differentiability, complex analysis 
I've been looking at this and have no idea where to start or how to solve this
 A: A map $f: D \to \mathbb C$ is complex differentiable if and only if it satisfies the Cauchy Riemann equations, that is, 
$$u_x = v_y$$ and 
$$u_y = - v_x$$ where $u$ is the real part of $f$ and $v$ is the imaginary part o $f$. 
Since $f^\ast (z) = u^\ast(x,-y) - iv^\ast (x,-y)$ we have
$u^\ast_x  = u_x(x,-y)$
$u_y^\ast = - u_y(x,-y)$
$v^\ast_x = -v_x(x,-y)$
$v_y^\ast = v_y(x,-y)$
And since $u_x = v_y$ we have $u_x(x,-y) = v_y(x,-y)$ and since $u_y = -v_x$ we have $u_y(x,-y) = -v_x(x,-y)$. Now putting things together:
$$ u^\ast_x = u_x(x,-y) = v_y(x,-y) = v^\ast_y$$
and
$$ u^\ast_y = -u_y(x,-y) = v_x(x,-y) = - v_x^\ast$$
which shows that $f^\ast$ satisfies the Cauchy Riemann equations.  
A: Hint: Let $f(z)= u(x,y) + i v(x,y)$ . Then $f^* (z) = u( x,-y) - i v(x, -y)$ . Now you can use C-R equation of $f$ to imply the C-R eqn of $f^*$.
Edited: 
Let $g(x,y) = (x, -y)$, which is differentiable and $g_x =1, g_y=-1$. Now note that $u^* = u\circ g , v^* = -v\circ g$ , where $f^* = u^* +i v^*$ . Now you see how the chain rule works to imply that $u^*_x = u_x =v_y= v^*_y, \text{and } u^*_y = -u_y = v_x= - v^*_x $
A: Hint
$$\frac{f^{*}(z)-f^{*}(z_0)}{z-z_0}=\overline{\frac{f(\bar z)-f(\bar{z_0})}{\bar{z}-\bar{z_0}}}$$
Now write every $z \in D^*$ as $z=\bar{w}$ with $w \in D$ and use that
$$\overline{\frac{f(\bar z)-f(\bar{z_0})}{\bar{z}-\bar{z_0}}}=\overline{\frac{f(w)-f(w_0)}{w-w_0}}$$
and that 
$$\lim_{w \to w_0}\frac{f(w)-f(w_0)}{w-w_0}=f'(w_0)$$
A: let $\phi$ be the conjugation involution restricted to a map:
$$
\phi:D^* \to D
$$
then set
$$
f^* =\phi^{-1}\circ f \circ \phi: D^* \to D^*
$$
regarding all the maps as transformations of subdomains of $\mathbb{R}^2$, with $f=(u,v)$ we can see that $f^*=(\psi_1,\psi_2)$ is differentiable. so we have:
$$
\begin{pmatrix} \frac{\partial\psi_1}{\partial x} & \frac{\partial\psi_1}{\partial y} \\ \frac{\partial\psi_2}{\partial x} & \frac{\partial\psi_2}{\partial y} \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & -1\end{pmatrix}
\begin{pmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{pmatrix} 
 \begin{pmatrix} 1 & 0 \\ 0 & -1\end{pmatrix}
$$
from which it is clear that $f^*$ satisfies the Cauchy-Riemann equations when $f$ does
