# How do I compute the square root of I?

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

 jcaron2 Posts: 983, Reputation: 1034 Senior Member #2 Jun 23, 2011, 08:17 PM
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!
 gmandelbraat Posts: 8, Reputation: 15 Junior Member #3 Jun 23, 2011, 08:32 PM
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?
 jcaron2 Posts: 983, Reputation: 1034 Senior Member #4 Jun 23, 2011, 08:36 PM
Quote:
 Originally Posted by gmandelbraat 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?
Yes! Very good!

And just to check, we can do

$(2.12 - 0.71i)^2=2.12^2-2(2.12)(0.71i)+(0.71i)^2 = 4.5 -3i -0.5 = 4 - 3i$.
 jcaron2 Posts: 983, Reputation: 1034 Senior Member #5 Jun 23, 2011, 08:43 PM
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

$\sqrt{4-3i}=\pm(2.12-0.71i)$
 gmandelbraat Posts: 8, Reputation: 15 Junior Member #6 Jun 23, 2011, 08:45 PM
This is all making sense now! Does that mean there are also two answers for cube root?
 jcaron2 Posts: 983, Reputation: 1034 Senior Member #7 Jun 23, 2011, 08:57 PM
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?
 gmandelbraat Posts: 8, Reputation: 15 Junior Member #8 Jun 23, 2011, 09:02 PM
Yes! Thank you so much! U are a genius!

Now when someone asks me what's cube root of 8, I will say -1 + or - sqrt(3)i!
 jcaron2 Posts: 983, Reputation: 1034 Senior Member #9 Jun 23, 2011, 09:04 PM
You did very well to pick up on this concept so quickly! Great job!
 Unknown008 Posts: 8,147, Reputation: 3745 Uber Member #10 Jun 24, 2011, 07:42 AM
Great walkthrough Josh

And congratulations gmandelbraat!

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