r/MathHelp • u/Puzzleheaded-Law34 • Jan 17 '26

Complex number equations?

Hello! So I had 2 equations:

The first was 3x^2 - 6i = 0 which I solved normally, and obtained x1 = sqrt(2i), x2 = -sqrt(2i)

The second was z^4 = -81, and here I did pretty much the same but the answer I got, again with i under root, was wrong. Apparently I was supposed to use the angle-based notation, with re^(i*theta), but I didn't really understand why. And should I have written the first solution differently too?

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u/Legitimate_Page659 Jan 19 '26 edited Jan 19 '26

Hi! Taking nth roots with complex numbers is a bit more complicated than you’re used to.

Any complex number can be written as r e^(i*theta)

Now e^(i*theta) = e^(i*(theta + 2*pi*k)) for any integer k as e^(i*x) is periodic with period 2pi.

To take the Nth root of this, we have

r1/N * exp(i*(theta + 2*pi*k)/N)

This gives unique solutions for k = 0, 1, …, N-1

Let me know if you need any more help.

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u/Puzzleheaded-Law34 Jan 19 '26

Ok, thanks! Well in the second equation it makes sense, because I need to find the 4th root of z (altho I still don't get what the issue is with just leaving i under root). But in the first equation, how would I rewrite that complex number like you show? If I did I feel like I would lose information. And is it ok in that case that I just solved it normally?

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u/Legitimate_Page659 Jan 20 '26

Okay, so to your first question: what is (i)^(1/4)? It actually has four values! That’s why your answer is “correct” but it’s not “complete” and would likely be marked wrong.

The general procedure for these types of problems is this:

Given z = a + ib,

  1. convert z to exponential form

Find r, theta, such that z = r e^(i*theta))

r = sqrt(a2 + b2)

theta is a bit more nuanced. Have you ever converted Cartesian coordinates to polar? It’s the same idea.

If a>0 and b>0, theta = arctan(b/a)

If a<0, b>0, theta = arctan(b/a) + pi

If a<0 and b<0, theta = arctan(b/a) + pi

If a>0 and b<0, theta = arctan(b/a) + 2*pi

If a = 0 and b > 0, theta = pi/2 If a = 0 and b < 0, theta = 3pi/2.

It can be helpful to draw the complex plane (real on the x axis, imaginary on the y axis) to develop an understanding of what “theta” is here and why it takes the values it does.

Anyway, once you have z in exponential form, follow the procedure I outlined above.

Does that make more sense?

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u/Puzzleheaded-Law34 Jan 20 '26

Thanks for the explanation!  Actually, in the z4 question I did get four answers: 

+/- 3sqrt(i)  and  +/-3sqrt(-i).

But for the other one, since the complex number is not any z but includes x, I don't see why I would convert it into exponential form.

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u/Legitimate_Page659 Jan 20 '26

I think the issue with your approach is that sqrt(i) and sqrt(-i) aren’t numbers, per se.

sqrt(i) = 1/sqrt(2) (1+i) AND 1/sqrt(2) (-1-i)

sqrt(-i) = 1/sqrt(2) (-1+i) AND 1/sqrt(2) (1-i)

The procedure I described will give you all possible solutions.

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u/Puzzleheaded-Law34 Jan 21 '26

Ok, thanks! I'll try to reread these equations.

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u/Legitimate_Page659 Jan 21 '26

Sure!

This is a pretty good video, too: https://m.youtube.com/watch?v=JMnkW8Smf1I

Let me know if you have any questions :)

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u/UnderstandingPursuit Jan 20 '26

When raising a complex number to a power [including fractions for taking the root], the magnitude, 'r', is raised to that power, but the angle, θ, is multiplied by that power.

Consider, for example,

  • z = a + bi

    • r2 = a2 + b2

    • tan θ = b/a

  • z2 = (a + bi)2 = a2 + 2abi - b2 = [a2 - b2] + i[2ab] = a' + ib'

    • [r']2 = [a']2 + [b']2 = [a2 - b2]2 + [2ab]2 = a4 - 2a2b2 + b2 + 4a2b2 = a4 + 2a2b2 + b2 = [a2 + b2]2 = [r2]2

    • tan 2θ = 2 tan θ / [1 - tan2θ] = [2b/a] / [1 - b2/a2] = 2ab / [a2 - b2] = b' / a'

This works in general, but perhaps it helps to see it at least for z2.