This question is similar to a question I posted earlier.
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?


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
  • 1
    $\begingroup$ 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

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


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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