What does $ e^{iz} $ mean? It was used to define complex sine in my book.
By James Williams
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?
$\endgroup$ 31 Answer
$\begingroup$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.
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