Prove that $\overline{f(z)}$ is differentiable at $a \in D(0;1)$ if and only if $f'(a)=0$ 
Let $f$ be holomorphic in $D(0;1)$ and define $k$ by $k(z)=\overline{f(z)}$. Prove that $k$ is differentiable at $a\in D(0;1)$ if and only if $f'(a)=0$.

What I tried was first, assuming $k$ is differentiable and letting $f=u+iv$ we have (first when $h \in \mathbb{R}$)
$$k'(z)= \lim_{h \to 0} \frac{u(x+h,y)-u(x,y)}{h} -i\frac{v(x+h,y)-v(x,y)}{h} = u_x -iv_x$$
and when $h=ik, \ k\in \mathbb{R}$
$$k'(z)=\lim_{k \to 0} \frac{u(x,y+k)-u(x,y)}{ik} -\frac{v(x,y+k)-v(x,y)}{h} = \frac{1}{i}u_y -v_y$$
And equating real and imaginary parts, we get that
$$u_x=-v_y, \; u_y=v_x$$
Since $f$ is holomorphic, it satisfies the Cauchy-Riemann equations and thus
$$u_x=v_y, \; u_y=-v_x$$
so
$$f'(a)=-f'(a)$$
and then $f'(a)=0$. I don't know if this works, so please correct me if I'm wrong. Besides that, I'm stuck in proving the other implication. So far I did
$$0=f'(a)=\lim_{h\to 0} \frac{f(a+h)-f(h)}{h}=\overline{\lim_{h\to 0}\frac{f(a+h)-f(h)}{h}}=\lim_{h\to 0}\frac{\overline{f(a+h)} -\overline{f(h)}}{\overline{h}}=\overline{f'(a)}=k'(a)$$
But again, I'm not sure if this is right. Any help will be highly appreciate, and thanks in advance!
 A: $\displaystyle \lim_{h\to0} |\dfrac{f(z+h)-f(z)}{h}|=0 \Leftrightarrow \displaystyle \lim_{h\to0} |\dfrac{\overline{f(z+h)-f(z)}}{h}|=0$
But the expression $\dfrac{\overline{f(z+h)-f(z)}}{h}= \dfrac{\overline{f(z+h)-f(z)}}{\overline{h}}\times\dfrac{\overline{h}}{h}$ does not tend to a limit if the first half of it tends to $w=\overline{f'(a)}\ne0$, because the second half can be made to have any complex value of unit length.
A: Yes, both directions of your proof are essentially correct, but some steps could be better justified. To see them done out in a bit more detail: 
For all $a \in D(0,1)$, $f$ is holomorphic, so the Cauchy-Riemann equations hold, i.e. 
$$u_x=v_y, \ \ \ \ u_y=-v_x$$
Assuming $k'(a)=0$, you have shown that the following equations must hold at $a$:
$$u_x=-v_y, \ \ \ \ u_y=v_x$$
Combining all of these, we get the following:
$$u_x=-u_x, \ \ \ \ u_y=-u_y, \ \ \ \ v_x=-v_x, \ \ \ \ v_y=-v_y$$
Now, $u_x=-u_x$ implies that $u$ is constant in $x$, $u_y=-u_y$ implies that $u$ is constant in $y$, etc. Hence, these equations imply $f'(a)=0$. 
Now, assume $f'(a)=0$. Then we certainly have $f'(a)=\overline{f'(a)}$. We also know that complex conjugation is additive, multiplicative, and that the complex conjugate of a continuous function is continuous. Hence, we have
$$0=\overline{\lim_{h\to 0}\frac{f(a+h)-f(h)}{h}}=\lim_{h\to 0}\overline{\frac{1}{h}(f(a+h)-f(h))}=\lim_{h\to 0}\frac{\overline{f(a+h)} -\overline{f(h)}}{\overline{h}}$$
Now, this last limit does evaluate exactly to $k'(a)$. To see what happens in detail, write $h=re^{i\theta}$. Then, we have
$$\lim_{h\to 0}\frac{\overline{f(a+h)} -\overline{f(h)}}{\overline{h}} = \lim_{r\to 0}\frac{\overline{f(a+re^{i\theta})} -\overline{f(h)}}{re^{-i\theta}} = e^{2i \theta}\lim_{r\to 0}\frac{\overline{f(a+re^{i\theta})} -\overline{f(h)}}{re^{i\theta}}=e^{2i \theta}k'(a)$$
Hence, we have $e^{2i \theta}k'(a)=0$. Note that, for all $\theta \in [0,2\pi]$, this implies $k'(a)=0$. Hence, $k'(a)=0$. 
