JEE Main · 2020 · 09 Jan 2020 · Shift 2 · Complex Numbers

Writing $z=x+iy$, the condition $|x|+|y|=4$ traces a square with corners at $(\pm 4,0)$ and $(0,\pm 4)$. The modulus $|z|=\sqrt{x^2+y^2}$ is smallest at the midpoints of the sides, such as $(2,2)$, where $|z|=2\sqrt{2}=\sqrt{8}$, and largest at the corners, where $|z|=4=\sqrt{16}$. So $|z|$ ranges over $[\sqrt{8},\sqrt{16}]$, and since $\sqrt{7}<\sqrt{8}$, the one value $|z|$ cannot take is $\sqrt{7}$.