Question : Find the real and imaginary part of e^(sin(x+iy))
Solution:
Let \[z = e^(sin(x+iy))\] We shall write it as \[z = e^{Sin\left(x+iy\right)}=e^{Sin\left(x\right)Cos\left(iy\right)+\cos{\left(x\right)}\sin{\left(iy\right)}}\] Since \[Cos\left(iy\right)=\cosh{\left(y\right)}\ ,\ \ \sin{\left(iy\right)}=isinh\left(y\right)\] we shall substitute here \[z =e^{Sin\left(x\right)Cosh\left(y\right)+i\cos{\left(x\right)}\sinh{\left(y\right)}}\] \[=e^{Sin\left(x\right)Cosh\left(y\right)}e^{i\cos{\left(x\right)}\sinh{\left(y\right)}}\] \[z=e^{Sin\left(x\right)Cosh\left(y\right)}\left(\cos{\left(\cos{\left(x\right)}\sinh{\left(y\right)}\right)}\ +i\sin{\left(\cos{\left(x\right)}\sinh{\left(y\right)}\right)}\right)\] Thus we obtained real and imaginary part of e^(sin(x+iy)) respectively as
\[e^{Sin\left(x\right)Cosh\left(y\right)}\cos{\left(\cos{\left(x\right)}\sinh{\left(y\right)}\right)},\ e^{Sin\left(x\right)Cosh\left(y\right)}\ \sin{\left(\cos{\left(x\right)}\sinh{\left(y\right)}\right)}\]