# complex variables

• Sep 27th 2008, 08:52 PM
PvtBillPilgrim
complex variables
How do I express sin(e^i) in terms of z = x + yi with x,y in R.

I say that e^i = cos1 + i*sin1 (since e^z = e^x(cosy + i*siny).
Then sin(cos1 + i*sin1) = [e^(i*(cos1 + i*sin1)) - e^(-i*(cos1 + i*sin1))]/2i, which equals (e^(i*cos1 - sin1) - e^(-i*cos1 + sin1))/2i. How do I write this in the form x + yi?

• Sep 27th 2008, 09:05 PM
Jhevon
Quote:

Originally Posted by PvtBillPilgrim
How do I express sin(e^i) in terms of z = x + yi with x,y in R.

I say that e^i = cos1 + i*sin1 (since e^z = e^x(cosy + i*siny).
Then sin(cos1 + i*sin1) = [e^(i*(cos1 + i*sin1)) - e^(-i*(cos1 + i*sin1))]/2i, which equals (e^(i*cos1 - sin1) - e^(-i*cos1 + sin1))/2i. How do I write this in the form x + yi?

note that if a complex number is given in the form $re^{i \theta}$ then we can find its rectangular form $x + iy$ using the following relationships:

$\cos \theta = \frac xr \implies \boxed{x = r \cos \theta}$ and $\sin \theta = \frac yr \implies \boxed{y = r \sin \theta}$
• Sep 28th 2008, 05:21 AM
Plato
Quote:

Originally Posted by PvtBillPilgrim
How do I express sin(e^i) in terms of z = x + yi with x,y in R.

$\sin (z) = \frac{1}{{2i}}\left( {e^{iz} - e^{ - iz} } \right)$
$\sin (x + yi) = \sin (x)\cosh (y) + i\left[ {\cos (x)\sinh (y)} \right]$