When calculating the real and imaginary parts of the complex number, do we take the angle as shown or the magnitude of it? I thought that we would just take the angle as shown, but apparently not, according to my textbook (unless its a typo):
$$z=2\sqrt{3}\operatorname{cis}\left(\frac{\pi }{3}\right)\:w=4\operatorname{cis}\left(\frac{-\pi }{6}\right)$$
Find $z + w$ in:
- Cartesian form
- Modulus-argument form
I worked out the complex number to be $3\sqrt{3}+i$:
$$4\:\cos\left(-\frac{\pi }{6}\right)\:+\:4i\:\sin\left(-\frac{\pi }{6}\right)\:=\:2\sqrt{3}-2i\\
2\sqrt{3}\cos\left(\frac{\pi }{3}\right)+2\sqrt{3}i\:\sin\left(\frac{\pi }{3}\right)=\:\sqrt{3}\:+3i\\
\sqrt{3}\:+3i\:+\:2\sqrt{3}-2i\:=\:3\sqrt{3}+i$$
(Following from this I got the mod-arg form to be $[2\sqrt{7},0.19]$)
The textbook says pretty much the same thing, except
$$4\:\cos\left(-\frac{\pi }{6}\right)\:+\:4i\:\sin\left(-\frac{\pi }{6}\right)\:$$ is taken to be: $4\cos\left(\frac{\pi }{6}\right)+4i\sin\left(\frac{\pi }{6}\right)$, resulting in the complex number $=3\sqrt{3}+5i$.
Best Answer
We take the angle as shown, not just its absolute value. Therefore, $4\operatorname{cis}\left(\frac\pi6\right)\neq4\operatorname{cis}\left(-\frac\pi6\right)$. It looks as if you are right and your textbook is wrong.