Complex geometric series

Problem

The geometric series satisfies $\displaystyle\sum_{k=0}^{n-1}z^{k}=\frac{1-z^{n}}{1-z}$. a) Show that if $z=e^{i\pi/6}$ and $n=12$ the sum is zero. c) What is the sum (same $n$) if $z=e^{i\pi/5}$? Write the result in algebraic form.

Solution

a) $z^{12}=e^{i\,12\pi/6}=e^{i\,2\pi}=1$, so the numerator $1-z^{12}=0$ and the sum is $0$ (the $12$ terms are the $12$ twelfth roots of unity, whose sum is zero). c) $z^{12}=e^{i\,12\pi/5}=e^{i\,2\pi/5}$, hence $S=\frac{1-e^{i\,2\pi/5}}{1-e^{i\,\pi/5}}=\frac{0{,}691-0{,}951i}{0{,}191-0{,}588i}\approx1{,}809+0{,}588\,i.$ $\boxed{\,S_a=0,\qquad S_c\approx1{,}809+0{,}588\,i\,}$