# Cube roots of a complex number

## Main Question or Discussion Point

hi,
is there any way to find the cube roots of a complex number WITHOUT converting it into the polar form? i am asking this because we can find the square root of a complex number without converting it. i was just wondering whether there is such a method for finding cube roots too.
i was trying to find the cube root of 110 + 74i

thanks in advance to anyone who can help.

matt grime
Homework Helper
There is a cubic formula, same as a quadratic, if that's what you mean. But it isn't worth using. There is also a quartic one, which is even less useful for practical formulae. There is no higher order general simple formula, though there are other methods. But they are much harder than simply putting it in polars.

to find the square roots of say 24+10i, we do the following:
Let $$\sqrt{24+10i} = a+bi$$

$$24+10i = a^2-b^2 + 2abi$$

$$a^2-b^2 = 24$$ and $$ab = 5$$

by solving these we can get a=5,-5 and b=1,-1. and thus we get the square roots of 24+10i.

i was wondering whether a similar method exists for finding the cube roots.

matt grime
Homework Helper
It would seem that it is very easy to try to repeat that for cubes. Did you try?

Office_Shredder
Staff Emeritus
Gold Member
It's incredibly ugly, since the form for (a+bi)3 is:

a3 + 3a2bi - 3ab2 - b3i

So if you're looking for the cube root of x + iy, you get

x = a3 - 3ab2
y = 3a2b - b3

And you want to solve for a and b. Maybe there's a neat trick, but I don't want to be the one to find it :P

Well I think cardano's method can be applied here, this might be some help

http://en.wikipedia.org/wiki/Cubic_equation.

I haven't applied it though.As Matt pointed out for higher orders it may becom tedious.Using the Polar form will be beneficial.

Office_Shredder
Staff Emeritus
Gold Member
I like this line:
Cubic equations were first discovered by Jaina mathematicians in ancient India sometime between 400 BC and 200 CE.
I can understand discovering, say, complex numbers, or 0 (not really discovering, but you get the point), but the idea of just sticking an x3 term doesn't really seem that impressive

matt grime said:
It would seem that it is very easy to try to repeat that for cubes. Did you try?
i tried. and it lead to the ugly form Office_Shredder wrote about. that is,

$$x = a^3 - 3ab^2$$
$$y = 3a^{2}b - b^3$$

here the only thing i can see is that $$x^2+y^2 = (a^2+b^2)^3$$

now how can i solve for a and b from here?

Last edited:
arildno
Homework Helper
Gold Member
Dearly Missed
Polar representation of a and b suggests itself readily..

robphy
Homework Helper
Gold Member
One could also do it geometrically... which, of course, strongly suggests the strength of a polar approach.

can anyone please be a bit more elaborate? i dont get it yet.

Office_Shredder
Staff Emeritus
Gold Member
murshid_islam said:
can anyone please be a bit more elaborate? i dont get it yet.
Nice and simple.... use polar form.

Polar form is my anti-drug

robphy
Homework Helper
Gold Member
murshid_islam said:
can anyone please be a bit more elaborate? i dont get it yet.
de Moivre's theorem
http://en.wikipedia.org/wiki/De_Moivre's_formula

see about 30% down the page [the Roots of Unity]