I have no problem accepting that $$i^i = e^{\frac{-\pi}{2}} $$ Here im assuming that $-\pi\leq \theta \leq \pi$
But what I am wondering is that, can someone give me some intuition behind understanding taking powers of a complex number where the power is also complex ? Like I am curious what you used to get an understanding of this property ? I chose $i^i$ because it seems like an easy example to give intuition behind but I am also looking at this from the point of view of also trying to understand say: $$(4+5i)^{(2-3i)}$$ Like how do you get an intuitive understanding of this ?

  • 2
    $\begingroup$ Look into Euler's formua, that should make things more clear: $$e^{i \theta } = \cos (\theta) + i \sin(\theta).$$ $\endgroup$ – Andrew Tawfeek Sep 26 '17 at 21:28
  • $\begingroup$ Also, $a^b = e^{b \ln{a}}$. Understanding complex exponentiation really means understanding the logarithm. And this is a basic topic in complex analysis. It also explains why some calculators give different answers, since definite log outside of the positive reals requires a choice of branch cut. $\endgroup$ – Tac-Tics Sep 26 '17 at 21:34
  • $\begingroup$ @Tac-Tics, while it is true that choosing a "branch cut" makes logarithm unambiguous, doing so is no different from making a fairly random choice, so really is not as explanatory as we'd wish. That is, there is no "canonical" branch cut (although there are traditions, these traditions do not so much reflect mathematical reality as they reflect the impulse to tradition together with an impulse to disambiguate). $\endgroup$ – paul garrett Sep 26 '17 at 22:24
  • $\begingroup$ Actually you should "have (some serious) problem(s) accepting that $i^i = e^{-\pi/2}$" since this is not true. $\endgroup$ – Did Sep 26 '17 at 22:42
  • $\begingroup$ @Did What I meant by my statement is that I don't have a problem with how you would get to that result, I am aware that I didn't include to periodicity, which is why I said in my question I am taking theta to be between $-\pi$ and $\pi$ I was more or less looking to understand complex powers more intuitively $\endgroup$ – Robert Sep 26 '17 at 22:48

I am afraid that the intuition you ask for gets lost somewhere on the way.

We may start by defining $$ \tag 1a^b:=\underbrace{a\cdot \ldots \cdot a}_b$$ for $b\in \Bbb N$ (or maybe define $a^b$ as the cardinality of the set of maps from a set of cardinality $b$ to a set of cardinality $a$, which allows us to use some infinite $b$-values, but poses a restriction on the $a$'s used; for this discussion. I'll stick with $(1)$ as a starting point) and then extend the definition step by step to larger domains, based on a permanence principle: For $(1)$ we prove rules such as $$\tag2a^{b+c}=a^ba^c$$ and extend the definition in a way that (among others) these rules remain valid. For example, this forces us to define $a^{-b}=\frac1{a^b}$ when extending to $b\in\Bbb Z$ (and which also shows that we cannot define $0^b$ for $b<0$).

Sooner or late, the only way of extending the domain is via the exponential function $$\exp(x)=\sum_{k=0}^\infty\frac{x^k}{k!} $$ which is fortunately convergent for all $c\in\Bbb C$ (and even in many situations where $x$ is not even a number, thus allowing us to define exponentiation even in some at the first glance strange areas). Among the many fascinating properties of $\exp$ is $$\exp(x+y)=\exp( x)\exp (y),$$ just as needed for $(2)$. But that requires us to use the inverse function $\ln$ and set $$\tag3a^b=\exp(b\ln a).$$ This is all fine, but $\ln a$ cannot be defined at for at $a=0$ (thus we better still use $(1)$ instead of $(3)$ in that case) and $\ln$ can only be defined "consistently" in a simply connected domain (not containing $0$). The most common convention(!) to pick such a domain as large as possible is to remove the non-positive reals from $\Bbb C$. That is, for all other $z$, $\ln z$ is a complex number $w$ with $\exp(w)=z$ and $-\pi<\arg w<\pi$. As $\exp$ is its own derivative, we quickly find that $\exp''(ix)+\exp(ix)=0$ and by comparing the initial conditions, find $\exp(ix)=\cos(x)+i\sin(x)$. Hence $\exp(i\frac\pi2)=i$, so that our chosen convention tells us $\ln(i)=i\frac\pi2$ and ultimately $$ i^i=\exp(i\cdot i\tfrac\pi2)=\exp(-\tfrac\pi2).$$ I think it is the forced use of some convention when picking a good domain in $\Bbb C$ which ultimately kills the intuition you are looking for. Note that I consequently wrote $\exp(x)$, not $e^x$ in order to avoid the idea that the exponential function "is" taking some number dubbed $e$ to some power.


