Check whether the function defined by $$f(z)=u+iv=\begin{cases} \Im(z^2)/\bar z& \text{if } z\neq 0\\ 0&\text{if} z=0 \end{cases}$$ is analytic or not.
My attempt
I tried to find the $\frac{\partial f}{\partial x}$ and $-i\frac{\partial f}{\partial y}$. But it is getting longer and it is not giving me the answer.
Using the identity $2\mathrm i\Im(w)=w-\bar w$, valid for every $w$ in $\mathbb C$, one sees that $$2\mathrm if(z)=g(z,\bar z),\qquad g(u,v)=\frac{u^2}v-v.$$ Now, the fact that $$\frac{\partial g}{\partial v}\ne0,$$ implies that $f$ is not analytic. Here is a related question.
If $f$ were analytic, the so would be $$ g(z)=\frac{1}{z}\cdot f(z)=\frac{\mathrm{Im}\, z^2}{\lvert z\rvert^2},\quad \text{for $z\ne0$,} $$ which takes only real values, and thus it has to be constant. But constant it is not. Contradiction.
Thus $f$ is not analytic.
If $f(z)=u(z)+uv(z)$ where $u$ and $v$ are real-valued, we have $$f(x+iy)=\frac{y^2}{x-iy}=\frac{xy^2}{x^2+y^2}+i\frac{y^3}{x^2+y^2}$$ that is $$u(x,y)=\frac{xy^2}{x^2+y^2}$$ $$v(x,y)=\frac{y^3}{x^2+y^2}$$
Now compute $$\frac{\partial v}{\partial x}=-\frac{2xy^3}{(x^2+y^2)^2}$$ $$\frac{\partial u}{\partial y}=\frac{2x^3y}{(x^2+y^2)^2}$$
Since $\partial v/\partial x\neq-\partial u/\partial y$, Cauchy-Riemann's equations are not satisfied and hence $f$ is not analytic.
