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       

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

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       

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

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       

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

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

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

and

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

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

Noah Dana-Picard 2007-12-24