# MATH3401

Lecture 8

### Inverse of $$\exp$$

Too much to hope for "$$?=\log = \log_e$$", due to $$\exp$$ being periodic (period $$2\pi i$$) in $$\C$$.

### Inverse of $$\exp$$

Note that $$e^w=z$$. Write $$z = r e^{i \Theta}$$, with $$r> 0$$ and $$\Theta = \Arg(z) \in (-\pi, \pi]$$.

Write $$w= u+iv$$. Then $$z = e^{w} = e^{u+iv} = e^u e^{iv}$$.

This implies \left\{ \begin{align*} e^u &= r \\ v &= \Theta + 2 k \pi ,\; k\in \Z \end{align*} \right.

### Inverse of $$\exp$$

So $$u = \ln r$$, meaning logarithm to the base $$e$$ of the positive real number $$r$$.

So \begin{align*} w &= u + iv \\ &= \ln r + i (\Theta + 2 k \pi),\; k\in \Z \\ &= \ln |z| + i \,\arg(z) \end{align*}

### Inverse of $$\exp$$

This defines the multi-valued function $$\log$$ on $$\C_*$$.

Note: $$\exp\left(\log z\right) = z$$ and $$\log\left(\exp z\right)= z+ 2 k \pi \,i$$

📝 Check: \left\{ \begin{align*} \log \left( z \,\xi \right) &= \log z + \log \xi \\ \log \left( \frac{z }{\xi} \right) &= \log z -\log \xi \end{align*} \right.

### Inverse of $$\exp$$

As with $$\Arg$$/$$\arg$$, define $\Log (z) = \ln |z| + i \Arg(z) \text{ on } \C_*.$

$$\Log$$ is a single valued, but discontinuous on the negative real axis $$\cup \{0\}$$, since $$\Arg$$ is.

Indeed, $$\Arg$$/$$\Log$$ are not even defined at $$0$$.

$$\Log$$ is called the principal logarithm. As with $$\Arg$$, there may hold: $\Log\left(z_1 z_2\right)\neq \Log z_1 + \Log z_2.$

### Complex exponent

Take $$c\in \C$$. Set $z^c \exp \left( c \log z\right), \quad z\neq 0.$

Note: corresponds to usual index laws for $$c=n\in \Z$$ ; and for $$c=1/n$$, we recover what we did in Lecture 4.

Remark: Brown & Churchill defines $$z^{1/n}$$ as a multi-valued function, and in particular defines the Principal Value, that is $\text{PV}\left(z^{1/n}\right) = |z|^{1/n}\exp \left(\frac{i \Arg(z)}{n}\right).$

### Complex exponent

Same procedure works for $$z\mapsto z^c$$: \begin{align*} \text{PV}\left(z^{c}\right) &= \exp \big(c\,\Log (z)\big) \\ &= \exp\big(c \ln |z| + i \,c\, \Arg (z)\big). \end{align*}

E. g. $$\text{PV}\left[(1-i)^{4i}\right]$$:

$$\qquad\qquad =\exp\big( 4i \left[ \ln |1-i| + i \Arg (1-i) \right] \big)$$

$$\qquad\qquad =\exp\big( 4i\ln\sqrt{2} -4(-\pi/4) \big)$$

$$\qquad\qquad =e^{\pi} \exp \left(4i\ln \sqrt{2}\right)$$

$$\qquad\qquad =e^{\pi} \cos \left(2\ln 2\right) + i\, e^{\pi} \sin \left(2\ln 2\right)$$.

### Complex exponent

Remark: Sometimes we need to use a different single-valued

$$\log\;$$ and $$\;\arg$$,

e. g. we may need to integrate around a contour like this:

### Remark

In this case, choose/define $${\Large\mathcal a}\mathcal{rg }(z)$$ such that $-\frac{\pi}{2}< {\Large\mathcal a}\mathcal{rg }(z) \leq \frac{3\pi}{2}$ where this $${\Large\mathcal a}\mathcal{rg }(z)$$ is a particlar value of $$\arg(z)$$: Leads to a new, single-valued $$\mathcal \Log$$, etc.