Complex power of a complex number

Can someone explain to me, step by step, how to calculate all infinite values of, say,

$(1+i)^{3+4i}$?

I know how to calculate the principal value, but not how to get all infinite values...and I'm not sure how to insert the portion that gives me the other infinity values.

Solution 1:

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

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

Solution 2:

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.

Solution 3:

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)