a) f(z)=g($\bar{z}$)
Ok, so I am trying to show using Cauchy Riemann equation that this function is either complex differentiable or not. The only problem is I'm not 100% sure if I took the partials of $u_y$ and $v_y$ correctly. The mess below is my work.
g(z)=u(x,y)+iv(x,y) $\ \ \ \ \ \ \ $f(z)=$\tilde{u}(x,y)+ i\tilde{v}(x,y)$
g($\bar{z}$)=u(x,-y)+iv(x,-y)
$\frac{\partial}{\partial x}[\tilde{u}(x,y)] = \frac{\partial}{\partial x}[u(x,-y)]=u_x(x,-y)$
$\frac{\partial}{\partial y}[\tilde{u}(x,y)] = \frac{\partial}{\partial y}[u(x,-y)]=u_y(x,-y)$ [or is it $-u_y(x,-y)$] <--- This is where I am confused
$\frac{\partial}{\partial x}[\tilde{v}(x,y)] = \frac{\partial}{\partial x}[v(x,-y)]=v_x(x,-y)$
$\frac{\partial}{\partial y}[\tilde{v}(x,y)] = \frac{\partial}{\partial y}[v(x,-y)]=v_y(x,-y)$ [or is it $-v_y(x,-y)$] <--- This is where I am confused
Then using Cauchy-Riemann equation we want to show if f(z) is analytic:
$\tilde{u}_x=\tilde{v}_y$ and $\tilde{u}_y=-\tilde{v}_x$
Any assistance would be excellent. If this question is confusing, then please message me.
By the chain rule, the bracket versions of your derivatives are the correct ones. I assume that $g$ is supposed to be holomorphic wherever these computations take place, otherwise the problem would make little sense.
Therefore you see that the Cauchy-Riemann equations are satisfied if and only if the derivatives both vanish everywhere. As a rule of thumb, anything that involves $\bar{z}$ is most likely not going to be holomorphic (obviousy unless the bars can be cancelled in some way); it is as far as you can be from being holomorphic. Such functions as your $f$ are called anti-holomorphic for that reason.