The principle amplitude of $\sqrt 2 \left [\cos\frac{5\pi}{3} +i\sin\frac{5\pi}{3}\right ]$ is

Question

I know that principe amplitude must lie between $-\pi and \pi$, but should I convert into? My guess was that since it’s beyond $\pi$ I should simply subtract $\pi$ from it to get $\frac{2\pi}{3}$, but the answer is $\frac{-\pi}{3}$.

Answer 1

We have that by $z=\sqrt 2 \left [\cos\frac{5\pi}{3} +i\sin \frac{5\pi}{3}\right]$

$$|z|^2=z\cdot \bar z=\sqrt 2 \left [\cos\frac{5\pi}{3} +i\sin \frac{5\pi}{3}\right]\cdot \sqrt 2 \left [\cos\frac{5\pi}{3} -i\sin \frac{5\pi}{3}\right]=2$$

and since $\Re(z)>0$

$$\arg (z)=\arctan\left(\frac{\sin \frac{5\pi}{3}}{\cos \frac{5\pi}{3}}\right)=-\frac{\pi}{3}$$

indeed recall that $\arctan(x)$ returns values in the range $(-\pi/2,\pi/2)$.

Note also that since $\frac{5\pi}{3}-2\pi=-\frac{\pi}{3}$ the two values represent the same angle.