# 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)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)## 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)And we also have that:

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

(5)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)## 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)Therefore:

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