What is the solution to (2-11i)^(1/3)?

[GregA]

Homework Statement

I need to find the solution to (2-11i)^{\frac{1}{3}}

Homework 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

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?

 

[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]

 

[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

 

[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.

 

[Gib Z]

I haven't done polar forms either, but I am assuming that the method you just described employs Eulers Formula?

 

[mjsd]

I haven't done polar forms either, but I am assuming that the method you just described employs Eulers Formula?

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

 

[Gib Z]

Yup I am 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 here's just one example.