# 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 $$\exp$$

$$\tiny\exp$$
$$\ra$$
$$\tiny h < 2\pi$$

#### Images under $$\exp$$

$$\tiny\exp$$
$$\ra$$
$$\tiny h \geq 2\pi$$

#### Properties of $$\exp$$

Many "laws" for $$\R$$-$$\exp$$ extend to $$\C$$-$$\exp$$:

1. $$e^0=1$$;
2. $$e^{-z}=1/e^{z};$$
3. $$e^{z_1+z_2}= e^{z_1}e^{z_2};$$
4. $$e^{z_1-z_2}= \dfrac{e^{z_1}}{e^{z_2}};$$
5. $$\left(e^{z_1}\right)^{z_2}= e^{z_1z_2}$$.

#### Properties of $$\exp$$

Some things do not extend:

1. $$e^x > 0$$ for all $$x\in \R$$, but e.g. $$e^{i\pi}=-1$$ and $$e^{i\pi/4}\in\C\setminus \R$$;
2. $$x \mapsto e^x$$ 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 $$\exp$$?