Definition (Branch): A branch of a multiple-valued function $f$ is $\color{teal}{\text{any}}$ single-valued function $F$ that is $\underline {analytic}$ in some domain at each point $z$ $\color{teal}{\text{of which}}$ the value $F(z)$ is one of the possible values of $f.$

Here I don't understand the word analytic

This is the definition of branch of multi valued functions.

Why does $$\operatorname{Log}(z)=\ln(r)+i\theta \quad (r>0,-\pi<\theta<\pi)$$ is a branch of the multi valued function $\log(z)?$

Is $\ln(r)+i\theta $ a possible value of $\log(z)$?


1 Answer 1


Consider the complex plane. Along the x axis you have the real numbers, along the y-axis, you have imaginary numbers, $[...-3i, -2i, -i, 0, i, 2i, 3i,...]$. The basic form of $z$ utilizes this cartesian system.
\begin{align*} z & = x + iy\\ \end{align*} When $z$ it's in polar form, the only values in polar coordinates are $r$ and $\theta$, where $r$ is the distance to a point in the plane, and $\theta$ is the angle to that point. We know that for an arc to travel the length of a circle about the origin (or any point). As you go further along either axis, the values become$[-\pi/2, 0, \pi/2, 2i\pi/2, 3i\pi/2, 4i\pi/2,...]$.

We also know that adding $2\pi$ to any $\theta$ maintains the value of that function. Thus a change of $2k\pi$ constitutes a branch of $z$. Limiting the branch only to only one change of $\pi$ such as $[-\pi,\pi]$ as you have, is a change of $2\pi$.

\begin{align*} z & = re^{i\theta} = e^x \cdot e^{iy} = r(\cos\theta + i\sin\theta) = r(\cos(\theta + 2k\pi) + i\sin(\theta + 2k\pi) = e^x \dot\ e^{i(\theta + 2k\pi)} = re^{i(\theta + 2k\pi)} \end{align*}


$Log(z) = Log(r) + Log(e^{i(\theta + 2k\pi)}) = Log(r) + i(\theta + 2k\pi)$

which is a multi-valued function of Log(z) as $Log(z)$ will be the same depending on what branch you choose of the $2k\pi$.

But if you limit $\theta$ to $[-\pi,\pi]$, then you will have

$Log(z) = Log(r) + Log(e^{i\theta}) = Log(r) + i\theta$ within the single branch of $Log(z)$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .