I'm stuck with a really basic problem, and would appreciate help to understand what I am doing wrong.
The task is to find imaginary and real part of $$\Big(\frac{-1+ i \sqrt{3}}{2}\Big)^3$$
Solution so far:
$$z^n=re^{i \theta n}$$
$$r=|z| = \sqrt{\frac{1}{4}+\frac{3}{4}}=1$$
$$\theta = \arg(z)=\arctan\Big(\frac{y}{x}\Big)=\arctan\Big(\frac{\sqrt{3}/2}{-1/2}\Big)=\frac{-\pi}{3}$$
Which gives me that $$\displaystyle z^n=re^{i \theta n} = e^{-i\frac{3\pi}{3}} = e^{-i\pi}$$
The correct answer is $$e^{i\frac{2\pi}{3}}$$
What in the calculation of $\theta$ am I doing wrong?
Thanks!
Best Answer
The angle is in the second quadrant because real part is negative and imaginary part is positive, so can be represented by a complex number
$$ r= e^ {i \frac{2 \pi}{3}}$$
$$ r^3 =(e^ {i \frac{2 \pi}{3}} )^3=e^{2\pi i}= 1= 1+0i $$ which is the pure real number unity by Euler formula with zero imaginary part.
The tip of radius vector is 1 unit away from origin on the positive real axis.
Geometrically if 3 copies are made of the angle rotation amount in counterclockwise direction the vector tip lands at $(1,0)$.