8
$\begingroup$

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?

$\endgroup$
3
  • 4
    $\begingroup$ That's correct. $\endgroup$
    – saz
    Mar 23, 2013 at 8:07
  • 1
    $\begingroup$ Yeah, you have it right. $\endgroup$ Mar 23, 2013 at 8:17
  • $\begingroup$ 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. $\endgroup$ Mar 23, 2013 at 9:53

1 Answer 1

8
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .