# Complex Logarithm Function

• May 22nd 2011, 01:37 PM
garunas
Complex Logarithm Function
Question:

Show using a suitable contour that the complex logarithm function is defined by:

$Log(z) = \int_{[1,z]} \frac{1}{\zeta}\hspace{2mm} d\zeta = ln|z| + i\hspace{1mm}Arg(z)\hspace{5mm} for\hspace{3mm} z \in \mathbb{C}_{\pi} = \{z : Arg(z) \neq \pi \}$

What I did was i used the contour $\gamma (t) = (1-t) + zt \hspace{3mm} 0 \leq t \leq 1$ so that:

$\int_{[1,z]} \frac{1}{\zeta} \hspace{2mm}d\zeta = \int_{0}^1 \frac{\gamma'(t)}{\gamma(t)} \hspace{2mm} dt = \int_{0}^1 \frac{z-1}{1-t+zt}\hspace{2mm} dt = ln|z|$

I've reached here and now I'm absolutely stuck. What do i do next?
• May 23rd 2011, 03:28 AM
Opalg
Quote:

Originally Posted by garunas
Question:

Show using a suitable contour that the complex logarithm function is defined by:

$Log(z) = \int_{[1,z]} \frac{1}{\zeta}\hspace{2mm} d\zeta = ln|z| + i\hspace{1mm}Arg(z)\hspace{5mm} for\hspace{3mm} z \in \mathbb{C}_{\pi} = \{z : Arg(z) \neq \pi \}$

What I did was i used the contour $\gamma (t) = (1-t) + zt \hspace{3mm} 0 \leq t \leq 1$ so that:

$\int_{[1,z]} \frac{1}{\zeta} \hspace{2mm}d\zeta = \int_{0}^1 \frac{\gamma'(t)}{\gamma(t)} \hspace{2mm} dt = \int_{0}^1 \frac{z-1}{1-t+zt}\hspace{2mm} dt = ln|z|$

I've reached here and now I'm absolutely stuck. What do i do next?

The formula $\textstyle\int\frac1\zeta\,d\zeta = \ln|\zeta|$ is valid for an integral over the positive real line or the negative real line, but it does not work for integrals along complex contours.

I think that the easiest way to do this question is to use a different contour, namely one that goes along the real axis from 1 to |z| and then around a circular arc from |z| to z. The integral along the straight line segment will then give you $\ln|z|$ without any difficulty. For the circular arc, write $z = re^{i\arg z}$ and parametrise the path by $\gamma(\theta) = re^{i\theta} \ (\theta\in[0,\arg z]).$ Then

$\int_0^{\arg z} \frac{\gamma'(\theta)}{\gamma(\theta)}\,d\theta = \int_0^{\arg z} \frac{ire^{i\theta}}{re^{i\theta}}\,d\theta = i\arg z.$
• May 23rd 2011, 05:46 AM
garunas
Absolutely amazing!! Cheers!