Complex exponential

Question

I know that the equation $e^{z}=-1$ has no solution had if been $z$ is a real number. So does the equation also has no solution when $z$ is complex?

Answer 1

On of the most popular formulas, if such a notion makes sense, is Euler's identity, which states that $$e^{\pi i}=-1.$$

Also compare this ;)

Answer 2

My answer may be a little bit formal. Regarding the equation $e^{z}=-1$, note that :

• $\exp$ is a group homomorphism whose kernel is $2i\pi \mathbb{Z}$. (the kernel is the set of all complex numbers $z$ such that $e^{z}=1$).
• You know that $e^{i\pi} = -1$, which gives you a particular solution to $e^{z}=-1$.

These two facts give you that

$$ \left\{ z \in \mathbb{C}, \; e^{z}=-1 \right\} = i\pi + 2i\pi \mathbb{Z} $$

Answer 3

Write $\;z=x+iy\;,\;\;x,y,\in\Bbb R\;$ , so by definition:

$$e^z=e^{x+iy}=e^xe^{iy}=e^x\left(\cos y +i\sin y\right)$$

It is now enough to use the periodicity of the trigonometric functions:

$$x=0\;,\;\;y=(2k+1)\pi\;,\;\;k\in\Bbb Z\implies e^x=1\;,\;\;\sin y=0\;,\;\;\cos y=-1\;\ldots$$