# Expanding Euler's Identity

1. Aug 5, 2015

### RyanTAsher

1. The problem statement, all variables and given/known data

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

2. Relevant equations

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

2. Aug 5, 2015

### SteamKing

Staff Emeritus
Remember, eix = cos (x) + i sin (x). If x = π/6, then eix = ?

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