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Today I was thinking about imaginary numbers and hypothesized the graph of $i^x$ would be sinusoidal based on the simple exponent rules of i. I think this can be explained with logarithms, which I've never studied in depth so cannot be sure.

I tried graphing real and imaginary parts of $i^x$ with sympy (a python symbolic maths package) and did end up with this sinusoidal curve.

I'm wondering if someone here can comment on this curve - if there is some formula involving a trigonometric function that is equal to $i^x$, how one would find the derivative of $i^x$, and any other related ideas. Also, links to relevant literature would also be great.

enter image description here

Update: I was encouraged to define $Re(i^x)$ which would make the answer clear to me.

$$\operatorname{Re}(i^x)=\operatorname{sign}(i^x)^x$$

Additionally, according to sympy, $\frac{d}{dx}[i^x]=\frac{i \pi}{2} i^{x}$. This just leads to more questions.

Where can I read more about these issues?

While I invite all criticism, please take into account that I'm in 10th grade and haven't any serious formal math schooling and so am prone to make beginners mistakes. Asking these questions is my attempt to simulate the rigorous education in math I hope to receive in college

Michael Hardy
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theideasmith
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    I’d be happier if you defined precisely what $i^x$ is, for real $x$. Once you do, I think your answer will pop out at you. After all, $i^{1/2}$ will usually be understood as a square root of $i$, and there are two equally good candidates there. – Lubin Mar 16 '16 at 01:17
  • @theideasmith $i^x$ is not a well-defined function $\Bbb R\to \Bbb C$, namely because $$i=e^{2ki\pi+\frac\pi2i}$$ for any $k\in\Bbb Z$, but you can easily see that the maps $$x\mapsto e^{\frac\pi2ix}\ x\mapsto e^{2\pi ix+\frac\pi2ix}$$ are different: for instance, they take different values at $x=\frac12$ (and for almost all the non-integer values). –  Mar 16 '16 at 01:21
  • The same issue carries on to real and imaginary parts: $$\cos\left(2\pi x+\frac\pi2x\right)\ne \cos\left(\frac\pi2x\right)\quad\text{for almost every }x\in\Bbb R$$ –  Mar 16 '16 at 01:28
  • However, I think the thing you are using to run the calcs is assuming the notation $$i^x:= e^{\frac{\pi}{2}ix}=\cos\left(\frac\pi2x\right)+i\sin\left(\frac\pi2x\right)$$ ($x\in\Bbb R$) which should also explain quite clearly the graphics you get. –  Mar 16 '16 at 01:30
  • @G.Sassatelli can you recommend specifically what learning path I can take to really understand these issues? – theideasmith Mar 16 '16 at 01:37

2 Answers2

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$$ \begin{array}{c|r} x & i^x \\ \hline 0 & 1 \\ 1 & i \\ 2 & -1 \\ 3 & -i \\ 4 & 1 \\ 5 & i \\ 6 & -1 \\ 7 & -i \\ 8 & 1 \\ \vdots & \vdots \end{array} $$ It goes around in circles. If you want a well-behaved function extending this to $x\in\mathbb R$, you just need something that goes around in circles at that same rate, thus $$ x\mapsto \cos\left(\frac \pi 2 x\right) + i \sin\left( \frac \pi 2 x \right) = e^{i(\pi/2)x} = e^{i\pi x/2}. $$

Michael Hardy
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    It might be instructive to mention that there are several branches of $i^x$ since $$i^x=e^{x\log(i)}=e^{i\pi x/2}e^{i 2n\pi x}$$ – Mark Viola Mar 16 '16 at 02:49
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As the commentators have noted, a (probably better) way of writing the function you are interested in is $$f(x)=e^{\pi ix/2},$$ which should make sense if we know $i=e^{\pi i/2}$ and apply properties of exponents. It is likely that your calculator has interpreted $i^x$ in this way for you in making your graphs.

The reason for the observation that $i=e^{\pi i/2}$ is the same one that will show that your graphs are in fact $\sin$ and $\cos$ functions. We have Euler's Formula,

$$e^{ix}=\cos x+i\sin x,$$

which holds in particular for real $x$. Thus $\Re [f(x)]=\cos(\frac {\pi x}{2})$. As is often the case, the relevant Wikipedia article is a great introduction and even a couple of good proofs. Knowing a little calculus will be helpful...

https://en.wikipedia.org/wiki/Euler%27s_formula

It should be noted that the equation above is one of great beauty and not necessarily obvious; I consider its discovery to have been a great accomplishment. Happy mathing

D. Wagner
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  • It might be instructive to mention that there are several branches of $i^x$ since $$i^x=e^{x\log(i)}=e^{i\pi x/2}e^{i 2n\pi x}$$ – Mark Viola Mar 16 '16 at 02:49