Lecture 7 — Exponential Maps

A note on coronavirus about the recent mail from Joanne Wright, the DVC(A).

Recall the Möbius transformation, and note that is is unique up to scaling for $\lambda > 0$. $w = \frac{az+b}{cz+d} = \frac{\lambda az+\lambda b}{\lambda cz+\lambda d}$

Remark: Any map from the inside of a (upper half) half-plane to the inside of a circle has the form $w = e^{-i\alpha} \frac{z-z_0}{z-z_0}\quad \text{ for some }\alpha \in \mathbb R, z_0 \in \mathbb C, \operatorname{Im} z_0 > 0.$

Exponential map

B.C. 103 (8Ed 104)

$z \mapsto e^z = \exp x = w, \quad \operatorname{dom} w = \mathbb C.$ Given a $z = x+iy$ for $x, y \in \mathbb R$, $w = e^z = e^{x+iy} = e^x e^{iy} = e^x (\cos y + i \sin y) = u+iv\\[0.7em] \begin{aligned} \text{ where }\quad u &= e^x \cos y\\ v &= e^x \sin y. \end{aligned}$ This is easier to see by writing $w = \rho e^{i\phi}$ where $\rho = e^x$, $\phi = y + 2k\pi$ for $k \in \mathbb Z$. This function is periodic in $\mathbb C$.

Images under exp

Properties

Many of the properties of the real $\exp$ extend to $\mathbb C$. Such as - $e^0 = 1$. - $e^{-z} = 1/e^z$. - $e^{z_1+z_2} = e^{z_1}e^{z_2}$. - $e^{z_1-z_2} = e^{z_1}/e^{z_2}$. - $(e^{z_1})^{z_2} = e^{z_1z_2}$.

However, some things do not extend: - $e^x > 0~\forall x \in \mathbb R$ but, for example, $e^{i\phi} = -1$. - $x \mapsto e^x$ is monotone increasing for $x \in \mathbb R$ but $z \mapsto e^z$ is periodic with period $2\pi i$.

Note: As in $\mathbb R$, $e^z = 0$ has no solution in $\mathbb C$. If there was some $z = x+iy$ such that $e^z = 0$, then $e^x e^{iy} = 0 \implies e^x = 0$ because $|e^{iy}| = 1$, contradiction.

Inverses

B.C. 31-33 (8Ed 30-32)

We have a function $f : \Omega \to \mathbb C$. Then, $g : \operatorname{Range}f \to \Omega$ is an inverse of $f$ if $g \circ f : \Omega \to \Omega$ is the identity. That is, $(g \circ f)(z) = z$ for all $z \in \Omega$.

Example: $z \mapsto z+1$ and $z \mapsto z-1$ are inverses for $\mathbb C \to \mathbb C$. $z \mapsto 1/z$ is its own inverse $\mathbb C_* \to \mathbb C_*$.