The logarithm of a complex number.

    

Definition 4.4.1       

$\forall z \in \mathbb{C}^*, \; \log z = \ln \vert z\vert + i \operatorname{arg}\nolimits (z)$ ,

where $\operatorname{arg}\nolimits (z)$ is defined up to an additive multiple of $2 \pi$ .

Example 4.4.2       

$$\log (1+i) = \ln \sqrt{2} + i \left( \frac {\pi}{4} + 2k \pi \right) = \frac 12 \ln 2 + i \left( \frac {\pi}{4} + 2k \pi \right).$$

$$\log (-1) = \ln 1 + i \operatorname{arg}\nolimits (1) = i (2k \pi).$$

$$\log (4i) = \ln 4 + i \operatorname{arg}\nolimits (4i) = 2 \ln 2 + i \left( \frac {\pi}{2} + 2k \pi \right).$$

We denote Log$(z)$ the principal value of $\log z$ , i.e. the value corresponding to the principal value $\operatorname{arg}\nolimits (z)$ of $\operatorname{arg}\nolimits (z)$ (recall that $- \pi < \operatorname{arg}\nolimits (z) \leq \pi$ ).

Example 4.4.3       

$$(i) = \ln 1 + i \frac {\pi}{2}.$$

$$(1+i) =\ln \sqrt{2} + i \frac {\pi}{4} = \frac 12 \ln 2 + i \frac {\pi}{4}.$$

Proposition 4.4.4       

1. $\log (z_1 z_2)= \log z_1 + \log z_2 + 2k i \pi$ .
2. $\log z^n = n \log z + 2k i \pi$ .
3. $\log e^z = z + 2k i \pi$ .

Proposition 4.4.5   The logarithmic function is analytic on its domain.

For a proof, use Cauchy-Riemann equations (v.s. 3).

Example 4.4.6  

We resume our work from Example 3.5.

Example 4.4.7  

Proposition 4.4.8       

$$\forall z \in \mathbb{C}, \; \log e^z = z + 2k \pi i.$$

The proof is simple: let $z=x+iy$ , where $x,y \in \mathbb{R}$ . Then we have:

$$e^z=e^x (\cos y +i \sin y)$$

and

$$\log e^z= \ln e^x + i (y + 2k \pi) = x + iy +2k \pi i.$$

Proposition 4.8 means that the $\log$ function is not exactly the inverse of the complex exponential function in basis $e$ .