Define the complex $\ln$-function on $\Bbb C\setminus\{0\}$ by $$\ln z = \ln|z| + i \arg (z)$$ where $\arg(z)$ is the unique angle $\theta \in [0,2\pi)$ so that $z = |z|(\cos(\theta) + i \sin(\theta))$.

Identify the points at which the above function is not continuous. Justify your answer using sequences.

I don't have idea how to start with this problem. Any hints and ideas will help me a lot.

Thank you.

  • $\begingroup$ $\ln z$ is the sum of two functions. If $\ln z$ is discontinuous, then at least one of the two summands is discontinuous. Is $\ln|z|$ continuous? Is $i\arg(z)$ continuous? $\endgroup$ – Michael Albanese Sep 23 '13 at 2:40
  • $\begingroup$ @MichaelAlbanese iarg(z) is discontinuous. $\endgroup$ – therexists Sep 23 '13 at 3:30
  • $\begingroup$ Correct. So you just need to focus on that part. Can you see where it is discontinuous? $\endgroup$ – Michael Albanese Sep 23 '13 at 3:47
  • $\begingroup$ @MichaelAlbanese Not really.... would you give me some more hints? $\endgroup$ – therexists Sep 23 '13 at 3:49
  • $\begingroup$ @MichaelAlbanese ln(z) must be positive so there must not continuities on negative axis. $\endgroup$ – therexists Sep 23 '13 at 3:51

Note that $\ln z$ is a sum of two functions. If $\ln z$ is discontinuous, at least one of them is discontinuous. As you have pointed out, $i\arg z$ is discontinuous, let's see why.

For any $z \neq 0$, $\arg z$ is the angle that $z$ makes with the positive $\operatorname{Re}(z)$-axis; as this angle could be described by infinitely many numbers (namely $\theta + 2k\pi$, $k \in \mathbb{Z}$), we need to choose a half-open interval of length $2\pi$ which we would like our angles to take values in; this interval is known as the principal value. In the question, you have chosen the interval $[0, 2\pi)$; it is worth noting that for the principal branch, it is common to choose $(-\pi, \pi]$.

The length of $z \in \mathbb{C}^*$ is not relevant here as $\arg z$ measures the angle, so we can always rescale and consider an element in the unit circle. Suppose now we are considering $z \in S^1 = \{z \in \mathbb{C} \mid |z| = 1\}$. One such point is $1$ which lies on the positive $\operatorname{Re}(z)$-axis. Whether you are using the interval $[0, 2\pi)$ or $(-\pi, \pi]$, we have $\arg 1 = 0$. Now consider what happens to $\arg z$ as we travel along the unit circle starting at $1$, first in a counter-clockwise direction, then in a clockwise direction. By doing this, you should be able to figure out where the discontinuity occurs; note, it will depend on your choice of principal value. The place where the discontinuity occurs is known as the branch cut.

Another way to think about the logarithm in the complex plane comes from the polar representation of a non-zero complex number. Recall, $z \in \mathbb{C}^*$ can be written in the form $re^{i\theta}$ where $r > 0$; as $e^{2k\pi i} = 1$, if $z = re^{i\theta}$ then $z = re^{i(\theta + 2k\pi)}$ for any $k \in \mathbb{Z}$. In order to make the polar representation unique, we need to choose a half-open interval of length $2\pi$ which we want $\theta$ to take values in; this is the principal value as before. Suppose this choice has been made and we choose $\arg$ to have the same principal value, then

$$\ln z = \ln re^{i\theta} = \ln |re^{i\theta}| + i\arg re^{i\theta} = \ln r + i\theta.$$

This is analogous to what would occur on the real line; that is, if $x = re^k$ then

$$\ln x = \ln re^k = \ln r + \ln e^k = \ln r + k.$$

If we have represented $z \in \mathbb{C}^*$ in polar form by $re^{i\theta}$ where $\theta$ is not in the principal value of $\arg$, then we have to shift by an appropriate multiple of $2\pi$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.