How can I compute sqrt(I)? How about the cube root or fourth root?

First, it helps to visualize complex numbers in a way you may not have done before. I'm sure you're familiar with a number line, where you write numbers on a horizontal line like.. -3, -2, -1, 0, 1, 2, 3,. It's simple enough to plot numbers on a number line; it's basically a one-dimensional graph.

In the case of complex numbers, you can visualize the real numbers on the normal horizontal number line and the imaginary numbers on a vertical number line, crossing at zero. In other words, picture a simple two-dimensional plot. Rather than getting plotted on a simple line, complex numbers are plotted on a plane. The real numbers correspond to the x-axis, and the imaginary correspond to the y-axis. To plot a number like 3-4i on this complex number plane, you'd go 3 units to the right and 4 units down and mark a dot. Again, just like plotting points on a simple cartesian graph.

So what does this have to do with finding the square root of a complex number? Well, you can think of any complex number as having a magnitude and an angle. The magnitude is the distance from the origin to the point, and the angle is the angle between the positive real axis and the arrow pointing from the origin to the point. When raising a complex number to a particular power (and remember that a square root is the same as raising to the 1/2 power), you have to do two things: First, you have to raise the magnitude to that power (in the old fashion sense, since the magnitude is just a positive real number). And second, you have to multiply the angle by that power.

So let's try it for your example: what's the square root of i?

First, what's the magnitude of i? On the complex number plane, i is 1 unit directly up from the origin. That means the distance from the origin to i is 1.

How about the angle? Well, since it's directly north of the origin, it's 90 degrees (pi/2 radians) away from the positive real axis.

So to find the square root of i, first we find the square root of the magnitude, which is just 1. Then we multiply the angle by 1/2 (since square root is the same as raising it to the 1/2 power), which results in 45 degrees.

So our answer is the point on the complex plane with a magnitude of 1 and an angle of 45 degrees from the positive real axis. So what are the coordinates of the point at the 45 degree mark on a circle with a radius of 1? The x-component (i.e. real part) is 1 * cos(45) = sqrt(2)/2, and the y-component (i.e. imaginary part) is 1 * sin(45) = sqrt(2)/2.

Thus \sqrt{i}=\frac{\sqrt 2+i \sqrt 2}{2}

How about the cube root? The magnitude is still 1, but now the angle is 30 degrees. So the answer is

\sqrt[3]{i}=1 \angle 30 = \frac{\sqrt 3 + i}{2}

For the fourth root it would be:

\sqrt[4]{i}=1 \angle 22.5 = \cos 22.5 + i \sin 22.5}

Note that this also works for exponents greater than 1. For example, we can use this technique to find i squared. Once again the magnitude is 1, now the angle is doubled to 180. Where does that put it on the complex number plane? At the point -1, exactly where you'd expect it to be!

Wow! Thanks!

So to compute sqrt(4-3i), I first find magnitude = sqrt(4^2+3^2)=5. Then I find angle = atan(-3/4) = -36.87. So answer is sqrt(5)*(cos(-18.44)+ I*sin(-18.44)) = 2.12 - 0.71i?

Wow! Thanx!

So to compute sqrt(4-3i), I first find magnitude = sqrt(4^2+3^2)=5. Then I find angle = atan(-3/4) = -36.87. So answer is sqrt(5)*(cos(-18.44)+ i*sin(-18.44)) = 2.12 - 0.71i?

Yes! Very good!

And just to check, we can do

(2.12 - 0.71i)^2=2.12^2-2(2.12)(0.71i)+(0.71i)^2 = 4.5 -3i -0.5 = 4 - 3i.

And one more thing: your calculator told you that the angle for 4-3i was -37 degrees. However, it also could have been 323 degrees. If you divide that in half to take the square root, you get 162 degrees, which is exactly 180 degrees away from what you calculated, resulting in an answer of -2.12 + 0.71i. So, just like in the case of square roots of real numbers, the true answer is

\sqrt{4-3i}=\pm(2.12-0.71i)

This is all making sense now! Does that mean there are also two answers for cube root?

I'm glad it's making sense for you. :)

Actually, for the cube root there are three answers. In the original example I said the angle for I was 90. However, it also could have been -270, in which case the answer would be -i. Or the angle could also be written as 450, in which case the answer would come out different still.

Notice that in the case of the square root, there were two answers 180 degrees apart. In the case of the cube root, there are three answers 120 degrees apart. For the fourth root, there would be four answers 90 degrees apart.

Hence for the nth root, there are n different roots, all mutually 360/n degrees apart. It makes sense, since when you raise the root to the nth power (to get back to the original number), all those increments of 360/n wash out as even multiples of 360 degrees.

Make sense?

Yes! Thank you so much! You are a genius!

Now when someone asks me what's cube root of 8, I will say -1 + or - sqrt(3)i!

You did very well to pick up on this concept so quickly! Great job!

Great walkthrough Josh :)

And congratulations gmandelbraat! :)

Thanks Jerry.

It's nice to see somebody grasp abstract concepts so quickly and easily. Gmandelbraat clearly has a pretty sharp mathematical mind!

Anyway, your method is simpler than mine, which I was taught as:

Let \sqrt{z} = a+bi

z is anything that you are taking the square root.

Then, z = (a+bi)^2

In this case...

i = (a+bi)^2

And,

i = a^2 - b^2 + 2abi

Equating coefficients;

a^2 - b^2 = 0

i = 2abi \rightarrow 1 = 2ab

And from there solve simultaneously. Of course, this gets more tedious as the we take larger values of n for the nth root :)

True. I suppose that would get pretty tedious for an 8th root or something, but it's still an effective technique. That's what I love about higher level math. There are lots of different techniques that can be applied to solve problems. It goes from being the epitome of left-brain subjects, into something where you get to apply creativity or even art!

Good explanation JC. The only quibble I would have is that it would help to let the OP know that you are describing "de Moivre's Theorem." You express a number in the complex plane using polar coordinates, and then:


[R(cos \theta + i \sin \theta)]^n = R^n (\cos (n \theta) + i \sin (n \theta))

Good point EB.

When I get some free time, I was also going to explain to the OP about raising a number to a complex exponent, at which point I was going to describe Euler's formula, from which de Moivre can easily be demonstrated.

I am very curious to learn about complex exponents.

I am very curious to learn about complex exponents.

Gmandelbraat, I probably won't be able to answer this morning, but ask this question in a separate thread. It's closely related to your original question, but I think it would be better if you asked it as a new question in case somebody has the same question in the future.

Meanwhile, Ebaines, Galactus, Unknown008 or another one of the other brilliant people that inhabit these forums may have time to answer for you. :)

There are lots of resources on the web - just Google "complex exponents." But basically Euler's formula:


e^{i \theta} = \cos \theta + i \sin \theta


leads to all sorts of interesting things, including logarithms of complex numbers.