How to prove that if $f(z)$ is complex function, then $\Delta Re(f)=0=\Delta Im(f)$? How to prove that if $f(z)$ is complex function, then $\Delta Re(f)=0=\Delta Im(f)$ and they are harmonic in area $U\subset \mathbb{C}$? 
For example if $f(z)=z^2$, then $\Delta Re(f)=0=\Delta Im(f)$
 A: It is not true for any complex function, but it is true for every holomorphic (and for even some more)
Take the Cauchy Riemann defiiferential equations and use that every  holomorphic function is 2 times continuous differentiable hence you can change the order of differention.
Using Cauchy Riemann differential equations:
\begin{align}
u_x(z_0)&=v_y(z_0)\\
u_y(z_0)&=-v_x(z_0)
\end{align}
so for the second derivative
\begin{align*}
\frac{\partial^2 u}{\partial x^2}&=\frac{\partial^2 v}{\partial x \partial y}
\end{align*}
Because of the derivatives are continuous this is equal to
\begin{align*}
&=\frac{\partial^2 v}{\partial y \partial x }\\
&=-\frac{\partial^2 u }{\partial y^2}
\end{align*}
Hence the real component of any (two times continuously differentiable) holomorphic
function is harmonic. That the complex component ist harmonic is analog.
A: Write $f(z)=u(z)+iv(z)$, that is, $Re(f)=u$, $Im(f)=v$. Assuming $f$ is holomorphic, if $z=x+iy$, then, by the Cauchy-Riemann equations,
$$
\dfrac{\partial u}{\partial x}=\dfrac{\partial v}{\partial y}
$$
and
$$
\dfrac{\partial u}{\partial y}=-\dfrac{\partial v}{\partial x}
$$
Therefore,
$$
\Delta u=\dfrac{\partial^2 u}{\partial x^2}+\dfrac{\partial^2 u}{\partial y^2}=\dfrac{\partial}{\partial x}\left( \dfrac{\partial v}{\partial y}\right)+\dfrac{\partial}{\partial y}\left( -\dfrac{\partial v}{\partial x}\right)=0
$$
By the Schwarz Theorem, the same being valid for $\Delta v$.
