# Euler's formula and $i^x = \cos(x \cdot \frac{\pi}{2})$

While playing around with a plotting software, i just found out that

$$f(x) = i^x = \cos(x·\frac{\pi}{2})$$

1. How does this connect to Euler's formula?
2. Obviously, here, the alternating sign change is responsible for periodicity and form of the cosine. Is this also true for Euler's formula?

Please don't beat me, i'm an engineering student.

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The link to Euler's formula is that $i = e^{i\pi/2}$, so $i^x = e^{i\pi x/2}$. –  froggie May 28 '12 at 13:07
Did you try to plot it for, say, $x=1/2$? Do you reckon $\sqrt i=\cos(\pi/4)$? –  Gerry Myerson May 28 '12 at 13:08
This formula is completely wrong... Or maybe you're saying $\Re(i^x) = \cos({\pi x \over 2})$? –  Najib Idrissi May 28 '12 at 13:11
Yes, right, it has to be just the real part! –  bijan May 28 '12 at 13:14
What about $$f(x) = i^x = \cos(x·\frac{\pi}{2}) + i\sin(x·\frac{\pi}{2})$$ ? –  bijan May 28 '12 at 13:18

The quantity $i^x$ by itself is not well-defined. The way one would like to define it is $i^x = e^{x\log i}$, and then use the Taylor series for the exponential to compute $e^{x\log i}$. The problem with this is that $\log i$ is not well-defined: there are infinitely many possible values of $\log i$, namely $$\log i = \frac{\pi i}{2} + 2\pi in$$ for any $n\in \mathbb{Z}$. Thus to define $i^x$, you have to make a choice as to which one of these logarithms you are using. The standard choice would be $\log i = \pi i/2$. In this case, $$i^x = e^{x\log i} = e^{i\pi x/2} = \cos(\pi x/2) + i\sin(\pi x/2).$$ However, if you had chosen $\log i = \pi i/2 + 2\pi in$ for some $n\neq 0$, then $$i^x = e^{x(\pi i/2 + 2\pi in)} = \cos(\pi x/2 + 2\pi nx) + i\sin(\pi x/2 + 2\pi nx).$$
@bijan No, a complex valued function is a mapping of plane to plane. There are different ways to visualize such mappings, e.g., plotting $\Re (f(z))$ or $|f(z)|$. –  Artem May 28 '12 at 14:07