## Sunday, February 7, 2021

Published February 07, 2021 by with 1 comment

# If the square root of -1 is i, what is the cube root of -1?

You probably learned at some point that the square root of -1 is i. What about the cubed root of it? There's the obvious answer of (-1)^3 = -1, but the answer isn't actually that simple.
To answer this, we'll need Euler's identity which is:
$e^{i\pi}=-1$
Just take the cubed root of each side:
$e^{i\pi*\frac{1}{3}}=-1^{\frac{1}{3}}$
$e^{\frac{i\pi}{3}}=-1^{\frac{1}{3}}$
Now we just need the following definition:
$e^{ix}=cos(x)+i*sin(x)$
Plugging in our value:
$e^{\frac{i\pi}{3}}=-1^{\frac{1}{3}}$
$cos(\frac{\pi}{3}) + i*sin(\frac{\pi}{3})=-1^{\frac{1}{3}}$
$\frac{1}{2} + i*\frac{\sqrt{3}}{2}=-1^{\frac{1}{3}}$
And that's it...there's another cube root of -1.

What does that actually mean? Consider this coordinate system:

With real numbers on the horizontal axis and imaginary numbers on the vertical axis, you can draw complex numbers as vectors. This has a cool property. We got pi/3 radians as our angle there. That's equal to 60 degrees, or one-sixth of a full rotation. Looking at that coordinate system, if r = 1:

• 0 degrees = 1
• 90 degrees = i
• 180 degrees = -1
• 270 degrees = -i
• 360 degrees = 1
• 450 degrees = i
• ...
It rotates around. Since an angle of pi/3 represents 60 degrees, cubing the value with r = 1 and angle = 60 degrees gives you the same thing as r = 1 and angle = 180 degrees, which is -1.

Thinking through it a bit more, that's not unique. What if we used 300 degrees instead? Rotating by 300 degrees 3 times gives you 900 degrees which is just 2 revolutions + 180 degrees. Will that give you -1 also?

$(\frac{1}{2} + i*\frac{\sqrt{3}}{2})*(\frac{1}{2} + i*\frac{\sqrt{3}}{2})*(\frac{1}{2} + i*\frac{\sqrt{3}}{2})$
$(\frac{1}{4} + i*\frac{\sqrt{3}}{2} - \frac{3}{4})*(\frac{1}{2} + i*\frac{\sqrt{3}}{2})$
$\frac{1}{8} + i*\frac{\sqrt{3}}{8} + i*\frac{\sqrt{3}}{4} - \frac{3}{4} - \frac{3}{8} - i*\frac{3*\sqrt{3}}{8}$
That adds up to -1 which is what we wanted.

$(\frac{1}{2} - i*\frac{\sqrt{3}}{2})*(\frac{1}{2} - i*\frac{\sqrt{3}}{2})*(\frac{1}{2} - i*\frac{\sqrt{3}}{2})$
$(\frac{1}{4} - i*\frac{\sqrt{3}}{2} + \frac{3}{4})*(\frac{1}{2} + i*\frac{\sqrt{3}}{2})$
$\frac{1}{8} - i*\frac{\sqrt{3}}{8} - i*\frac{\sqrt{3}}{4} + \frac{3}{4} + \frac{3}{8} + i*\frac{3*\sqrt{3}}{8}$
That also adds up to -1 which is what we wanted. Finally, we have the -1^3 = -1 answer which is just the 180 degree one.

Thus, we found three cubed roots of -1: 0.5 + 0.866i, 0.5 - 0.866i, and -1.

For the one we all learned...'square root of -1 is i'...is that really the only answer? Doing a similar exercise, you want to end up at m*360 + 180 degrees after n rotations where n is the root and m is an integer. Here, n = 2. That means 2*rotation = m*360 + 180, or rotation = 180*m + 90. Start with m = 0. rotation = 90 which means i is an answer which we know. Try m = 1. rotation = 270 which means -i is answer. Trying that out...-i * -i = i^2 = -1. That works. Try m = 2. rotation = 450 which is just 90 + 1 full cycle, so we're repeating now. i and -i are our square roots of -1.

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