Differentiability on $Re(z)^2$ Suppose that $l(z)=Re(z)^2$. Is $l$ differentiable at $z_0 \in \mathbb{C}$? What is $f'(z_0)$? Where is $l$ differentiable?
Write $w=u+iv$ and $z_0=x_0 +iy_0$. So $$\lim  \limits_{w \rightarrow 0} \frac{l(z_0 +w)-l(z_0)}{w} =\lim  \limits_{w \rightarrow 0} \frac{(x_0+u)^2-x_0^2}{w} =\lim  \limits_{w \rightarrow 0} \frac{u^2}{w} + \lim  \limits_{w \rightarrow 0} \frac{2x_0 u}{w}$$
What next? The first limit is apparently zero but I don't know how...

Use the Cauchy-Riemann equations for $f(x+iy) = x^2$, with $u(x,y) = x^2$ and $v(x,y) = 0$ in the usual notation. Then: $$\begin{cases} u_x = v_y \\ u_y = -v_x\end{cases} \implies \begin{cases} 2x = 0 \\ 0 = 0\end{cases}$$ The equations are only satisfied for $x = 0$ (and any $y$). The function is differentiable only in the imaginary axis.
