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


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.

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    $\begingroup$ You may find this of assistance: math.stackexchange.com/questions/9776/… $\endgroup$ Aug 27, 2013 at 2:24
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    $\begingroup$ It may help to separate the complexity into simpler parts: $(1+i)^3(1+i)^{4i}$ as a start... $\endgroup$
    – abiessu
    Aug 27, 2013 at 2:26

5 Answers 5


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

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

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


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

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

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


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)
  • $\begingroup$ I think you meant to write, “then so is $w\cdot\exp(\alpha+2n\pi i)$”, which also affects your last bulleted statement. $\endgroup$
    – Lubin
    Aug 27, 2013 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.


let's to establish general formula.Suppose we have $(c+di)^{(a+bi)}$,Then we proceed as fellow

we take as $Z=(c+di)^{(a+bi)}$ Taking $\log$ on both side $$ \log Z=\log(c+di)^{(a+bi)}$$ $$=(a+bi)log(c+di)$$ $$=(a+bi)(lnr+i\theta)$$ where $$r=\sqrt{c^{2}+d^{2}},\quad\theta=\tan^{-1}(\frac{d}{c})$$ so lastly $$ Z=\exp^{(a+bi)(\ln r+i\theta)}$$


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