Question

Show that the function omega = sin z satisfies the Cauchy-Riemann and Laplace
equations

Answer 1

The Cauchy-Riemann equation:

Let $f(x,y)=u(x,y)+iv(x,y)$ , where $u$ and $v$ are real.

Then u=sinxcoshy and v=cosxsinhy st.

$u_x=\cos x\cosh y = v_y,~~~~~~~v_x=-\sin x\sinh y = - u_y,$

i.e., CR conditions hold.

$f(z)=\frac{(e^{ix}-e^{-ix})(e^y+e^{-y})}{4i}+i\frac{(e^{ix}+e^{-ix})(e^y-e^{-y})}{4}=\frac{e^{ix}e^{-y}-e^{-ix}e^{y}}{2i}= \frac{e^{i(x+iy)}-e^{-i(x+iy)}}{2i}=\sin z;$

The Laplace equation:

$\Omega_{x}=\cos(x+iy),\Omega_{xx}=-\sin(x+iy),$

$\Omega_{y}=\cos (x+iy)*i, \Omega_{yy}=-\sin(x+iy)*i^2=\sin(x+iy).$

Thus, $\Omega_{xx}+\Omega_{yy}=0$