# complex calculation

• Jan 2nd 2011, 05:50 PM
Nati
complex calculation
Hi, can some one help me please solve that?
Let $v=1-\sqrt{3}i , u=i-1$
compute $v^6u^5$

So i did:
$\phi = tan^{-1}(\frac{-\sqrt{3}}{1})=-\frac{1}{3}\pi$
$r=\sqrt{1^2+\sqrt{3}^2} =2$
$\Rightarrow v=2e^{\frac{\pi}{3}i}$

$\phi_2 = tan^{-1}(\frac{1}{-1})=-\frac{1}{4}\pi = \frac{3}{4}\pi$
$r_2=\sqrt{1^2+1^2} =\sqrt{2}$
$\Rightarrow u=\sqrt{2}e^{\frac{3\pi}{4}i}$
now i did:
$(2e^{\frac{\pi}{3}i})^6 (\sqrt{2}e^{\frac{3\pi}{4}i})^5$

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

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

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

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

am i right?
I think I have a mistake somewhere...
If not, how can i go from here?

Thanks!!!
• Jan 2nd 2011, 06:00 PM
dwsmith
23Pi/4 can be simplified to 7Pi/4 or -Pi/4

Where do you want to go with it?

$\displaystyle re^{i\theta}=r(\cos{\theta}+i\sin{\theta})$
• Jan 2nd 2011, 07:24 PM
Nati
to $r=a+bi$
i did not did mistakes?
• Jan 2nd 2011, 07:26 PM
dwsmith
It is correct. Is there a certain form you want to represent it in? You can simplify the angle (post #2) but that is about it.
• Jan 2nd 2011, 07:27 PM
dwsmith
Quote:

Originally Posted by Nati
to $r=a+bi$
i did not did mistakes?

$r=\sqrt{a^2+b^2}$
• Jan 2nd 2011, 07:40 PM
pickslides
You mean in the form $z = a+bi$

I get $\displaystyle (1-\sqrt{3}i)^6\times (-1+i)^5 = 64\times (4-4i)$
• Jan 2nd 2011, 08:34 PM
Nati
Quote:

Originally Posted by pickslides
You mean in the form $z = a+bi$

I get $\displaystyle (1-\sqrt{3})^6\times (-1+i)^5 = 64\times (4-4i)$

• Jan 2nd 2011, 08:50 PM
pickslides
I use De Moivre's theorem

$\displaystyle (x+yi)^n= r^n(\cos n\theta +i \sin n\theta)$

Give it a go!
• Jan 2nd 2011, 08:57 PM
SammyS
Quote:

Originally Posted by Nati
Hi, can some one help me please solve that?
Let $v=1-\sqrt{3}i , u=i-1$
compute $v^6u^5$

So i did:
$\phi = tan^{-1}(\frac{-\sqrt{3}}{1})=-\frac{1}{3}\pi$
$r=\sqrt{1^2+\sqrt{3}^2} =2$
$\Rightarrow v=2e^{\frac{\pi}{3}i}$

The above should be: $\displaystyle v=2e^{-\frac{\pi}{3}i}.$
• Jan 2nd 2011, 09:25 PM
mr fantastic
Quote:

Originally Posted by Nati
to $r=a+bi$
i did not did mistakes?

You were told that your calculation ws correct but that the agument could be simplified. Now it looks like you want to give your answer in cartesian form. You should know how to convert from polar to cartesian form. Where are you stuck?