Hi, can some one help me please solve that?

Let $\displaystyle v=1-\sqrt{3}i , u=i-1$

compute $\displaystyle v^6u^5$

So i did:

$\displaystyle \phi = tan^{-1}(\frac{-\sqrt{3}}{1})=-\frac{1}{3}\pi$

$\displaystyle r=\sqrt{1^2+\sqrt{3}^2} =2$

$\displaystyle \Rightarrow v=2e^{\frac{\pi}{3}i}$

$\displaystyle \phi_2 = tan^{-1}(\frac{1}{-1})=-\frac{1}{4}\pi = \frac{3}{4}\pi $

$\displaystyle r_2=\sqrt{1^2+1^2} =\sqrt{2}$

$\displaystyle \Rightarrow u=\sqrt{2}e^{\frac{3\pi}{4}i}$

now i did:

$\displaystyle (2e^{\frac{\pi}{3}i})^6 (\sqrt{2}e^{\frac{3\pi}{4}i})^5$

$\displaystyle 2^62^{\frac{5}{2}}(e^{\frac{\pi}{3}i})^6(e^{\frac{ 3\pi}{4}i})^5 $

$\displaystyle 2^62^{\frac{5}{2}}(e^{\frac{6\pi}{3}i})(e^{\frac{1 5\pi}{4}i})$

$\displaystyle 2^{6+\frac{5}{2}}(e^{\frac{6\pi}{3}i+\frac{15\pi}{ 4}i})$

$\displaystyle 2^{\frac{17}{2}}e^{\frac{23\pi}{4}i}$

am i right?

I think I have a mistake somewhere...

If not, how can i go from here?

Thanks!!!