MATH3401
Complex Analysis
Lecture 7
Final remarks on
Möbius transformations (MT)
continued
Remark 2:
\(w=\dfrac{az+b}{cz+d}\)
\(=\dfrac{\left(\lambda a\right)z+\left(\lambda
b\right)}{\left(\lambda c\right)z+\left(\lambda d\right)}\)
for every \(\lambda \in \C_*\).
Final remarks on MT
continued
Remark 3: Any Möbius transformation that maps the upper half plane \(\Im(z) > 0\) (UHP) onto the open disk \(|w|<1\) and the boundary \(\Im(z)=0\) of the half plane onto the boundary \(|w|=1\) of the disk, has the form \[ w = e^{i\alpha}\frac{z-z_0}{z-\conj{z_0}} \] for some \(\alpha \in \R\) and \(z_0\in \C\), with \(\Im(z_0)>0\).
Conversely: Any Möbius transformation of this form maps the UHP onto the inside of the unit circle.
Exponential map
\(z \mapsto e^z = \exp(z)=w\), with dom\((w)=\C\).
For \(z=x+iy\) with \(x,y\in \R\),
\(w= e^z=e^{x+iy}=e^xe^{iy}\)
\(\quad = e^x\left( \cos y + i \sin y \right) = u+iv \)
where \(u=e^x\cos y, v=e^x\sin y\).
Exponential map
Write \(w=\rho e^{i\phi}\), where: \[ \left\{ \begin{array}{rl} \rho &= e^{x} \\ \phi &= y + 2 k \pi, \; k\in \Z \end{array} \right. \]
Images under \(\exp\)
Vertical line \(x=c_1\).
|
\(\tiny\exp\) \(\ra\) |
|
|---|
Images under \(\exp\)
Vertical line \(y=c_2\).
|
\(\tiny\exp\) \(\ra\) |
|
|---|
Images under \(\exp\)
How about:
|
\(\tiny\exp\) \(\ra\) \(\tiny h < 2\pi\) |
|
|---|
Images under \(\exp\)
How about:
|
|
\(\tiny\exp\) \(\ra\) \(\tiny h \geq 2\pi\) |
|
|---|
Properties of \(\exp\)
Many "laws" for \(\R\)-\(\exp\) extend to \(\C\)-\(\exp\):
- \(e^0=1\);
- \(e^{-z}=1/e^{z};\)
- \(e^{z_1+z_2}= e^{z_1}e^{z_2};\)
- \(e^{z_1-z_2}= \dfrac{e^{z_1}}{e^{z_2}};\)
- \(\left(e^{z_1}\right)^{z_2}= e^{z_1z_2}\).
Properties of \(\exp\)
Some things do not extend:
- \(e^x > 0\) for all \(x\in \R\), but e.g. \(e^{i\pi}=-1\) and \(e^{i\pi/4}\in\C\setminus \R\);
- \(x \mapsto e^x\) is monotone increasing on \(R\)
but
\(z\mapsto e^z\) is periodic,
with period \(2\pi i\): \[ \begin{array}{rl} e^{z+2\pi i} &= e^{z}e^{2\pi i} \\ &= e^z\left( \cos 2\pi + i \sin 2 \pi \right)\\ &= e^z. \end{array} \]
Note: as in \(\R\), \(e^z=0\) has no solution in \(\C\). If exists \(z=x+iy\) such that \(e^z=0\), then \(e^xe^{iy} = 0 \implies e^x=0\). Contradiction!
Inverses
\(f:\Omega \ra \C\). Then
\(g : \text{Range}\big(f\big) \ra \Omega\) is an inverse of \(f\)
says that
\( f\circ g: \Omega \ra \Omega \) is the identity,
i.e. \(\left(f \circ f \right)(z) = z\) for every \(z\in \Omega\).
E.g. \(z\mapsto z+1\), \(z\mapsto z-1\) are inverses
\(\C\ra \C\);
\(z\mapsto 1/z\) is its own inverse,
\(\C_*\ra \C_*\).
Inverse of \(\exp\)?