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

Does it even exist?

Keith McNulty


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

## 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)
## (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(2kπ) = 1 and sin(2kπ) = 0. (This is basically the cosine of zero and every multiple of a full rotation of 2π radians clockwise or anti-clockwise).

Therefore, for any integer value of k,



Keith McNulty

Pure and Applied Mathematician. LinkedIn Top Voice in Tech. Expert and Author in Data Science and Statistics. Find me on LinkedIn, Twitter or keithmcnulty.org