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authorbnewbold <bnewbold@alum.mit.edu>2010-06-30 06:23:15 +0000
committerbnewbold <bnewbold@adelie.robocracy.org>2010-06-30 06:23:15 +0000
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@@ -161,11 +161,12 @@ 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}{}ee^{iy} & = & 1+iy+\frac{(iy)^{2}}{2!}+\frac{(iy)^{3}}{3!}+\frac{(iy)^{4}}{4!}+\frac{(iy)^{5}}{5!}+\cdots\\
+
+$$\begin{array}{ccc}
+ee^{iy} & = & 1+iy+\frac{(iy)^{2}}{2!}+\frac{(iy)^{3}}{3!}+\frac{(iy)^{4}}{4!}+\frac{(iy)^{5}}{5!}+\cdots\\
& = & 1+iy-\frac{y^{2}}{2!}-i\frac{y^{3}}{3!}+\frac{y^{4}}{4!}+i\frac{y^{5}}{5!}+\cdots\\
& = & (1-\frac{y^{2}}{2!}+\frac{y^{4}}{4!}+\cdots)+i(y-\frac{y^{3}}{3!}+\frac{y^{5}}{5!}-\cdots)\\
- & = & \cos y+i\sin y\end{array}$
-
+ & = & \cos y+i\sin y\end{array}$$
Substituting $y=\pi$, we recover Euler's famous identity,
$e^{i\pi}=-1.$