From d4481a0d023cacf22a1dcf9a35324ba5a42c63f6 Mon Sep 17 00:00:00 2001
From: joshuab <>
Date: Tue, 29 Jun 2010 15:23:39 +0000
Subject: tex

---
 ClassJune26.page | 7 ++++---
 1 file changed, 4 insertions(+), 3 deletions(-)

diff --git a/ClassJune26.page b/ClassJune26.page
index b28d40e..faa516c 100644
--- a/ClassJune26.page
+++ b/ClassJune26.page
@@ -152,9 +152,10 @@ We can raise complex numbers to powers, divide by the real denominators,
 and add them up just fine, so we can exponentiate complex values of
 $z$. We know what happens to real values, what happens to pure imaginary
 ones? Let $y\in\mathbb{R}$. Then  
-$\begin{array}e^{iy} & = & 1+iy+\frac{(iy)^{2}}{2!}+\frac{(iy)^{3}}{3!}+\frac{(iy)^{4}}{4!}+\frac{(iy)^{5}}{5!}+\cdots
- & = & 1+iy-\frac{y^{2}}{2!}-i\frac{y^{3}}{3!}+\frac{y^{4}}{4!}+i\frac{y^{5}}{5!}+\cdots
- & = & (1-\frac{y^{2}}{2!}+\frac{y^{4}}{4!}+\cdots)+i(y-\frac{y^{3}}{3!}+\frac{y^{5}}{5!}-\cdots)
+$\begin{array}
+e^{iy} & = & 1+iy+\frac{(iy)^{2}}{2!}+\frac{(iy)^{3}}{3!}+\frac{(iy)^{4}}{4!}+\frac{(iy)^{5}}{5!}+\cdots\\
+ & = & 1+iy-\frac{y^{2}}{2!}-i\frac{y^{3}}{3!}+\frac{y^{4}}{4!}+i\frac{y^{5}}{5!}+\cdots\\
+ & = & (1-\frac{y^{2}}{2!}+\frac{y^{4}}{4!}+\cdots)+i(y-\frac{y^{3}}{3!}+\frac{y^{5}}{5!}-\cdots)\\
  & = & \cos y+i\sin y\end{array}$
 
 
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