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