From 439d6846c04f52a84c93c1d66fd45dd9c6b40c5b Mon Sep 17 00:00:00 2001
From: joshuab <>
Date: Tue, 29 Jun 2010 15:39:58 +0000
Subject: tex

---
 ClassJune26.page | 5 ++---
 1 file changed, 2 insertions(+), 3 deletions(-)

diff --git a/ClassJune26.page b/ClassJune26.page
index fb826d9..77376f7 100644
--- a/ClassJune26.page
+++ b/ClassJune26.page
@@ -152,8 +152,7 @@ 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  
-$\begin{array}
- e^{iy} & = & 1+iy+\frac{(iy)^{2}}{2!}+\frac{(iy)^{3}}{3!}+\frac{(iy)^{4}}{4!}+\frac{(iy)^{5}}{5!}+\cdots\\
+$\begin{array}ee^{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}$
@@ -236,7 +235,7 @@ holomorphic. Holomorphic functions are slightly more general, as the
 Jacobian can vanish; $df$ can be the complex number $0$. This is
 what happens at the origin for $f=z^{n}$. We can now check that $e^{z}$
 is conformal, and that $z^{n}$ is holomorphic, so conformal away
-from $0$.
+from $0$.  
 **Example:** $z^{2}$ is holomorphic. $z^{2}=(x+iy)^{2}=x^{2}-y^{2}+i(2xy).$ So
 $u=x^{2}-y^{2}$ and $v=2xy$.
 
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