JEE Main 2024 (Online) 27th January Evening ShiftQuestion 1
Let the complex numbers $\alpha$ and $\frac1{\bar\alpha}$ lie on the circles $|z-z_0|^2=4$ and $|z-z_0|^2=16$ respectively, where $z_0=1+i$. Then, the value of $100|\alpha|^2$ is _____.
Complex Numbers
Answer$20$
Solution
Let $r=|\alpha|^2$, so $\frac1{\bar\alpha}=\frac\alpha r$. $|\alpha-z_0|^2=r-2\mathrm{Re}(\alpha\bar z_0)+2=4$ and $\left|\frac\alpha r-z_0\right|^2=16$ give $1-2\mathrm{Re}(\alpha\bar z_0)+2r=16r$. Eliminating: $3+r=16r$, $r=\frac15$, so $100|\alpha|^2=20$.