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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
2 years, 10 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! Daniel Liu · 2 years, 10 months ago

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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. Anqi Li · 2 years, 10 months ago

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Good job!Keep these kind of posts coming! Bogdan Simeonov · 2 years, 10 months ago

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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 Rishabh Tiwari · 5 months, 3 weeks ago

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

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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 !!!! Vishal Sharma · 2 years, 6 months ago

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@Vishal Sharma Yeah it is principal root Ashish Siva · 5 months, 2 weeks ago

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