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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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3  
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
1  
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
2  
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

2 Answers 2

up vote 4 down vote accepted

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 $$i^x = e^{x(\pi i/2 + 2\pi in)} = \cos(\pi x/2 + 2\pi nx) + i\sin(\pi x/2 + 2\pi nx).$$

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My answer is not about your questions, which are wonderfully addressed in the answer and comments before me, but about your discovery. This should be a good warning: a lot of computer systems plot only the real part of a given expression. For example, it is true for Maple. Therefore, students should be careful because sometimes what is shown in the plot is not exactly in the original expression.

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1  
Yes, exactly! Ideally, in this case, the program should switch to 3d space and add an imaginary axis, right? –  bijan May 28 '12 at 13:56
    
@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

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