Extra (optional) exercises – Short Day

These exercises are optional extra material for students who have finished the exercises of the day and have the time and interest for more.

Extra exercise 1: The complex logarithm

In this exercise we will study how to define the logarithm of a non-zero complex number. More precisely, the function \(\mathrm{Log}: \mathbb{C}\setminus \{0\} \to \mathbb{C}\) is defined by the expression \(\mathrm{Log}(z)=\ln(|z|)+i \mathrm{Arg}(z)\), where \(\mathrm{Arg}\) denotes the principal argument and \(\ln\) the natural logarithm.

Question a

Check that \(\mathrm{Log}(z)=\ln(z)\) for \(z \in \mathbb{R}_{>0}\). Also determine \(\mathrm{Log}(z)\) for \(z \in \mathbb{R}_{<0}\).

Answer

If \(z\in \mathbb{R}_{>0}\), then \(\mathrm{Log}(z)=\ln(z)+i0=\ln(z)\). If \(z\in \mathbb{R}_{<0}\), then \(\mathrm{Log}(z)=\ln(-z)+i \pi\).

Question b

Show that \(\mathrm e^{\mathrm{Log}(z)}=z\) for all \(z \in \mathbb{C}\setminus \{0\}\). Conclude that \(w=\mathrm{Log}(z)\) is a solution to the equation \(\mathrm e^w=z\) for all \(z \in \mathbb{C}\setminus \{0\}\).

Question c

The natural logarithm \(\ln\) is the inverse of the (real) exponential function. Find \(z \in \mathbb{C}\) such that \(\mathrm{Log}(\mathrm e^z) \neq z\). Conclude that \(\mathrm{Log}\) is not the inverse function of the complex exponential function.

Hint

The definition of \(\mathrm{Log}(z)\) uses the principal argument of \(z\). However, a complex number has infinitely many arguments.

Question d

The natural logarithm \(\ln\) satisfies \(\ln(x_1x_2)=\ln(x_1)+\ln(x_2)\) for all positive real numbers \(x_1\) and \(x_2\). Show that it is not true that \(\mathrm{Log}(z_1z_2)=\mathrm{Log}(z_1)+\mathrm{Log}(z_2)\) for all non-zero complex numbers \(z_1\) and \(z_2\).

Hint

In order to show that the formula \(\mathrm{Log}(z_1z_2)=\mathrm{Log}(z_1)+\mathrm{Log}(z_2)\) is not true for all non-zero complex numbers, a single counterexample is enough.

Answer

The formula does not hold in general since \(\mathrm{Log}((-1)^2) \neq \mathrm{Log}(-1)+\mathrm{Log}(-1)\).