JEE Main 2016 (15)

Let $z = r\cdot e^{i\phi} ;r>0, \phi\in (0,2\pi)$ .

$\frac{1}{z} = \frac{e^{-i\phi}}{r}$

$Im(z+\frac{1}{z}) = r \sin\phi + \frac{-\sin\phi}{r}=0$

$\displaystyle \Rightarrow \sin\phi\frac{(r^2-1)}{r} =0$

$\displaystyle \Rightarrow \sin\phi=0 \text{ or } r=1$

$\sin\phi =0$ represents the real axis, i.e., $\boxed{y=0}$ .

$r=1$ represents the unit circle with centre at origin.

Hence, the locus of points satisfying $Im(z+\frac{1}{z})=0$ is both (a) and (b).