Let $f:\mathbb{C}\rightarrow\mathbb{C}$ be a function defined by $f(z)=2|z|^2-\bar{z}^2$. At which points is $f$ complex differentiable? At which points is $f$ analytic?
Im not too sure where or how to start this. But I know that if $z\in\mathbb{R}$ then $z=\bar{z}$ and hence $f=z^2$ which is differentiable. If this is the only points where f is differentiable, then f is not analytic since there is no neighbourhood around these points in $\mathbb{R}$ that are complex differentiable everywhere in these neighbourhoods.
Use the Cauchy-Riemann equations. If $x,y\in\mathbb R$, then\begin{align}f(x+yi)&=2(x^2+y^2)-(x^2-y^2+2xyi)\\&=x^2+3y^2-2xyi.\end{align}So, let $u(x,y)=x^2+3y^2$ and let $v(x,y)=-2xy$. Then the Cauchy-Riemann equations are$$\left\{\begin{array}{l}2x=-2x\\6y=2y\end{array}\right.$$They have one and only one solution: $(0,0)$. So, and since $\frac{\partial u}{\partial x}$, $\frac{\partial u}{\partial y}$, $\frac{\partial v}{\partial x}$, and $\frac{\partial v}{\partial y}$ are continuous functions, $f$ is differentiable at $0$ and only there. In particular, $f$ is analytic nowhere.