# What does $e^{iz}$ mean? It was used to define complex sine in my book.

I have been trying to understand the notion of complex sine that was defined in my book. The book first starts out defining $e^{z}$ as

$$\text{If } z = x + iy, \text{ then } e^z = e^{x}\cos y + ie^x\sin y$$

Next, the book states that for any $y \in \mathbb{R}$:

$$\begin{eqnarray} e^{iy} &=& \cos y + i \sin y\\ e^{-iy} &=& \cos y - i \sin y\\ \implies \sin \ y &=& \frac{1}{2i}(e^{iy} - e^{-iy}) \end{eqnarray}$$

I followed up to this point, but then they generalized this to define $\sin z \text{ for } z \in \mathbb{C}$. This is the definition they gave:

$$\sin z = \frac{1}{2i}(e^{iz} - e^{-iz})$$

I do not understand want $e^{iz}$ means in this equation. If $z = x + iy$ then does $e^{iz} = e^{-y + ix} = e^{-y}\cos x + ie^{-y}\sin x$. Is this correct, or does it mean something else?

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That's correct. – saz Mar 23 '13 at 8:07
Yeah, you have it right. – Jesse Madnick Mar 23 '13 at 8:17
Thanks for the confirmation. – Max Mar 23 '13 at 8:33
This is basically an analytic continuation. You have a complex analytic (entire) function that happens to coincide with the sine on the real line. By analytic function uniqueness theorems, this formula is the only sensible way to extend the sine to the complex plane. – Yoni Rozenshein Mar 23 '13 at 9:53

Let $z=x+i y$, where $x$ and $y$ are real. Then
$$e^{i z} = e^{i (x+i y)} = e^{i x} e^{-y} = (\cos{x}+i \sin{x}) e^{-y}$$
So $e^{i z}$ is a complex number with real part $e^{-y} \cos{x}$ and imaginary part $e^{-y} \sin{x}$, as you state. The important thing to recognize is that $x$ and $y$ are the real and imaginary parts of $z$, respectively.