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Does the function $e^{z}=\sum\limits_{n=0}^{\infty}\frac{z^{n}}{n!}$ where $z$ is complex have all of the same properties of the real exponential function?

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    There some differences. For example $e^x+1$ is alway non-zero. On the other hand $e^z+1$ has infinitely many solutions.2012-11-13

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The biggest properties are

  • It is well-defined for all $z\in\mathbb C$
  • $e^{w+z}=e^{w}e^{z}$
  • $e^{0}=1$
  • It is continuous and (complex) differentiable, and its derivative is the same function.
  • For all $z$, $e^z\neq 0$.

Additionally:

  • It is equal to $\lim_{n\to\infty} (1+\frac{z}{n})^n$

What other properties might you want?

As other have pointed out in comments, there are a few properties it doesn't have. It is not $1-1$, so its inverse (the natural logarithm) is not as "nice" a function. You have to either leave it undefined or deal with multi-valued functions.

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    @Stefan Harald's comment is actually somewhat difficult to phrase in complex numbers. On the real line, $e^x$ grows arbitrarily large if you are traveling to the right towards infinity, while it grows arbitrarily close to zero as $x\to -\infty$. But in complex number, you have lots of additional directions to travel.2012-11-14