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@@ -146,26 +146,26 @@ look like? PICTURE.
 Another key example is the exponential map. Recall the power series
 for $e^{z}$:
 
-\[
-e^{z}=1+z+\frac{z^{2}}{2!}+\frac{z^{3}}{3!}+\frac{z^{4}}{4!}+\frac{z^{5}}{5!}+\cdots\]
+$e^{z}=1+z+\frac{z^{2}}{2!}+\frac{z^{3}}{3!}+\frac{z^{4}}{4!}+\frac{z^{5}}{5!}+\cdots$
+
 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{eqnarray*}
-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{eqnarray*}
+ones? Let $y\in\mathbb{R}$. Then  
+$\begin{array}e^{iy} & = & 1+iy+\frac{(iy)^{2}}{2!}+\frac{(iy)^{3}}{3!}+\frac{(iy)^{4}}{4!}+\frac{(iy)^{5}}{5!}+\cdots
+ & = & 1+iy-\frac{y^{2}}{2!}-i\frac{y^{3}}{3!}+\frac{y^{4}}{4!}+i\frac{y^{5}}{5!}+\cdots
+ & = & (1-\frac{y^{2}}{2!}+\frac{y^{4}}{4!}+\cdots)+i(y-\frac{y^{3}}{3!}+\frac{y^{5}}{5!}-\cdots)
+ & = & \cos y+i\sin y\end{array}$
 
 
-Substituting $y=\pi$, we recover Euler's famous identity,\[
-e^{i\pi}=-1.\]
+Substituting $y=\pi$, we recover Euler's famous identity,
+$e^{i\pi}=-1.$
  Given a real argument $y$, the $e^{iy}$ gives the unit vector with
 that argument. Given an arbitrary complex number $z=x+iy$, it's exponnt
 $e^{z}$ is the complex number with magnitude $e^{x}$ and angle $y$
 radians.
 
-What does $z\rightsquigarrow e^{z}$ look like? Periodic in the $i$
+What does $z\mapsto e^{z}$ look like? Periodic in the $i$
 direction with period $2\pi$. Takes a horizontal strip and wraps
 it around, forming an annulus. PICTURE.