I have a problem with the exercise below. Please help.
Let $f:A\rightarrow C$ and $z_0=(x_0,y_0)$ an interior point of $A$. We write
$$f(x,y)=u(x,y)+i\upsilon (x,y).$$ If first partial derivatives of $u,\upsilon$ with respect to $x,y$ exist on an open disk centered at $z_0$ and are continuous at $z_0$ then prove the following statement:
$$\lim_{z\rightarrow z_0}{\rm Re}\frac{f(z)-f(z_0)}{z-z_0}\in {\mathbb R}\quad\Rightarrow\quad {f}'(z_0) \ {\rm exists.}$$
Thank you in advance.
For simplicity we assume $x_0=y_0=0$ and $f(0)=0$. Furthermore we write $x=r\cos\phi$, $y=r\sin\phi$ when appropriate. The assumptions made in the question then imply $$u(z)=ax+by+o(r),\quad v(z)=cx+dy+o(r)\qquad(r\to0)\tag{1}$$ for certain constants $a$, $b$, $c$, $d$. We then have $${\rm Re}{f(z)\over z}={\rm Re}{u+iv\over z}={1\over |z|^2}{\rm Re}\bigl((u+iv)(x-iy)\bigr)={1\over r^2}\bigl(u(z)x+v(z)y\bigr)\ .$$ Plugging in $(1)$ here we obtain $${\rm Re}{f(z)\over z}=\bigl(a\cos^2\phi+(b+c)\cos\phi\sin\phi+d\sin^2\phi\bigr)+o(1)\qquad(r\to0)\ .$$ Since by assumption the limit on the LHS exists when $r\to0$, and is independent of $\phi$, it follows that $\>a\cos^2\phi+(b+c)\cos\phi\sin\phi+d\sin^2\phi\>$ should be independent of $\phi$. This enforces $a=d$ and $b=-c$, hence the pair $(u,v)$ fulfills the CR equations at $z=0$. This in turn implies that $f'(0)$ exists as complex derivative.