The Complex Exponential Function Examples 1

Recall from The Complex Exponential Function page that if $z = x + yi \in \mathbb{C}$ then we defined the complex exponential function to be:

(1)

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

We will now look at some example problems regarding this function.

Example 1

Write $e^{3 + \pi i}$ in the form $a + bi$.

We have that:

(2)

\begin{align} \quad e^{3 + \pi i} = e^3 (\cos \pi + i \sin \pi) = e^3 (-1) + e^3 (0)i = -e^3 + 0i \end{align}

Example 2

Find all values $z \in \mathbb{C}$ for which $\overline{e^{iz}} = e^{i\overline{z}}$.

Let $z = x + yi$. Then $iz = -y + xi$, and so:

(3)

\begin{align} \quad \overline{e^{iz}} = \overline{e^{-y + xi}} = \overline{e^{-y} (\cos x + i \sin x)} = \overline{ e^{-y} \cos x + i e^{-y} \sin x} = e^{-y} \cos x + i (-e^{-y} \sin x) \quad (*) \end{align}

And we also have that:

(4)

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

Set $(*)$ equal to $(**)$. Then we must simultaneously solve the following:

(5)

\begin{align} \quad (1) & \: e^{-y} \cos x = e^y \cos x \\ \quad (2) & \: -e^{-y} \sin x = e^y \sin x \end{align}

First assume that $\cos x = 0$. Then $x = \frac{\pi}{2} + k\pi$ for some $k \in \mathbb{Z}$. If $x = ..., -\frac{3\pi}{2}, \frac{\pi}{2}, \frac{5\pi}{2}, ...$ for some $m \in \mathbb{Z}$ then $\sin x = 1$, and so $-e^{-y} = e^y$ which has no solutions. If $x = ..., -\frac{\pi}{2}, \frac{3\pi}{2}, \frac{7\pi}{2}, ...$ then $\sin x = -1$ and so $e^{-y} = -e^y$ which has no solutions.

So we may assume that $\cos x \neq 0$. Dividing equation (1) by $\cos x$ on both sides gives us that $e^{-y} = e^y$ which implies that $y = 0$. Substituting this into equation (2) gives us that $- \sin x = \sin x$. So $2 \sin x = 0$ and $\sin x = 0$ which happens when $x = k\pi$ for some $k \in \mathbb{Z}$.

Therefore the values of $z = x + yi \in \mathbb{C}$ for which $\overline{e^{iz}} = e^{i\overline{z}}$ are:

(6)

\begin{align} \quad z = k\pi + 0i \:, \: k \in \mathbb{Z} \end{align}

Example 3

Write $e^{z^2}$ in the form $a + bi$. What is $\mid e^{z^2} \mid$? What is $\arg (e^{z^2})$?

Let $z = x + yi$. Then:

(7)

\begin{align} \quad z^2 = (x + yi)(x + yi) = (x^2 - y^2) + 2xyi \end{align}

Therefore:

(8)

\begin{align} \quad e^{z^2} = e^{(x^2 - y^2) + 2xyi} = e^{x^2 - y^2} (\cos 2xy + i \sin 2xy) = (e^{x^2 - y^2}\cos 2xy) +i(e^{x^2 - y^2} \cos 2xy) \end{align}

We have that $\mid e^{z^2} \mid = e^{x^2 - y^2}$, and $\arg (e^{z^2}) = 2xy$.