What is $i^{1/i}$

Question

What is $i^{1/i}$?

Step by step walk through would be appreciated.

Tried searching, but didn't find much about this.

Answer 1

There are infinitely many possible values of $i^{1/i}$. They are all real numbers of the form

$$z_k= e^{\pi(4k+1)/2}$$

This is because of the multiple-valuedness of the complex logarithm; the computation explicitly is, for any integer $k$,

$$i^{1/i}=i^{-i} =e^{-i\log i} = e^{-i(\pi i/2+2\pi ik)}=e^{\pi(4k+1)/2}$$

The principal value (taking $k=0$) is $e^{\pi/2}\approx 4.81048$.

Answer 2

As has already been explained, $\dfrac{1}{i} = -i$. So we will evaluate $i^{-i}$.

Let $z = i^{1/i} = i^{-i}$. We'll take the complex logarithm, denoted $\operatorname{Log}$, of both sides. So we have $\operatorname{Log} z = \operatorname{Log} i^{-i} = -i \operatorname{Log } i$. So what is $\operatorname{Log} i$?

The complex logarithm of the complex number $z = a + bi$ is defined to be $\operatorname{Log} z = \ln |z| + i\theta$, where $|z| = \sqrt{a^2 + b^2}$ is the modulus and $\theta$ (also called the "argument" of $z$) is the angle that the point $(a,b)$ makes with the positive $x$-axis. For example, $\theta$ for $-3+3i$ is $3\pi/4$ radians because the point $(-3,3)$ makes an angle of $3\pi/4$ radians with the positive $x$-axis. Note that there are infinitely many values of $\theta$ that could work (just keep adding/subtracting $2\pi$ as often as you'd like) but by convention we take the "branch cut" $-\pi < \theta < \pi$. We can take other branch cuts, but that's beyond the scope of this discussion. Anyway, this $\theta$ is the same $\theta$ you have when you convert a complex number in rectangular form to polar form.

Since $i = 0 + 1i$, then the corresponding point is $(0,1)$. The $\theta$ for this is $\pi/2$ radians. Note also that $|i| = \sqrt{0^2 + 1^2} = 1$. So then $$ \operatorname{Log} i = \ln|i| + i\theta = \ln 1 + i\pi/2 = \frac{i\pi}{2}, $$ because $\ln 1 = 0$.

Putting this together, we have $$ \operatorname{Log} z = -i \operatorname{Log} i = -i \cdot \frac{i\pi}{2} = \frac{\pi}{2}. $$

Finally, $\operatorname{Log} z = \pi/2$ means $z = e^{\pi/2}$. Therefore: $$i^{1/i} = e^{\pi/2}$$

Answer 3

Here's the naive answer (which I admit, was my original answer!)

$$i^{1/i} = i^{-i} = (e^{i\pi/2})^{-i} = e^{-i\cdot i\pi/2} = e^{\pi/2} \approx 4.81$$

The second equality utilizes Euler's identity, $e^{i\theta} = \cos\theta +i\sin\theta$, evaluated at $\theta=\pi/2$.

But we needn't have chosen $\theta=\pi/2.$ Any angle of the form $\theta=\pi/2 + 2\pi n$ with $n\in\mathcal{Z}$ would work. Then, in actuality, the result is multivalued:

$$i^{1/i} = i^{-i} = (e^{i\pi/2 + i2\pi n})^{-i} = e^{-i\cdot i(\pi/2+2\pi n)} = e^{\pi/2+2\pi n}$$

If we had restrictions on $\theta$, then we could confine the answer to be single valued. Note that $n=0$ gives the naive answer in the first line.

Answer 4

Basically, this is a combination of two more well-known problems, 1/i=i and i^i equals about .2.

i^(1/i) is actually i^-i because 1/i is equal to -i. https://www.math.hmc.edu/funfacts/ffiles/20013.3.shtml shows how i^i is in fact equal to a real number. i^(1/i)=i^-i=(i^i)^-1 is equal to the inverse of that real number, which in turn is equal to another real number, about 4.8. If you are looking for a proof of the 1/i=-i thing, go here: https://forums.penny-arcade.com/discussion/140370/1-i-i