I am currently studying complex numbers. Recently I saw terms like this: $(a+ib)^{c+id}$ , actually, I was simplifying them.

But it was okay till I arrived at the following: $$i^i=(e^{i\pi/2})^i=e^{-\pi/2}$$ I think it's a real number, Wolfram Alpha says it's a transcendental and $$e^{-\pi/2}\approx0.2078797...$$ I guess it's strange.

So, My questions are

1. What exactly is meant by expressions like $(a+bi)^{c+di}$ ?

2. How come some imaginary number raised to some imaginary power can yield a real number ??

  • 2
    $\begingroup$ mathworld.wolfram.com/ComplexExponentiation.html $\endgroup$ – Yiyuan Lee Oct 20 '14 at 13:15
  • $\begingroup$ @YiyuanLee Thanks, but see, $x^2$ is basically $x\times x$, can you explain it that way? $\endgroup$ – user171358 Oct 20 '14 at 13:19
  • $\begingroup$ In complex exponentiation $w^z$ can be defined as $e^{z\ln w}$, where $\ln w$ is a logarithm of $w$. Note that the logarithm of a complex number is multi-valued. That is, raising a complex number to a complex exponent can yield more than one value. $\endgroup$ – Yiyuan Lee Oct 20 '14 at 13:26

Fisrt, you have to define the logarithm of a complex number.

The basic idea is identical to the real logarithm, that is:

$$\log z=w\iff e^w=z$$

The problem is that, unlike the real exponential function, the complex exponential function is not bijective. In fact, we have: $$e^z=e^w\iff \exists n\in\Bbb Z:z-w=2\pi ni$$

Thus, we could say that a complex number has infinitely many logarithms. But if we fix an interval (being its length $2\pi$) for the imaginary part of the logarithm, then the logarithm is unique, with this restriction. If we fix the interval $(-\pi, \pi]$, this is the main branch of the logarithm.

For example: $e^{i9\pi/4}=\sqrt2/2+i\sqrt2/2$. But $\log(\sqrt2/2+i\sqrt 2/2)=i\pi/4$, because $9\pi/4>\pi$, but $9\pi/4-2\pi=\pi/4$ is in the proper interval.

Now that we have a logarithm we can define: $$z^w=e^{w\log z}$$ for all $z,w\in\Bbb C$.

Then: $$i^i=e^{i\log i}=e^{i(i\pi/2)}=e^{-\pi/2}$$

  • $\begingroup$ Don't you mean 9π/4>2π? $\endgroup$ – Brad Oct 20 '14 at 13:43
  • $\begingroup$ @Brad No, the main branch has the imaginary part between $-\pi$ and $\pi$ $\endgroup$ – ajotatxe Oct 20 '14 at 13:44

In general, a complex exponent is ill-defined. One exception is $e$, where Taylor expansions can be used to prove that $e^{a + bi} = e^a(\cos b + i\sin b)$.

For other numbers, complex logarithms are used, and complex logarithms are tricky animals, since there often isn't one unique answer to "What power of $e$ gives me x?" If you get the answer $\ln(z) = a + bi$, then the number $a + (b + 2\pi)i$ is just as valid.

That said, let's try to calculate $i^i$. We get $$ i^i= e^{i\ln i}\\ = e^{i\cdot \pi i(2n + 1/2)}\\ = e^{-\pi(2n + 1/2)} $$ which means that any integer value of $n$ gives a consistent value of $i^i$. It could be $e^{-\pi/2}$. But it could also be $e^{3\pi/2}$, or $e^{115\pi/2}$.

  • $\begingroup$ $e$ is not "the one exception", but it is a notable exception. $\endgroup$ – Omnomnomnom Oct 20 '14 at 13:39
  • $\begingroup$ I think that $e$ isn't really an exception. You can still say that $e^2=e^{2\log e}=e^{2(1+2\pi ni)}$. The problem is that $z^w$ is an ambiguous notation if $z$ is a real positive number. But when this occurs, every log branch gives the same result when computing powers. $\endgroup$ – ajotatxe Oct 20 '14 at 13:48

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