How To Take The Square Root Of A Complex Number

How To Take The Square Root Of A Complex Number // Adding, subtracting, multiplying, and dividing complex numbers - not too hard, right? Agreed. But taking the square root of complex numbers - not so easy!

In this video I show you step by step how to simplify complex radicals by simply setting them equal to a + bi and then squaring both sides. It’s a cool trick that is not often taught in precalculus. We end up with simultaneous equations and a good bit of algebra, but it’s all good fun and has a satisfying solution.

I should mention that Steve from black pen red pen (bprp) did a similar video:

in which he takes the square root of i. Here I take it one step further and show how to take the square root of complex numbers (a + bi), not just purely imaginary numbers (bi).

This video should hit like a laser if you’re looking for good problems in the following categories:

✔ How to simplify imaginary numbers
✔ How to solve imaginary numbers
✔ 11th grade math problems,
✔ 12 grade math problems,
✔ Math competition problems for MathCounts and Math Olympiad
✔ SAT math
✔ ACT math

If you enjoy hard algebra problems like these that involve complex radicals, then you’ll love the inverse to this problem: powers of i. Check out my i^i^i video:

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🎸🎸Music: Feeling Free by Martin Riopel 🎸🎸

Still reading this? Then write "I got 5 + 6i and -5 - 6i" in the comments!

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[RisetotheEquation]

Which method is better for this problem: DeMoivre's formula or the method in this video? LMK below!!!

[claudiocosta]

Yes, use the polar form of imaginary numbers, was my first ideia. Again, good vídeo, thanks for sharing, and keep going.
10k will come soon

[narfharder]

First! Here from SyberMath, instant sub. Looking forward to the slow consumption of your back catalog. ☺

I've seen this method of splitting by dissimilar coefficients (like R vs I, or "synthesizing constants" +C => Ax+B) on many math channels, but never learned it in school. I was confused at first as to what they were even doing. Does it have a name?

[cristinaferraro]

I also would have used the polar representation z=r*exp(i*theta), and in this case the initial numbers are "kind" to us, so 60i-11 translates to z = 61*exp(i*arctan(-60/11).
And then z^(1/2) = r^(1/2)*exp(i*theta/2)

In numbers, z^(1/2) = 61^(1/2)*exp(i*arctan(6/5)

[redroach]

What happens if the quadratic yields an imaginary result?

[mathematurques]

The morning I was doing this kind of things during lesson, but with complex number like (5+i)/(2i-3), and we look for the algrebic form x+iy.

I've never seen something like this before, but I will definitly propose an extra question with a square root to my kids :D

[kristoferkrus]

1:18 Oh, great, now I know how to calculate the square root of 60i - 11.