What are the square roots of i?

Now let us turn our attention to this exercise in complex numbers, to find the square roots of a complex number. It is actually a rather simple matter, as we'll see. To keep the matter even simpler, we will start with the simplest imaginary numbers, i and -i.

Notation: sqrt(n) means the principal (i.e., positive) square root of a real number n.

First I will derive that [sqrt(2)/2 + i sqrt(2)/2] and its negative, -[sqrt(2)/2 + i sqrt(2)/2], are the two square roots of i. This problem can of course be restated as that of finding a value for x such that x2 = i. Which is the same as finding the roots of the second degree polynomial x2 - i = 0. There are a few different approaches to solve this problem, I will present two obvious ways.

Method I: Let the complex number a + bi be a square root of i, where a and b are two real numbers. Then,

(a + bi)2 = i.

Expand the left-hand side to get (a + bi)2 = a2 + 2abi + b2i2 = (a2 - b2) + 2abi. Equating this to the right-hand side, which can be written as 0 + 1i, by comparing the real part against the real part and the imaginary part against the imaginary part. So,

(a2 - b2) + 2abi = 0 + 1i.

We see that the real parts give the equation a2 - b2 = 0, and the imaginary parts give 2ab = 1. The first of the two equations says that a2 = b2, i.e. a = b or a= -b. Substitute these two relations into the second equation to get 2a2 = 1 and -2a2 = 1, respectively. The last relation, -2a2 = 1, or a2 = -1/2, should be discarded since a is a real number, therefore its square will never be negative. Thus we only need to solve:

a = b, and 2a2 = 1.

There are two solutions: a = b = sqrt(2)/2, or a = b = -sqrt(2)/2. Respectively, they give the two square roots of i described above: [sqrt(2)/2 + i sqrt(2)/2] and -[sqrt(2)/2 + i sqrt(2)/2].

Method I above is a straight forward attempt to solve the equation x2 = i. We assumed the form of the solution, a + bi, and then plug it in to see what specific values the coefficients a and b have to be. Earlier we have also noted that the problem is also equivalent to finding the roots of the polynomial x2 - i = 0. So one might reasonably think that we can also use the trusty quadratic formula to solve this problem. Unfortunately, using the quadratic formula here will only give you the already-known information that x = (+/-) sqrt(i), without giving the more explicit description that we desire. The next method relies on slightly more advanced mathematical knowledge, but it is ultimately more useful than the first one.

Method II: This method uses the exponential function exp(x) = ex, and the Euler's Formula:

exi = cos(x) + i sin(x).

We see that the number i is equal to e(pi/2)i, because e(pi/2)i = cos(pi/2) + i sin(pi/2) = 0 + 1i. It is also a fact that any complex number can be represented as rexi = rcos(x) + ri sin(x). Where the value of r and x are just the radius and the angle in polar coordinates, on the Cartesian plane, given the usual mapping that represents a complex number a + bi on the complex plane by the point (a,b).

Therefore, the problem is reduced to finding a complex number in the form rexi such that its square (rexi)2 is equal to i, or e(pi/2)i. Put this formulation into an equation, we get:

(rexi)2 = e(pi/2)i.

Expand the left-hand side (recall the laws of exponential function, that (ez)2 = e2z),

r2e2xi = e(pi/2)i.

We can see easily that r has to be either 1 or -1. Then we compare the two sides' exponents. We have 2x = pi/2, or x = pi/4. Therefore, the square roots of i are (+/-) e(pi/4)i. Using the Euler's formula again to change this back into the form a + bi, and we get that the desired square roots are e(pi/4)i = cos(pi/4) + i sin(pi/4) = [sqrt(2)/2 + i sqrt(2)/2], and -e(pi/4)i = -cos(pi/4) - i sin(pi/4) = -[sqrt(2)/2 + i sqrt(2)/2].

Alternatively, at the step r2e2xi = e(pi/2)i we can restrict the value to be the principal square root only when solving r2 = 1, i.e., r = 1. We then find the second square root by examining the possible angles: what are the possible values of the angle x so that 2x is equivalent to pi/2? The answers are pi/4 (which gives our first square root) and 5pi/4. The second angle, in the third quadrant, yields the second square root. While this might seem unnecessary, given the ease of the original way, it is actually the more useful approach, usable in more general problems of similar type. One such generalized problem will be explored next time.

Lastly, using either of the above methods we can calculate that the square roots of -i are [sqrt(2)/2 - i sqrt(2)/2] and [- sqrt(2)/2 + i sqrt(2)/2]. The detail will be left as a simple exercise.

Next time: How to find the n-th roots of a complex number?