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$ z = e^{\frac{2i\pi}{5}} $

Write $ z^2, z^3 $and $z^4$ in the form $e^{i\theta} $

Explain why $ z + z^2 + z^3 +z^4 = -1 $

Is there an easy way to explain using the $e^{i\theta} $ form (and not the sum of roots explanation or by using cis form?)

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    You could always check that $1 + z + z^2 + z^3 + z^4 = 0$ using the partial geometric series formula. – 2017-02-11
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    Just multiply $e^{i\theta}+e^{2i\theta}+\ldots$ by $e^{i\theta}$ and see what happens – 2017-02-11
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    Basic exponent laws for the powers. – 2017-02-11
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    Well, by symmetry, the sum of the roots is 0. Because e^0 is 1, the other 4 roots must total to -1 – 2017-02-11

2 Answers 2

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Notice that $z^5=e^{2i\pi}=1$, hence $z$ is a root of the polynomial $x^5-1$. But $x^5-1=(x-1)(x^4+x^3+x^2+x+1)$, and obviously $z\neq 1$. So...

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$z$ is a root of $1-z^5=(1-z)(1+z+z^2+z^3+z^4)$, and it is is not a root of $1-z$, hence it is a root of the other factor.