Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
3  
That's correct. –  saz Mar 23 '13 at 8:07
1  
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

1 Answer 1

up vote 2 down vote accepted

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.