Hey guys wondered if someone could help me out again, if you let Z=root 3 - i what is Z in polar form?
Then how do I find the real and imaginary parts of (z^7)/8?
$\displaystyle x= r cos(\theta)$ and $\displaystyle y= r sin(\theta)$ so [tex]x^2+ y^2= r^2 cos^2(\theta)+ r^2 sin^2(\theta)= r^2[/itex]: $\displaystyle r= \sqrt{x^2+ y^2}$. Dividing y by x, $\displaystyle y/x= sin(\theta)/cos(\theta)= tan(\theta)$ and [tex]\theta= arctan(y/x)[/itex].
Here, $\displaystyle x^2+ y^2= 3+ 1= 4$ so r= 2. [tex]y/x= -1/\sqrt{3}[tex] so [tex]tan(\theta)= -1/\sqrt{3}[/itex] and $\displaystyle \theta= -\pi/6$.
Once you have Z in polar form, $\displaystyle Z^8$ has $\displaystyle r= 2^8= 256$ and [tex]\theta= 8(-\pi/6)= -4\pi/3[/itex]. The real part of $\displaystyle Z^8$ is $\displaystyle 256 cos(4\pi/3)$ and the imaginary part is $\displaystyle - 256 sin(4\pi/3)$.