The Complex Natural Logarithm Function

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.

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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}
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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 $y_0 \leq \arg (z) < y_0 + 2\pi$.
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