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

## Does it even exist?

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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, piexp(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(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,

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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