1. ## Complex number problem!

Find $\displaystyle Re(z)$ and $\displaystyle Im(z)$ if:

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

My idea was to simplify the expression to the normal form $\displaystyle x+iy$ (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!

2. I think you might do better to use the "polar form", $\displaystyle r e^{i\theta}= r(cos(\theta)+ i sin(\theta))$. That way $\displaystyle z^3= r^3 e^{3i\theta}= \frac{2}{-\sqrt{\frac{2}}{-\sqrt{2+ i\sqrt{2}}}$

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

Put that last number into polar form: its magnitude is $\displaystyle \sqrt{\frac{16}{9}+ \frac{4}18}= \sqrt{\frac{36}{18}$$\displaystyle = \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)$\displaystyle = -\frac{\pi}{4}$.

Now solve $\displaystyle r^6= \sqrt{2}$ and $\displaystyle 6\theta= -\frac{\pi}{4}$. The real and imaginary parts are $\displaystyle r cos(\theta)$ and $\displaystyle r sin(\theta)$, respectively.

3. Should it not be $\displaystyle \sqrt{\frac{16}{9} + \frac{8}{9}}= \sqrt{\frac{24}{9}}=\frac{2\sqrt{6}}{3}$

4. Originally Posted by 11rdc11
Should it not be $\displaystyle \sqrt{\frac{16}{9} + \frac{8}{9}}= \sqrt{\frac{24}{9}}=\frac{2\sqrt{6}}{3}$
yes it should, that's just typo

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

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