# Thread: complex calculation

1. ## complex calculation

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!!!

2. 23Pi/4 can be simplified to 7Pi/4 or -Pi/4

Where do you want to go with it?

$\displaystyle \displaystyle re^{i\theta}=r(\cos{\theta}+i\sin{\theta})$

3. to $\displaystyle r=a+bi$
i did not did mistakes?

4. 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.

5. Originally Posted by Nati
to $\displaystyle r=a+bi$
i did not did mistakes?
$\displaystyle r=\sqrt{a^2+b^2}$

6. You mean in the form $\displaystyle z = a+bi$

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

7. Originally Posted by pickslides
You mean in the form $\displaystyle z = a+bi$

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

8. I use De Moivre's theorem

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

Give it a go!

9. Originally Posted by Nati
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:
$\displaystyle \phi = tan^{-1}(\frac{-\sqrt{3}}{1})=-\frac{1}{3}\pi$
$\displaystyle r=\sqrt{1^2+\sqrt{3}^2} =2$
$\displaystyle \Rightarrow v=2e^{\frac{\pi}{3}i}$
The above should be: $\displaystyle \displaystyle v=2e^{-\frac{\pi}{3}i}.$

10. Originally Posted by Nati
to $\displaystyle r=a+bi$
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?