Find all $z \in \mathbb{C}$ such that the function $f(z) = {\bar z}\sin(z)$ satisfies the Cauchy-Riemann equations.
Own work: Let $z = x+iy$ where $x$ and $y$ are real numbers
Thus, ${\bar z} = x-iy$, and $\sin(z) = \frac{\exp(iz)-\exp(-iz)}{2i}$
Hence $f(x+iy)= (x-iy)\frac{\exp(i(x+iy))-\exp(-i(x+iy))}{2i}$
Im struggling to simplify this in order to do the partial derivatives to see that it satisfies the Cauchy-Riemann equations.
Any help will be appreciated.
If you know the complex Cauchy–Riemann equation namely: $$\frac{\partial f}{\partial \bar{z}}=0,$$ you have $$\frac{\partial }{\partial \bar{z}} (\bar{z} \sin z) = \left(\frac{\partial }{\partial \bar{z}} \bar{z}\right) \sin z + \bar{z} \left(\frac{\partial}{\partial \bar{z}} \sin z\right)=\sin z$$ as $\sin z$ is holomorphic. Hence $f$ satisfies the Cauchy-Riermann equations when $\sin z$ vanishes.
sin(z)=sin(x+iy)=sin(x)cos(iy)+cos(x)sin(iy)
now multiplying with z bar and taking partial derivatives becomes easier