Consider Euler's Formula: $$e^{i\theta}=cos(\theta)+isin(\theta)$$ Using this, the value of $i$ is $e^{{i\pi}/2}$. Following this: $$i^i=e^{i^2\pi/2}$$ The accepted principal result of this is that $$i^i=e^{-\pi/2}$$ I appreciate that the result given is one of many solutions - but for my question here, I'm only focussing on this case.

Instead of simply using $i^2=-1$, I will separate the exponent for $i^i=e^{i*i\pi/2}$.

Going back to Euler's Formula, and inserting ${\theta=i\pi/2}$ we get: $$i^i=\cos (i\pi/2) + i\sin (i\pi/2)$$ Which gives $$cosh(\pi/2)+i*sinh(pi/2)$$ Or roughly $2.509+2.301i$ Again - I know there are many different solutions here, I'm just focussing on the principal case.

Why is there a contradiction here? What exactly causes my method to not yield the same result? Originally my answer was marked as duplicate - however the thread I was linked to did not contain this method - and the reasons for its failure was said to be due to $ln(i)$ not being well defined over complex space (I used $ln(i)$ in my original question) . Considering that I can now show my method without the use of $ln(i)$, I am wondering if there is another reason why my method is incorrect.

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    $\begingroup$ Because $z^s$ is not well-defined for $z,s$ general complex numbers, and this is because $ln(z)$ is not well-defined for a generic complex number $z$, and that's because a single complex number has many arguments $\endgroup$ Nov 11, 2017 at 17:40

2 Answers 2


The great tragedy of the complex numbers is that, while $e^x$ works out perfectly and even more beautifully than before, we lose its inverse function, $\ln x$. You cannot write $\ln x$ and expect to mean something for complex $x$. Always remember this when working with complex numbers.

The best you can come to defining natural logarithm for all complex numbers is what is called a multi-valued function -- "multi-valued" meaning multiple possible answers. The usual definition is $$ \ln (r e^{i\theta}) := \ln r + i(\theta + 2 \pi k), $$ where $r$ and $\theta$ are real numbers -- and $k$ is any integer. In other words, there are infinitely many answers, one for each value of $k$.

Let's apply this to your question.

Recently, I found something intriguing that stated that for some $k$, due to $e^{ln(k)}=k$, $k^i=cos(ln(k))+i*sin(ln(k))$

This is correct enough, while keeping in mind that on both sides there are many possible values!

I decided to apply this logic to find a value for $i^i$; and so $(e^{ln(i)})^i=i^i$.

There is not just one value; there are multiple values in fact.

Upon following the steps outlined; I achieved $$i^i=cos(i*\pi/2)+i*sin(i*\pi/2)$$

Which gives $$cosh(\pi/2)+i*sinh(pi/2)$$

Or roughly $2.509+2.301i$

That's one possible value :) But there are many possible values. Every time you take a $\ln$, you must add $2 \pi i k$ to the result, choosing a new value of $k$ each time.

Actually I think that you introduced some extraneous solutiions during your calculation, which are not actually solutions. (Every time you take a natural log it is possible to introduce extra solutions.) Usually it is understood that $$ i^i = e^{i \ln i} = e^{-\frac{\pi}{2} + 2 \pi k}, $$ for some $k$, so there is no imaginary part. See here for a derivation of this.


The logarithm of a complex number $z\in \mathbb C$, $\ln(z)$ is not well-defined over the complex space, which occurs because any given complex number has multiple arguments $\arg (z)$.


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