To calculate $z^u=(4+5i)^{2-3i}$ you have to use three formulas

  • convert $z=(4+5i)$ from rectangular form to polar form $z=re^{i\theta}$

  • calculate $\ln(z)=\ln(r)+i\theta+2ik\pi\qquad k\in\mathbb Z$

  • calculate $\displaystyle z^u=e^{u\ln(z)}$

Let's go:

$r^2=|z|^2=4^2+5^2=41\implies r=\sqrt{41}$

Since $z$ is in the first quadrant then $\theta=\tan^{-1}(\frac 54)$

$\begin{array}{ll}z^u &=\exp\bigg((2-3i)\left(\ln(r)+i\theta+2ik\pi\right)\bigg)\\&=\exp\bigg(\left(2\ln(r)+3\theta+6k\pi\right)+i\left(-3\ln(r)+2\theta+4k\pi\right)\bigg)\\ &=r^2\ e^{3\theta}\ e^{6k\pi}\bigg(\cos\big(2\theta-3\ln(r)\big)+i\sin\big(2\theta-3\ln(r)\big)\bigg) \end{array}$

And from there you can calculate a numerical value by replacing $r,\theta$ by their respective value:

$(4+5i)^{2-3i}\approx e^{6k\pi}\left(-484.77 + 358.42\,i\right)$

Note that $\ln(z)$ is multivalued since it is the reciprocal function of exponential which is $2i\pi$ periodic.

So in the end you get a factor $e^{6k\pi}$ and a principal value $[z^u]_{k=0}=-484.77+358.42i$

You need to keep this factor for things like $z^u\times z^v=z^{u+v}$ to stay true.

If you only consider the principal value then $[z^u]_{k=0}\times[z^v]_{k=0}=[z^{u+v}]_{k=0}$ may be false.

You have to see it like $\exists (k_1,k_2,k_3)\in\mathbb Z^3\mid\ [z^u]_{k=k_1}\times [z^v]_{k=k_2}=[z^{u+v}]_{k=k_3}$

Now that you understood the principle, let's do it for $i^i$ :

$\displaystyle i^i=\exp\bigg(i\ln(e^{i\frac{\pi}2})\bigg)=\exp\bigg(i(\ln(1)+i\frac{\pi}2+2ik\pi)\bigg)=\exp\bigg(-\frac{\pi}2-2k\pi\bigg)=e^{-\frac{\pi}2}e^{-2k\pi}$

So the principal value is $[i^i]_{k=0}=e^{-\frac{\pi}2}$ and the factor is $e^{2k\pi}$

[note that since $k$ is arbitrary, $2k\pi$ or $-2k\pi$ does not matter].

To conclude let's calculate $i^2$. Should it also have multiple values ?

$\displaystyle i^2=\exp\bigg(2\ln(e^{i\frac{\pi}2})\bigg)=\exp\bigg(2(\ln(1)+i\frac{\pi}2+2ik\pi)\bigg)=\exp\bigg(i\pi-2ik\pi\bigg)=\underbrace{e^{i\pi}}_{-1}\underbrace{e^{2ik\pi}}_{1}=-1$

This time it has only one value as expected $i^2=-1$, because the factor $e^{2ik\pi}=1$ for any value of $k$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.