# 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.

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.

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

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

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)
• I think you meant to write, “then so is $w\cdot\exp(\alpha+2n\pi i)$”, which also affects your last bulleted statement. – Lubin Aug 27 '13 at 4:41

Let us find in general $w^z$ where $w$ and $z$ are complex. This expression is by definition equal to $$\exp\{z\ln w\}$$ where $\ln w$ is one of the complex logarithms of $w$. That is, it is $w'$ where $$e^{w'} = w.$$ Suppose $w = re^{i\theta}$. Then $$w' = \ln r + i\theta +2ik\pi$$ where $k$ is an arbitrary integer and the $\ln$ is the ordinary real-valued logarithm. (Since $r\ge 0$ this is well-defined everywhere except for $r=0$, in which case we are dealing with $0^z$, which really is ambiguous.)

Putting this back into the original formula we have the answer, that \begin{align} (re^{i\theta})^z & = \exp\{z (\ln r + i\theta + 2ik\pi)\}\tag{\star} \\ & =\exp\{z(\ln r + i\theta)\}\cdot \exp\{2ik\pi\cdot z\} \end{align} where $k$ is an integer.

Now observe that although $(\star)$ seems to list an infinite number of solutions, they are not always distinct. For example, when $z$ is a real integer, the second factor, $\exp\{2ik\pi\cdot z\}$ part is 1 for every choice of $k$, and so can be disregarded.

To find out how many values of $(\star)$ are distinct, one needs to ask about the values of $e^{2ik\pi \cdot z}$. When $z$ has nonzero imaginary part, or is a real irrational, these are all distinct and there are an infinite family of values of $w^z$, given by different choices of $k$. But when $z$ is a real rational number with (lowest-terms) denominator $n$, there are exactly $n$ distinct values.