If $f(z)=( \mathfrak{Re}z)(\mathfrak {Im}z)(z^2+1)$. Find all complex numbers $z$ for which $f$ is differentiable. If $f(z)=( \mathfrak{Re}z)(\mathfrak {Im}z)(z^2+1)$. Find all complex numbers $z$ for which $f$ is differentiable.
As there is no point already given so I am little bit confused. After some calculation I think that the function is differentiable only at $0$. Is it true? Can anyone give any short process of doing it?
 A: You have $$f(z)=( \mathfrak{Re}z)(\mathfrak {Im}z)(z^2+1)=\frac{z+\overline{z}}{2}\frac{z-\overline{z}}{2i}(z^2+1)=\frac{1}{4i}(z^2-\overline{z}^2)(z^2+1)$$ and $$\frac{\partial f}{\partial \overline{z}}=-\frac{1}{2i}\overline{z}(z^2+1)$$ $f$ is $\mathbb C$-differentiable if and only if $\frac{\partial f}{\partial \overline{z}}=0$ which means $z \in \{0,i,-i\}$.
A: The quickest way is to use Cauchy-Riemann's equations. Since $f$ is smooth (as a function $\mathbb{R^2} \to \mathbb{R}^2$), we don't have to worry: $f'(z)$ exists if and only if Cauchy-Riemann's equations are satisfied at $z$.
Here, we get 
$$f(x+iy) = \underbrace{xy(x^2+y^2-1)}_{=u} + \underbrace{2x^2y^2}_{=v} i$$
omitting some tedious computation. Hence, we want to see when
$$
\begin{cases} u'_x = v'_y \\ u'_y = - v'_x \end{cases}
\iff
\begin{cases}
y(x^2+y^2-1) = 0 \\
x(x^2+y^2+1) = 0
\end{cases}
$$
From the second equation, $x=0$ (since $x$ and $y$ are real), so the first equation gives $y=0$ or $y = \pm 1$.
So, in fact your function is complex differentiable at exactly three points: $0$ and $\pm i$.
(If you know Wirtinger calculus, the computations are a little bit shorter.)
A: Suppose $f$ is complex differentiable (CD) at some $z_0\in \mathbb {C}\setminus \{i,-i\}.$ Then $xy = f(z)/(1+z^2)$ is CD at $z_0.$ But the Cauchy-Riemann equations show that a real $C^1$ function is CD at a point iff the gradient of that function is $0$ at that point. For $xy,$ that only happens at $0.$ Thus in $\mathbb {C}\setminus \{i,-i\},$ $f$ is CD only at $0.$ At $i,$ $(f(z)-f(i))/(z-i) = xy(z+i) \to 0$ as $z\to i.$ Same thing at $-i.$ Therefore $f$ is CD only at $0,i,-i,$ with $f'=0$  at these points.
