The Complex Natural Logarithm Function

If $x \in \mathbb{R}$ and $x > 0$ then we know that the real-valued natural logarithm function is denoted $\ln x$ and is the inverse function to the real-valued exponential function $e^x$. That is, $e^{\ln x} = x$ and $\ln (e^x) = x$ for all valid $x$. The graph of the real-valued natural logarithm function is given as reference below.

We would like to extend this function to the domain of complex numbers. There is one problem that we will encounter though. Notice that the real-valued exponential function $e^x$ is injective, and so the inverse function $\ln x$ is well defined. The complex-valued exponential function $e^z$ is not injective on all of $\mathbb{C}$ though. For example, for all $k \in \mathbb{Z}$ we have that:

(1)

\begin{align} \quad e^{2k \pi i} = 1 \end{align}

Instead consider the domain $A_{y_0} \subset \mathbb{C}$:

(2)

\begin{align} \quad A_{y_0} = \{ z = x + yi \in \mathbb{C} : x \in \mathbb{R}, y \in [y_0, y_0 + 2\pi) \} \end{align}

Then $f : A_{y_0} \to \mathbb{C} \setminus \{ 0 \}$ defined for all $z \in A_{y_0}$ by $f(z) = e^z$ is injective (actually bijective). To show this, let $z_1, z_2 \in \mathbb{C} \setminus \{ 0 \}$ and suppose that $f(z_1) = f(z_2)$. Then $e^{z_1} = e^{z_2}$. We know that $e^z \neq 0$ for all $z \in \mathbb{C}$ so $e^z \neq 0$ for all $z \in A_{y_0}$ so:

(3)

\begin{align} \quad \frac{e^{z_1}}{e^{z_2}} = e^{z_1 - z_2} = 1 \end{align}

From the Properties of the Complex Exponential Function page we know that then $z_1 - z_2 = 2k\pi i$ for some $k \in \mathbb{Z}$. But since $z_1, z_2 \in A_{y_0}$ this can only be accomplished if $k = 0$ so $z_1 - z_2 = 0$, i.e., $z_1 = z_2$. So $f$ is injective.

We now show that $f$ is surjective. Let $w = r (\cos \theta + i \sin \theta) \in \mathbb{C} \setminus \{ 0 \}$. We want to show that there exists a $z \in A_{y_0}$ such that $e^z = w$, i.e.:

(4)

\begin{align} \quad e^x(\cos y + i \sin y) = r(\cos \theta + i \sin \theta) \end{align}

Therefore $e^x = r$ ($x = \log r$) and $y = \theta + 2k\pi i$ for some $k \in \mathbb{Z}$. Note that $log r \in \mathbb{R}$ and $y \in [y_0, y_0 + 2\pi)$ for a suitably chosen $k$, i.e., $y = \arg (z)$ where $y_0 \leq \arg (z) \leq y_0 + 2\pi$. So $f$ is indeed surjective.

From the comments made above we can appropriately define the complex natural logarithm function of the branch $A_{y_0}$ for all $z \in \mathbb{C} \setminus \{ 0 \}$ by:

(5)

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

Where $y_0 \leq \arg (z) < y_0 + 2\pi$. Note that if $z = x + yi = r (\cos \theta + i \sin \theta)$, $\mid z \mid = r$ and $\arg (z) = \theta$ then:

(6)

\begin{align} \quad e^{\log (z)} = e^{log r + i \theta} = e^{\log r} e^{i \theta} = re^{i\theta} = r(\cos \theta + i \sin \theta) = z \end{align}

(7)

\begin{align} \quad \log(e^z) = \log \mid e^z \mid + i \arg (e^z) = \log (e^x) + i \arg(e^x (\cos y + i \sin y)) = x \log (e) + i y = x + yi = z \end{align}

So our definition of the complex natural logarithm for the branch $A_{y_0}$ is an inverse to the complex exponential function.

If we remove the restriction on $\arg (z)$ above, then $\log (z)$ is still a well-defined function on $\mathbb{C} \setminus \{ 0 \}$ (though it is no longer an inverse to $e^z$), and for every $z \in \mathbb{C} \setminus \{ 0 \}$, $\log (z)$ will have infinitely many values. Regardless, we define this multivalued function to be the general complex natural logarithm function.

Definition: The Complex Natural Logarithm Function in the branch $A_{y_0} = \{z = x + yi : x \in \mathbb{R}, y_0 \leq y < y_0 + 2\pi \}$ is the function $\log : \mathbb{C} \setminus \{ 0 \} \to A_{y_0}$ defined for all $z \in \mathbb{C} \setminus \{ 0 \}$ by $\log (z) = \log \mid z \mid + i \arg(z)$ where $y_0 \leq \arg (z) < y_0 + 2\pi$.