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authorjoshuab <>2010-06-29 15:39:58 +0000
committerbnewbold <bnewbold@adelie.robocracy.org>2010-06-29 15:39:58 +0000
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@@ -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 $v=2xy$.