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# What is i to the power of i?

Disclaimer: knowledge of complex numbers and trigonometry is needed to fully appreciate this note.

Some of you may know the definition of $$i$$, the imaginary unit. For those who don't, $$i=\sqrt{-1}$$. But what is $$i^i$$? Is it an imaginary number? Perhaps even one step deeper than imaginary, a quaternary number?

In fact, $$i^i$$ is a real number! It approximately equals $$0.20788$$.

But how can that be? How can an imaginary number to the power of an imaginary number be real? How did I arrive at that inexplicable answer?

To find the value of $$i^i$$, we'll first need to know Euler's Formula.

Euler's Formula states that $$e^{ix}=\cos{x}+i\sin{x}$$. The right hand side can be shortened to $$\text{cis }{x}$$ (just notation, don't get confused), and I'll be using this shorthand for brevity.

Note that when $$x=\dfrac{\pi}{2}$$, then $$\text{cis }{x}=\cos{\dfrac{\pi}{2}}+i\sin{\dfrac{\pi}{2}}\implies\text{cis }{x}=i$$. Therefore, to find $$i^i$$, we can simply take $$\text{cis }{\dfrac{\pi}{2}}$$ to the $$i$$th power. However, since $$\text{cis }{\dfrac{\pi}{2}}=e^{i\frac{\pi}{2}}$$, then

\begin{align*}i^i&=(\text{cis }{\dfrac{\pi}{2}})^i\\ &=e^{i^2\frac{\pi}{2}}\\ &=e^{-\frac{\pi}{2}}\\ &\approx 0.20788\end{align*} and we are done.

However, the astute reader would notice something wrong with my argument. When I said $$x=\dfrac{\pi}{2}$$ gives $$\text{cis }{x}=i$$, I was only considering one scenario of the infinite possible $$x$$ that satisfy $$\text{cis }{x}=i$$. What about $$x=\dfrac{5\pi}{2}$$? How about $$x=\dfrac{-3\pi}{2}$$? Any $$x$$ satisfying $$x=\dfrac{\pi}{2}+2\pi n$$ for some integer $$n$$ would work, and every single $$n$$ gives a different value for $$i^i$$. Which one is the correct answer?

The short answer is: all of them are equally correct. But how can that be? Shouldn't $$i^i$$, a constant to the power of a constant, yield only one value? One reason that $$i^i$$ might not give what is expected is the fact that it is to the power of $$i$$. What does it mean to take an $$i$$th power? The fundamental meaning of taking "to the power" breaks down when you consider this.

You can make this weird fact more acceptable to yourself by asking yourself: What are all the possible values of $$4^{\frac{1}{2}}$$? There are, in fact, two values: $$-2$$ and $$2$$.

Therefore along those lines of reasoning, $$i^i=0.20788$$, given by $$n=0$$. Also, $$i^i=111.31778$$, where $$n=-1$$. $$i^i=0.00039$$, where $$n=1$$.

But much like the principle value of $$4^{\frac{1}{2}}=2$$, we also have the principle value of $$i^i$$, which is when $$n=0$$. So, finally, we have shown that $$i^i$$ does indeed equal $$0.20788$$. $$\Box$$

Got any questions? Feel free to ask below.

Note by Daniel Liu
3 years, 3 months ago

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Thanks for reading through my second #CosinesGroup post, this time about imaginary powers (specifically $$i^i$$). I hope you all found it quite intriguing and informative. If you have any questions or feedback, feel free to post.

Cheers! · 3 years, 3 months ago

I know this probably isn't fit for #CosinesGroup, but I'll post it anyways. Just recall that for complex $$x,y$$, we can define $$x^y = \exp(y \log x)$$. Also, remember that the complex logarithm is defined as $$\log x = \log|x| + i \arg|x|$$. So if we fix the range of the argument then we can perfectly well define $$x^y$$ as we wish.

Just sub in the values of $$x,y = i$$ to get:

$$i^i = \exp(-\arg i)$$.

And we are done. · 3 years, 3 months ago

Good job!Keep these kind of posts coming! · 3 years, 3 months ago

Thank you very much..amazing note..I was discussing this concept with my friends in my coaching classes..now I understand it completely..thank u sir..@Daniel Liu · 10 months, 3 weeks ago

@Daniel Liu Can you add this to a suitable skill in the Complex Numbers Wiki? Thanks Staff · 2 years, 5 months ago

I don't think that √4 is equal to both 2 & -2 Rather its actually just 2 since √x is always pos. For real x.

Otherwise , nice thoughts !!!! · 2 years, 11 months ago