Find the real and imaginary terms of the function $z=\sin(x-iy)$ If I have the function:
$$z=\sin(x-iy)$$
How do I find the real and imaginary parts of this function without relying on the hyperbolic identities?  In my attempt of this problem, I first used the trig identities to simplify this to: $$\sin(x)cos(iy)-cos(x)sin(iy)$$ but I am unsure how to continue from here without using the identities: 
$$ \cos(iy) = \cosh(y), \quad    \sin(ib) = \sinh(y)$$
Thanks in advance.
 A: Hint: Use Eulers Formula for $\sin(x)$
$$\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$$
Derivation from Eulers Formula $e^{ix}=\cos(x)+i\sin(x)$:
$$e^{ix}=\cos(x)+i\sin(x)$$
Equally we can plug in $-x$ for $x$ and use oddness of sine and evenness of cosine.
$$e^{-ix}=\cos(-x)+i\sin(-x)=\cos(x)-i\sin(x)$$
Now subtract both equations to get the formula for $\sin(x)$. Adding both formulas will give you an analogue formula for cosine:
$$\cos(x)=\frac{e^{ix}+e^{-ix}}{2}$$
Now let us apply this formula to $z=sin(x-iy)$
$$z=\frac{e^{i(x-iy)}-e^{-i(x-iy)}}{2i}=\frac{e^{ix+y}-e^{-ix-y}}{2i}$$
$$=\frac{e^ye^{ix}-e^{-y}e^{-ix}}{2i}=\frac{e^y[\cos(x)+i\sin(x)]-e^{-y}[\cos(x)-i\sin(x)]}{2i}$$
From here it should be easy to calculate the real and imaginary terms.
A: Notice, we can simplify as follows $$z=\sin(x-iy)$$
$$=\frac{e^{i(x-iy)}-e^{-i(x-iy)}}{2i}$$
$$=-i\left(\frac{e^{y+ix}-e^{-(y+ix)}}{2}\right)$$
$$=i\left(\frac{e^{-y}e^{-ix}-e^{y}e^{ix}}{2}\right)$$
$$=i\left(\frac{e^{-y}(\cos x-i\sin x)-e^{y}(\cos x+i\sin x)}{2}\right)$$
$$=i\left(\frac{(e^{-y}-e^{y})\cos x-i(e^{y}+e^{-y})\sin x}{2}\right)$$
$$=i\frac{(e^{-y}-e^{y})\cos x}{2}-i^2\frac{(e^{y}+e^{-y})\sin x}{2}$$
$$z=\frac{(e^{y}+e^{-y})}{2}\sin x+i\frac{(e^{-y}-e^{y})}{2}\cos x$$
Now, the real & the imaginary parts are as follows $$Re(z)=\color{red}{\frac{(e^{y}+e^{-y})}{2}\sin x}$$ 
$$Im(z)=\color{red}{\frac{(e^{-y}-e^{y})}{2}\cos x}$$ 
