If both $f(z)$ and $−\overline{f(z)}$ are analytic, what can you say about $f(z)$? Prove your claim. I am struggling with this question. If both $f(z)$ and its negative conjugate $-\overline{f(z)}$ are analytic, what can you say about $f(z)$? Prove your claim.
I know that the sum on difference of two analytic functions is analytic but I can't see how that will help. Does anyone know how to approach this question or a theorem that would lead to an argument. Please provide at least a start to the argument so I may understand your approach well enough.
 A: 
I know that the sum on difference of two analytic functions is analytic

Perfect! So, what can we tell about the difference $f(z)-(-\overline{f(z)})$? You say you know it's analytic. Here is one more thing we can tell: It is real-valued, since it's equal to $2\operatorname{Re}(f(z))$.
An analytic function that always takes real values, and thus has constant imaginary part, must be constant. So $f$ has constant real part, which means that $f$ itself must be constant.
A: If $f(z)$ and $-\overline{f(z)}$ are both analytic, then $f(z) - \overline{f(z)} = 2i \Im(f(z))$ is analytic ($\Im(h)$ being imaginary part of $h$), and thus the imaginary part of $f(z)$ is analytic.
The imaginary part of $f(z)$ is a real-valued analytic function. Any real-valued analytic function must be constant. For suppose $g$ is a real-valued analytic function and $g'(z) \neq 0$. Then consider a sufficiently small real $\delta$. Then $g(z + i \frac{h}{g'(z)}) = g(z) + ih + o(h)$, which is not a real number.
Thus, the imaginary part of $f(z)$ is constant. Then the real part of $f(z)$ is also a real-valued analytic function, hence is constant. Then $f$ is constant.
A: Here's another approach: Complex functions are complex differentiable iff they are real differentiable and their Jacobian is of the form
$$\mathrm Df=\begin{pmatrix}a&-b\\b&a\end{pmatrix}$$
That's essentially the content of the criterion for complex differentiability using the Cauchy-Riemann equations (it requires that $f$ be real differentiable and that its partial derivatives satisfy the CR-equations, which is the same as saying that the Jacobian has the above form). Now the function $g:z\mapsto -\bar z$ is also real differentiable with Jacobian
$$\mathrm Dg=\begin{pmatrix}-1&0\\0&1\end{pmatrix}.$$
Now the chain rule says that $g\circ f$ is also real differentiable with Jacobian
$$\mathrm D(g\circ f)=\mathrm Dg(f)\mathrm Df=\begin{pmatrix}-1&0\\0&1\end{pmatrix}\begin{pmatrix}a&-b\\b&a\end{pmatrix}=\begin{pmatrix}-a&b\\b&a\end{pmatrix}.$$
By assumption, $g\circ f$ is analytic, so this Jacobian must also satisfy that the diagonal elements are equal, while the off-diagonal elements differ by sign, which gives us $-a=a$ and $b=-b$. But that means $a=b=0$, so our function has vanishing derivative, and is thus constant (on connected sets).
