Imaginary number solution?

1. Dec 13, 2009

rpardo

I'm doing some practice problems for my mechanics exam tomorrow (good ol' SHM) and I can't solve this for the life of me:

Determine: (-1+i)^(1/3)

Any help would be greatly appreciated.

2. Dec 13, 2009

nicksauce

Write (-1 + i) in polar form (rexp(itheta)). Then multiply the angle by 1/3, and take the cube root of the radius.

3. Dec 13, 2009

rpardo

Thanks very much for the answer. IT WORKS!
However, I don't understnad why/how. Whats is the reasoning behind multiplying then angle and then cube rooting the radius?

4. Dec 13, 2009

rochfor1

Think about the laws of exponents. If $$-1 + i = r e^{ i \theta }$$, then $$(-1 + i)^{ \frac{ 1 }{ 3 } } = ( r e^{ i \theta } )^{ \frac{ 1 }{ 3 } } = \sqrt[3]{r} ( e^{ i \theta } )^{ \frac{ 1 }{ 3 } } = \sqrt[3]{r} e^{ i \frac{ \theta }{ 3 } }$$.

5. Dec 13, 2009

nicksauce

Taking the cube root of the radius should be obvious. The radius is a real number, and the cube root satisfies (AB)^1/3 = A^1/3 * B^1/3. Diving the angle by 3 is an application of De Moivre's formula:

http://en.wikipedia.org/wiki/De_Moivre's_formula

6. Dec 14, 2009

zgozvrm

To multiply two complex numbers in polar form $$a \angle b^\circ$$ and $$c \angle d^\circ$$, you multiply the radii (a x c) and add the angles (b + d), so the product is $$ac \angle (b+d)^\circ$$

Cubing a complex number $$a \angle b^\circ$$ is then $$(aaa) \angle(b+b+b)^\circ = a^3 \angle(3b)^\circ$$

So, to find the cube root of a complex number, you can see that you take the cube root of the radius and divide the angle by 3.