# What is the i-th root of the imaginary number i?

## Does it even exist?

For many, the idea of an imaginary number *i = √-1* is a bit mind bending. But even more mind bending is the question of roots of *i*. Probably the most mind-bending question of this nature is to work out the value of the *i*-th root of *i*.

The answer to this question is that there are, in fact, infinitely many *i*-th roots of *i, *all of them positive real numbers ranging from the infinitesimally small to the infinitely large. Here are a few examples, approximated to a few significant figures: 0.00898, 4.8104, 2575.97, 1379411

## What? How?

The idea that the *i*-th root of *i *can be such precise real numbers is a bit counterintuitive, right? But let’s look at how we derive the *i*-th root of *i* and it will make more sense.

First, let’s move from root to index notation:

Now we rationalize the index by multiplying it by *i/i = 1, *and hence we derive the following (noting that *i² = -1):*

## Use of Euler’s Formula

Now, taking Euler’s formula, we know that for any given real value *x*:

So we can conclude from that that when *x = π/2 *radians*, *we know that sin*(π/2)=1 *and cos(*π/2)=0, *so we have:

So putting all this together we have

And there we have it. Calculating this in Python, we have:

`from math import exp, pi`

exp(pi/2)

## 4.810477380965351

We can also verify in Python that raising our answer to the power *i *will return *i *as expected (excusing the round-off error in the real part).

`z = complex(0, 1)`

exp(pi/2)**z

## (6.123233995736766e-17+1j)

## But wait — didn’t you say there were infinitely many solutions?

Yes, I did. Let’s go back to Euler’s formula and note that, for any integer value of *k*, cos(2*kπ) = 1 *and sin(2*kπ) = 0*. (This is basically the cosine of zero and every multiple of a full rotation of 2*π *radians clockwise or anti-clockwise).

Therefore, for any integer value of *k,*