First slide

De-moivre's theorem

Question

Let $z = a (\cos \frac{π}{5} + i \sin \frac{π}{5}), a \in R, | a | < 1,$ then $S = z^{2015} + z^{2016} + z^{2017} + \dots$ equals

Moderate

A

$\frac{a^{2015}}{z - 1}$
B

$\frac{a^{2015}}{1 - z}$
C

$\frac{a^{2015}}{1 - a}$
D

$\frac{a^{2015}}{a - 1}$

Solution

We have $|z| = |a| < 1,$ thus
$S = \frac{a^{2015}}{z - 1}$
But $z^{2015}$ $= a^{2015} [\cos (403 π) + i \sin (403 π)] = a^{2015}$
$∴ S = \frac{a^{2015}}{z - 1}$

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