Proof of the Week for 20120827

Theorem: The natural logarithm of a complex variable: $\ln z = \ln r + i\theta + i 2\pi k, k \in \mathbb{Z}$

Proof: Let $z = r e^{i\theta} = r e^{i\left(\theta + 2 \pi k \right)}$, then take natural logs of both sides to yield, $\ln z = \ln \left(r e^{i\left(\theta + 2 \pi k \right)} \right) = \ln r + \ln e^{i\left(\theta + 2 \pi k \right)}$. Then we have $\ln z = \ln r + i\theta + i 2\pi k, k \in \mathbb{Z}$.

Notes:

1. This is an infinitely valued function. To recover the single valued function familiar for real numbers, we designate $z = 0$ as a branch point and the negative real axis as a branch cut. This allows us to define the principal branch of this multivalued function as $Ln z = \ln r + i Arg z, -\pi < Arg z \le \pi$. If $z$ is real, then $Arg z = 0$ and $Ln z = \ln r$.
2. We have used the fact that the complex exponential function has a period of $2 \pi$. Thus $e^{i\theta} = e^{i\left(\theta + 2 \pi k \right)}, k \in \mathbb{Z}$