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| -rw-r--r-- | ClassJune26.page | 5 |
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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 $v=2xy$. |