The Graph of the Complex Exponential Function

The Graph of the Complex Exponential Function

Recall from The Complex Exponential Function page that if $z = x + yi$ then we defined:

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

We noted that in this form that the modulus of $e^z$ is $\mid e^z \mid = e^x$ and that the argument of $e^z$ is $\arg (e^z) = y$.

We will now investigate the behaviour of this complex-valued function. One way to do this is graphically by determining what the graph of $e^z$ looks like by taking various paths in the domain. We will look at the behaviour of $e^z$ along vertical and horizontal lines in $\mathbb{C}$

We first begin by looking at the behaviour of $e^z$ along vertical lines. First, recall that $e^{iy} = \cos y + i \sin y$. We can represent this as a function with parametric equation $e^{iy} = (\cos y, \sin y)$ as $y$ varies. So, as $y$ varies, i.e., as we move northward starting at the origin, then for every length of $2\pi$ we go through the domain, the unit circle starting at the point $1$ is traced counterclockwise once in the range as depicted below:


If we fix any $x_0 \in \mathbb{R}$ and move northward along the vertical line passing through $x_0$ then $e^z = e^{x_0 + yi} = e^{x_0}(\cos y + i \sin y) = (e^{x_0} \cos y, e^{x_0} \sin y)$ and $e^{z}$ traces out the circle centered at the origin with radius $e^{x_0}$.

Now if we fix any $y_0 \in \mathbb{R}$ and move right along the horizontal line passing through the point $0 + y_0i$ in the complex plane then $e^{z}$ traces out a half line starting at (but not ever equalling to) $0$ and that has angle $\arg z = y$ with the positive $x$-axis as depicted below:

Unless otherwise stated, the content of this page is licensed under Creative Commons Attribution-ShareAlike 3.0 License