Let $\Omega=\{z\in \mathbb{C}:0<|z|<r,0\leq \text{arg}(z)\leq \pi/2\}$ and $f:\Omega\rightarrow \mathbb{C}$ bounded in $\Omega$, continuous in $\arg(z)=0,\pi/2$ and holomorphic in the interior of $\Omega$. Besides $\text{Im}(f(z))=0$ when $\text{arg(z)}=\pi/2,0$. Prove that $f$ has an analytic extension to the whole open disk $D(0,r)$ and has a power series of the form $$f(z)=\sum_{n=0}^\infty a_{2n}z^{2n},a_{2n}\in \mathbb{R}.$$
Supposing the analytic extension exists I know how to prove the power series representation above. I haven't been able to prove the analytic extension. I though about reflecting the image of $f$ in the other quadrants of the disk, but if I reflect in the 2nd quadrant the function is $f(-\overline{z})$, which is not necessarily holomorphic. Any suggestions?
Update: If I define $$g(z)=\begin{cases} f(z)&z\in[0,\pi/2)\\ \overline{f(-\overline{z})}&z\in[\pi/2,\pi)\\ f(-z)&z\in[\pi,3\pi/2)\\ \overline{f(\overline{z})}&z\in[\pi,2\pi) \end{cases}.$$ Then since $f$ is holomorphic in $\Omega$, has a series representation $$f(z)=\sum_{n=1}^\infty c_n(z-z_0)^n$$ for every $z_0\in \Omega$. Therefore $$\overline{f(-\overline{z})}=\sum_{n=1}^\infty \overline{c_n}(-z-\overline{z_0})^n$$ where $\overline{z_0}$ is in the 2nd quadrant. So $g$ is holomorphic in the second quadrant, and so on is in all the quadrants. Since the operations over $z$ are reflections of the 1st quadrant then $g$ is continuos on the borders. Therefore is holomorphic in $D(0,r)\backslash\{0\}$. Is it right?
You are going well but it is not enough. If you have a function defined on the first quadrant ($\Omega$) then you can extend it for $z$ on the fourth quadrant by defining $f(z) = \overline{f(\overline{z})}$ (holomorphicity can also be checked by means of the Cauchy-Riemann equations) note that this extended function is continuous for $z$ real and positive. Now you have a function defined on the right plane ($\Omega$ and it's reflection). Now we want to extend this function to the left part of the plane, note that $f(ix) = f(-ix)$ for $x$ real, so what if we take a $z$ on the left part (some reflection of $\Omega$) and translate it to the right part by multiplying it by $-1$ (we are rotating $\pi$ radians), then defining $f(z) = f(-z)$ on the left plane should do the work (here is an abuse of notation, the left $f$ is a new function while the right $f$ is the function we have already extended to the right plane), since the note we made ensures the continuty on the imaginary axis. So far, we have a function that is defined on all the reflections of $\Omega$, but it is not necessarily holomorphic on the four axis of the plane, right? I noticed that you assumed it, but it must be proven, you can do it so by using Moreras' Theorem (showing that the integral along any triangle centered in any point in any axis, and contained in the domain in question, shall vanish), then you will get that the constructed function is holomorphic in all the unit disk. Then notice that our construction forces $f$ to be a par function, hence it follows the desired form of the series (boundedness allows us to say that $f$ has a removable singularity at the origin, hence a Taylor series).