Complex power of a complex number

$Let (1+i)^{3+4i}=k$

Taking ln on both sides gives us $(3+4i)log_e{(1+i)}=log_ek\cdots(1)$

also $(1+i)=\sqrt{1^2+1^2}e^{\frac{i\pi}4}=\sqrt2e^{\frac{i\pi}4}\cdots(2)$

$log_e(1+i)$=$log_e$($\sqrt2e^{\frac{i\pi}4}$)

Substituting $(2)$ in $(1)$ we get

$(3+4i)(log_e\sqrt2+{\frac{i\pi}4}) = log_ek$

or $k = e^{(3+4i)(log_e\sqrt2+{\frac{i\pi}4})}$

NOTE :

GENERALISATION: To evaluate numbers of the form $(a+ib)^{c+id}$

Let $\sqrt{a^2+b^2}=r$ and argument of $a+ib$ be $\theta$

Then $(a+ib)=re^{i\theta}$ = $e^{log_e(r)+i\theta}$

Hence, $(a+ib)^{c+id}=e^{{log_e}{(r)(c+id)+i\theta}(c+id)}$

When you write your complex number as an e-power, your problem boils down to taking the Log of $(1+i)$. Now that is $\ln\sqrt{2}+ \frac{i\pi}{4}$ and here it comes: + all multiples of $2i\pi$. So in your e-power you get $(3+4i) \times (\ln\sqrt{2} + \frac{i\pi}{4} + k \cdot i \cdot 2\pi)$ I would keep the answer in e-power form. You can now work it out.

Let's suppose you've already defined $\log r$ for real $r > 0$, say, using Taylor series. Then given $z, \alpha \in \mathbb{C}$, you can define $$z^{\alpha} = \exp(\alpha \log z)$$ where

$$\exp(w) = \displaystyle \sum_{j=0}^{\infty} \dfrac{z^j}{j!} \qquad \text{and} \qquad \log(w) = \log |w| + i \arg(w)$$

This is not well-defined - it relies on a choice of argument, which is well-defined only up to adding multiples of $2\pi$. It's these multiples of $2\pi$ which give you new values of $z^{\alpha}$.

Explicitly, if $w$ is one value of $z^{\alpha}$, then so is $$w \cdot e^{2n \pi \alpha i}$$ for any $n \in \mathbb{Z}$.

Fun facts ensue:

• if $\alpha$ is an integer then $z^{\alpha}$ is well-defined
• if $\alpha$ is rational then $z^{\alpha}$ has finitely many values
• if $\alpha$ is pure imaginary then $z^{\alpha}$ is real (but not well-defined)