# Thread: Solve with de moivres formula

1. ## Solve with de moivres formula

a) (sqrt3 - i)^6 b) (2-2i)^5

2. ## Re: Solve with de moivres formula

Originally Posted by kalemale
a) (sqrt3 - i)^6 b) (2-2i)^5
$\sqrt3-i=2\exp\left(-\frac{i\pi}{6}\right)$ and $2-2i=\sqrt8(\exp\left(\frac{-i\pi}{4}\right)$

3. ## Re: Solve with de moivres formula

Originally Posted by Plato
$\sqrt3-i=2\exp\left(-\frac{\pi}{3}\right)$
OP, even experts make tiny mistakes sometimes.

$\sqrt3-i=2\exp\left(-\frac{\pi}{6} i \right)$

I hope i got that right.

4. ## Re: Solve with de moivres formula

sorry if I dont understand fully but why is it
-pi/4 and -pi/6

5. ## Re: Solve with de moivres formula

Originally Posted by kalemale
sorry if I dont understand fully but why is it
-pi/4 and -pi/6
If you plot the complex number $\sqrt 3 - i$ on the complex plain and convert to polar coordinates you get $R = 2$ and $\theta= \frac {- \pi}6$. Stated another way: $\sqrt 3 - i = 2(\cos (\frac {-\pi} 6}) + i \sin (\frac {-\pi} 6}) ) = 2 cis (\frac {-\pi} 6})$. Now to raise it to the 6th power apply de Moivre's theorem: $(R cis \theta) ^n = r^n cis(n\theta)$

6. ## Re: Solve with de moivres formula

so we get -pi / 6 becouse its negativ and the angle is 30 degress?

7. ## Re: Solve with de moivres formula

Originally Posted by kalemale
so we get -pi / 6 becouse its negativ and the angle is 30 degress?
Yes - the angle is negative 30 degrees.