# How do you solve e^x=0?

Jun 30, 2016

There is no $x$ such that ${e}^{x} = 0$

#### Explanation:

The function ${e}^{x}$ considered as a function of Real numbers has domain $\left(- \infty , \infty\right)$ and range $\left(0 , \infty\right)$.

So it can only take strictly positive values.

When we consider ${e}^{x}$ as a function of Complex numbers, then we find it has domain $\mathbb{C}$ and range $\mathbb{C} \text{\} \left\{0\right\}$.

That is $0$ is the only value that ${e}^{x}$ cannot take.

Note that ${e}^{x + y i} = {e}^{x} {e}^{y i} = {e}^{x} \left(\cos y + i \sin y\right)$

We have already noted that iof $x \in \mathbb{R}$ then ${e}^{x} > 0$.

For pure imaginary exponents the result is on the unit circle, specifically:

${e}^{y i} = \cos y + i \sin y \ne 0$

So ${e}^{x + y i} \ne 0$ for all $x , y \in \mathbb{R}$