The Complex Natural Logarithm Function Examples 1

Recall from The Complex Natural Logarithm Function page that the complex natural logarithm function is defined for all $z \in \mathbb{C} \setminus \{ 0 \}$ by:

(1)

\begin{align} \quad \log (z) = \log \mid z \mid + i \arg (z) \end{align}

Where $\arg (z)$ is specified in some branch $A_{y_0} = \{ z = x + yi \in \mathbb{C} : x \in \mathbb{R}, y_0 \leq y < y_0 + 2\pi \}$, i.e., $y_0 \leq \arg (z) < y_0 + 2\pi$.

We will now look at some example problems regarding the complex natural logarithm function.

Example 1

Simplify $\log (1 + i)$ for the principal branch of the logarithm function, i.e., for all $z \in \mathbb{C} \setminus \{ 0 \}$, $0 \leq \mathrm{Arg}(z) < 2\pi$.

We have that:

(2)

\begin{align} \quad \log (2 + 3i) &= \log \mid 1 + i \mid + \mathrm{Arg} (1 + i) \\ & \log \sqrt{2} + \frac{\pi}{4} \end{align}

Example 2

Find all possible values for $\log (1)$.

We have that:

(3)

\begin{align} \quad \log(1) = \log \mid 1 \mid + i \arg (1) = i arg(1) \end{align}

Note that $\arg(1) = 2k \pi$ where $k \in \mathbb{Z}$, and so for each $k \in \mathbb{Z}$:

(4)

\begin{align} \quad \log(1) = 2k \pi i \end{align}

Example 3

Find all possible values for $\log (-i)$.

We have that:

(5)

\begin{align} \quad \log (-i) = \log \mid -i \mid + i\arg (-i) = i \arg (-i) \end{align}

Note that $\arg(-i) = \pi + 2k\pi$ where $k \in \mathbb{Z}$, and so for each $k \in \mathbb{Z}$:

(6)

\begin{align} \quad \log (-i) = (2k + 1)\pi i \end{align}