How do I add the complex numbers $z^4$ and $z$?

Question

This question is similar to a question I posted earlier.
$$z=\cos\frac{\pi}{3}+j\sin\frac{\pi}{3}$$
This time I have to do the sum $z^4+z$

I have used the approach I was shown in my previous question. Here is what I've done: $$\left(\cos\frac{\pi}{3}+j\sin\frac{\pi}{3}\right)^4+\left(\cos\frac{\pi}{3}+j\sin\frac{\pi}{3}\right)$$ $$\cos\frac{4 \pi}{3}+j\sin\frac{4 \pi}{3}+\cos\frac{\pi}{3}+j\sin\frac{\pi}{3}$$ collecting like terms... $$\cos\frac{5\pi}{3}+2j\sin\frac{5\pi}{3}$$ I verified this with wolframalpha but the answer it gave was zero. Is this approach I'm using appropriate for this problem?

Answer 1

The answer is $0$ because$$\cos\left(\frac{4\pi}3\right)=\cos\left(\pi+\frac\pi3\right)=-\cos\left(\frac\pi3\right)\text{ and }\sin\left(\frac{4\pi}3\right)=\sin\left(\pi+\frac\pi3\right)=-\sin\left(\frac\pi3\right).$$

Answer 2

Via de Moivre's Theorem, $z^4=\cos\big(\frac{4\pi}{3}\big)+j\sin\big(\frac{4\pi}{3} \big)=-\frac{1}{2} -\frac{\sqrt{3}}{2}j$ $$\cos{\frac{\pi}{3}}+j\sin{\frac{\pi}{3}}=\frac{1}{2}+\frac{\sqrt{3}}{2}j$$

Adding those together yields $0$.