How do you solve $e^{x} = 0$ $e^x=0$ ?

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Answer 1 · 2016-06-30T06:31:51

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

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

So it can only take strictly positive values.

When we consider $e^{x}$ $e^x$ as a function of Complex numbers, then we find it has domain $CC$ and range $CC "\" { 0 }$ .

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

Note that $e^(x+yi) = e^x e^(yi) = e^x(cos y+i sin y)$

We have already noted that iof $x in RR$ then $e^x > 0$ .

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

$e^(yi) = cos y + i sin y != 0$

So $e^(x+yi) != 0$ for all $x, y in RR$

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