The Derivatives of the Complex Exponential and Logarithmic Functions

# The Derivatives of the Complex Exponential and Logarithmic Functions

We will now look at some elementary complex functions, their derivatives, and where they are analytic. We begin by looking at the complex exponential function which we looked at on The Complex Exponential Function page and the complex logarithmic function which we looked at on The Complex Natural Logarithm Function page.

## The Derivative of the Complex Exponential Function

Theorem 1: Let $f(z) = e^z$. Then $f$ is analytic on all of $\mathbb{C}$ and $f'(z) = e^z$. |

**Proof:**Let $z = x + yi$. Then:

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

- So for $f = u + iv$ we have that:

\begin{align} \quad u(x, y) = e^x \cos y \end{align}

(3)
\begin{align} \quad v(x, y) = e^x \sin y \end{align}

- The partial derivatives of $u$ and $v$ clearly exist and are continuous on all of $\mathbb{C}$, so all that remains to show is that the Cauchy-Riemann equations are satisfied. We compute the partial derivatives of $u$ and $v$:

\begin{align} \quad \frac{\partial u}{\partial x} = e^x \cos y \quad , \quad \frac{\partial v}{\partial y} = e^x \cos y \quad , \quad \frac{\partial u}{\partial y} = -e^x \sin y \quad , \quad \frac{\partial v}{\partial x} = e^x \sin y \end{align}

- So indeed the Cauchy-Riemann equations are satisfied on all of $\mathbb{C}$ since $\displaystyle{\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}}$ and $\displaystyle{\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}}$ on all of $\mathbb{C}$. So by the Cauchy-Riemann theorem, $f$ is analytic on all of $\mathbb{C}$. $\blacksquare$

## The Derivative of the Complex Logarithmic Function

Theorem 2: Let $f(z) = \mathrm{Log} (z)$ where $\mathrm{Log(z)} = \log \mid z \mid + i \mathrm{Arg} (z)$ where $\mathrm{Arg} (z)$ is such that $0 < \mathrm{Arg} (z) < 2\pi$ (i.e., the "Principal" branch of the logarithmic function). Then $f$ is analytic on all of $\mathbb{C} \setminus \{ x + yi \in \mathbb{C} : x \geq 0, y = 0 \}$ and $f'(z) = \frac{1}{z}$ on this set. |

**Proof:**Recall that if $f(z) = e^z$ where $z \in \{ z = x + yi \in \mathbb{C} : x \in \mathbb{R}, 0 < y < 2\pi \}$ then $f$ has an inverse, namely $f^{-1} (z) = \mathrm{Log} (z)$.

- Recall that the Inverse Function Theorem for Complex Functions says that if $f : A \to \mathbb{C}$ is analytic (and $f'$ is continuous which is redundant as we will see later) and $f'(z_0) \neq 0$ then there exists open sets $U$ of $z_0$ and $V$ of $f(z_0)$ such that $f : U \to V$ is a biection and that $f^{-1}$ is analytic on $V$.

- In particular, when $f(z) = e^z$ is restricted such that $z \in \{ z = x + yi \in \mathbb{C} : x \in \mathbb{R}, 0 < y < 2\pi \}$ then $f^{-1}$ exists (as mentioned above) and more importantly, $f(z) \neq 0$ (since recall that $e^z = 0$ and $e^z \neq 0$ on this set. So by the inverse function theorem we have that if $w = f(z)$ then:

\begin{align} \quad \frac{d}{dw} f^{-1} (z) = \frac{1}{f'(z)} = \frac{1}{w} \end{align}

- Thus $\frac{d}{dz} \mathrm{Log}(z) = \frac{1}{z}$ and $\mathrm{Log}(z)$ is analytic on all of $\mathbb{C} \setminus \{ x + yi \in \mathbb{C} : x \geq 0, y = 0 \}$. $\blacksquare$