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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?

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closed as unclear what you're asking by Lord Shark the Unknown, Christopher, Nosrati, Don Thousand, Parcly Taxel Oct 16 '18 at 23:10

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  • $\begingroup$ Note that $\cos\left(\frac{4\pi}{3}\right)+\cos\left(\frac\pi3\right)\color{red}\neq\cos\left(\frac{5\pi}{3}\right)$ $\endgroup$ – Teh Rod May 7 '18 at 18:17
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    $\begingroup$ Alternatively, using that $\,z^3=-1\,$ it follows that $\,z+z^4=z(z^3+1)=0\,$. $\endgroup$ – dxiv May 7 '18 at 18:21
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The answer is $0$ because$$\cos\left(\frac{4\pi}3\right)=\cos\left(\pi+\frac\pi3\right)=-\cos\left(\frac\pi3\right)\text{ and }\sin\left(\frac{4\pi}3\right)=\sin\left(\pi+\frac\pi3\right)=-\sin\left(\frac\pi3\right).$$

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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 $0$.

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