum_{\substack {0 \mathop \le j \mathop \le k \\ \text {$j$ even} } } \paren {-1}^{j / 2} \dbinom k j a^{k - j} b^j} + i \paren {\sum_{\substack {0 \mathop \le j \mathop \le k \\ \text {$j$ odd} } } \paren {-1}^{j / 2 - 1} \dbinom k j a^{k - j} b^j}$

from which it is to be shown that:
 * $\ds z^{k + 1} = \paren {\sum_{\substack {0 \mathop \le j \mathop \le k + 1 \\ \text {$j$ even} } } \paren {-1}^{j / 2} \dbinom {k + 1} j a^{k + 1 - j} b^j} + i \paren {\sum_{\substack {0 \mathop \le j \mathop \le k + 1 \\ \text {$j$ odd} } } \paren {-1}^{j / 2 - 1} \dbinom {k + 1} j a^{k + 1 - j} b^j}$

Induction Step
This is the induction step:

From :

where $z^k = u_k + i v_k$.

Taking the real part and imaginary part separately:

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\forall n \in \Z_{\ge 0}: \ds z^n = \paren {\sum_{\substack {0 \mathop \le j \mathop \le n \\ \text {$j$ even} } } \paren {-1}^{j / 2} \dbinom n j a^{n - j} b^j} + i \paren {\sum_{\substack {0 \mathop \le j \mathop \le n \\ \text {$j$ odd} } } \paren {-1}^{j / 2 - 1} \dbinom n j a^{n - j} b^j}$