# Complex number problem!

• August 12th 2010, 04:50 AM
GreenMile
Complex number problem!
Find $Re(z)$ and $Im(z)$ if:

$z=\sqrt[3]{ \frac{2}{-\sqrt{2+i\sqrt{2}}}}$

My idea was to simplify the expression to the normal form $x+iy$ (The problem is that i don't know how). And then, by De Moivre's formula I can find the roots.

Any idea on how to simplify the formula and reach the normal form?

Thank you!
• August 12th 2010, 05:57 AM
HallsofIvy
I think you might do better to use the "polar form", $r e^{i\theta}= r(cos(\theta)+ i sin(\theta))$. That way $z^3= r^3 e^{3i\theta}= \frac{2}{-\sqrt{\frac{2}}{-\sqrt{2+ i\sqrt{2}}}$

$z^6= r^6e^{6i\theta}= \frac{4}{2+ i\sqrt{2}}$ $= \frac{4}{2+ i\sqrt{2}}\frac{2- i\sqrt{2}}{2- i\sqrt{2}}$ $= \frac{8- 4i\sqrt{2}}{6}= \frac{4}{3}- \frac{2}{3}\sqrt{2}$.

Put that last number into polar form: its magnitude is $\sqrt{\frac{16}{9}+ \frac{4}18}= \sqrt{\frac{36}{18}$ $= \frac{6}{3\sqrt{2}}= \sqrt{2}$ and its argument is [tex]arctan\left(-\frac{6\sqrt{2}}{12}\right)= arctan\left(-\sqrt{2}{2}}\right) $= -\frac{\pi}{4}$.

Now solve $r^6= \sqrt{2}$ and $6\theta= -\frac{\pi}{4}$. The real and imaginary parts are $r cos(\theta)$ and $r sin(\theta)$, respectively.
• August 12th 2010, 10:14 AM
11rdc11
Should it not be $\sqrt{\frac{16}{9} + \frac{8}{9}}= \sqrt{\frac{24}{9}}=\frac{2\sqrt{6}}{3}$
• August 12th 2010, 10:19 AM
yeKciM
Quote:

Originally Posted by 11rdc11
Should it not be $\sqrt{\frac{16}{9} + \frac{8}{9}}= \sqrt{\frac{24}{9}}=\frac{2\sqrt{6}}{3}$

yes it should, that's just typo :D

$\displaystyle |z| == r == \rho == \sqrt{(\Re^2+\Im^2)}$

$\displaystyle \theta == \varphi == \tan^{-1} \frac {\Im}{\Re}$