# 3rd root of a complex number

1. Jan 28, 2007

### GregA

1. The problem statement, all variables and given/known data
I need to find the solution to $$(2-11i)^{\frac{1}{3}}$$

2. Relevant equations
If $$(2-11i)^{\frac{1}{3}}$$ were to equal (a + bi) for some real numbers a and b then $$2 - 11i = a^3 +3a^2bi-3ab^2-b^3i$$

3. The attempt at a solution

From above $$a^3-3ab^2 = 2$$ and $$3a^2b - b^3 = -11$$
I can factorise (but only slightly) as follows:
$$a(a^2-b^2) = 2 + 2ab^2$$
$$b(a^2-b^2) = -11 - 2a^2b$$

after losing the a^2-b^2 I'm left with $$2b + 2ab^3 = -11a - 2a^3b$$ and I can see no other useful factorisations or substitutions :(

The actual values I need to find here are simple and with not so much guess work found that a = 2 and b = -1. My problem is that I'm not so sure that guess-work is the correct method to be using. I could probably plot both functions and find where there is a point of intersection but is there an algebraic method I can employ?...If so can anyone throw me any pointers?

Last edited: Jan 28, 2007
2. Jan 28, 2007

### mjsd

it appears to me that you were trying to express $$2-11i$$ as a perfect cube $$(a+ib)^3$$ and want to find $$a,b$$.

firstly, note that since you are taking a cube root, there will be THREE answers (possibly all complex). The "proper" way to do this is to turn your number $$2-11i$$ into "polar form" or "exponential form"
$$x+iy \rightarrow r e^{i\theta}$$
where $$r=|x+iy|$$ absolute value and $$\theta=\text{Arg}(x+iy)$$

you may now evaluate
$$(2-11i)^{1/3} = r^{1/3} e^{i\theta/3}$$

note you get three answers because
$$\text{Arg}(x+iy) +2n\pi$$ is also a valid angle where $$n=1,2,\ldots$$
in this case $$\theta/3$$ should give you "three" valid angles lie in the range $$\left(-\pi,\pi\right]$$

3. Jan 28, 2007

### GregA

cheers for that mjsd...the book I'm using introduces polar co-ordinates and argand diagrams in the next chapter, (though unfortunately for me, later than the question) I'll return to this question at a later date

4. Jan 28, 2007

### drpizza

I think the book is just showing you how much easier these problems become later - the answer to "why do I need to learn a new way to do something I already can do." I love the look on students' faces when I teach them a neat trick that knocks 5 minutes off a 6 minute problem.

5. Jan 29, 2007

### Gib Z

I havent done polar forms either, but im assuming that the method you just described employs Eulers Formula?

6. Jan 29, 2007

### mjsd

yes: $$e^{ix}=\cos x +i \sin x$$

7. Jan 29, 2007

### Gib Z

Yup Im familiar with the formula :) Truly a miraculous formula, when I tell people about it they wonder what possible use it could have beyond its beauty, and heres just one example.