How Does e^(iπ) Expand in Euler's Identity?

[Destroxia]

Homework Statement

## y^({9}) + y''' = 6 ##

Homework Equations

The Attempt at a Solution

## y^({9}) + y''' = 6 ##

## r^9 + r^3 = r^{3}(r^{6}+1)=0 ##

## r = 0, m = 3 ##

## r^6 + 1 = 0 = e^{(i(\pi + 2k\pi)} ##

## r = -1 = e^{i(\frac {\pi +2k\pi} {6})} ##

## k = 0 , r = e^{i(\frac {\pi} {6})} ##

## k = 1 , r = e^{i(\frac {\pi} {2})} ##

## k = 2 , r = e^{i(\frac {5\pi} {6})} ##

My question is, when doing the ##k = 0,1,2, ... ## How does the ##e^{i\pi}## expand? My teacher has the answer for ## k = 0 ## as:

## r = e^{i(\frac {\pi} {6})} = \frac {\sqrt{3}} {2} + \frac {i} {2}##

I don't understand how the exponential works out to the ##\frac {\sqrt{3}} {2} + \frac {i} {2}##

 

[SteamKing]

Homework Statement

## y^({9}) + y''' = 6 ##

Homework Equations

The Attempt at a Solution

## y^({9}) + y''' = 6 ##

## r^9 + r^3 = r^{3}(r^{6}+1)=0 ##

## r = 0, m = 3 ##

## r^6 + 1 = 0 = e^{(i(\pi + 2k\pi)} ##

## r = -1 = e^{i(\frac {\pi +2k\pi} {6})} ##

## k = 0 , r = e^{i(\frac {\pi} {6})} ##

## k = 1 , r = e^{i(\frac {\pi} {2})} ##

## k = 2 , r = e^{i(\frac {5\pi} {6})} ##

My question is, when doing the ##k = 0,1,2, ... ## How does the ##e^{i\pi}## expand? My teacher has the answer for ## k = 0 ## as:

## r = e^{i(\frac {\pi} {6})} = \frac {\sqrt{3}} {2} + \frac {i} {2}##

I don't understand how the exponential works out to the ##\frac {\sqrt{3}} {2} + \frac {i} {2}##

Remember, e^ix = cos (x) + i sin (x). If x = π/6, then e^ix = ?

https://en.wikipedia.org/wiki/Euler's_identity