# Question 6abef

Jan 27, 2018

$\cos \left(\frac{2015 \pi}{3}\right) + i \sin \left(\frac{2015 \pi}{3}\right) = \frac{1}{2} - i \frac{\sqrt{3}}{2}$

#### Explanation:

${\left(1 + \cos \left(\frac{2 \pi}{3}\right) + i \sin \left(\frac{2 \pi}{3}\right)\right)}^{2015}$

First we need to include the 1 in the complex number's polar form so:

$1 + \cos \left(\frac{2 \pi}{3}\right) + i \sin \left(\frac{2 \pi}{3}\right)$

$= 1 - \frac{1}{2} + i \frac{\sqrt{3}}{2}$

$\frac{1}{2} + i \frac{\sqrt{3}}{2}$

So now transforming back to polar form:

The modulus of this complex number is:

$\sqrt{{\left(\frac{1}{2}\right)}^{2} + {\left(\frac{\sqrt{3}}{2}\right)}^{2}} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1$

And the argument will be:

${\tan}^{- 1} \left(\frac{\left(\frac{\sqrt{3}}{2}\right)}{\left(\frac{1}{2}\right)}\right) = {\tan}^{- 1} \left(\sqrt{3}\right) = \frac{\pi}{3}$

Hence: $\frac{1}{2} + i \frac{\sqrt{3}}{2} = \cos \left(\frac{\pi}{3}\right) + i \sin \left(\frac{\pi}{3}\right)$

So it follows that:

${\left(1 + \cos \left(\frac{2 \pi}{3}\right) + i \sin \left(\frac{2 \pi}{3}\right)\right)}^{2015} = {\left(\cos \left(\frac{\pi}{3}\right) + i \sin \left(\frac{\pi}{3}\right)\right)}^{2015}$

In modulus argument form, raising a complex number to some power is given by:

(r(cos(x)+isin(x))^n=r^n(cos(nx)+isin(nx))#

(In our case the modulus $r$ is $1$). Therefore:

${\left(\cos \left(\frac{\pi}{3}\right) + i \sin \left(\frac{\pi}{3}\right)\right)}^{2015}$

$= \cos \left(\frac{2015 \pi}{3}\right) + i \sin \left(\frac{2015 \pi}{3}\right)$

$= \frac{1}{2} - i \frac{\sqrt{3}}{2}$