• Mar 23rd 2011, 05:45 AM
Biscaim
Hello everybody!

I am trying to solve the following question: Find $\displaystyle (-11-2i)^{1/3}$

I did the following: $\displaystyle (-11-2i)^{1/3}=(a+bi)$ and I'll get $\displaystyle a^{3}-3ab^{2}=-11$ and $\displaystyle 3a^{2}b-b^{3}=-2$

If I put $\displaystyle a=1$ I can solve this question. But I would like to know how to solve the system above.

• Mar 23rd 2011, 06:40 AM
Plato
Quote:

Originally Posted by Biscaim
I am trying to solve the following question:
Find $\displaystyle (-11-2i)^{1/3}$

There is no neat solution to that question.
Let $\displaystyle \theta=\arctan\left(\frac{2}{11}\right)-\pi$.
Then $\displaystyle \left(-11-2\imath\right)^{\frac{1}{3}}=\exp\left(\frac{\ln(5 )}{2}+\dfrac{\imath\theta}{3}\right)$

There is a good deal work to get to that answer.
• Mar 23rd 2011, 07:09 AM
Prove It
$\displaystyle \displaystyle |-11 - 2i| = \sqrt{(-11)^2 + (-2)^2}$

$\displaystyle \displaystyle = \sqrt{121 + 4}$

$\displaystyle \displaystyle = \sqrt{125}$.

$\displaystyle \displaystyle \arg{(-11 - 2i)} = \pi + \arg{(11 + 2i)}$

$\displaystyle \displaystyle = \pi + \arctan{\left(\frac{2}{11}\right)}$.

So $\displaystyle \displaystyle -11 - 2i = \sqrt{125}\,e^{i\left[\pi + \arctan{\left(\frac{2}{11}\right)}\right]}$

and therefore

$\displaystyle \displaystyle (-11 - 2i)^{\frac{1}{3}} = \left\{\sqrt{125}\,e^{i\left[\pi + \arctan{\left(\frac{2}{11}\right)}\right]}\right\}^{\frac{1}{3}}$

$\displaystyle \displaystyle = \sqrt[6]{125}\,e^{i\left[\frac{\pi}{3} + \frac{1}{3}\arctan{\left(\frac{2}{11}\right)}\righ t]}$.