Maths Question 2 – JEE-MAIN 2025

The given equation is $\frac{2+kz}{k+z}=kz$. Given $|z|=1$ and $k\in \mathbb{R}$. Rearranging the equation, we get a quadratic in $z$: $k{z}^2+(k^2-k)z-2=0$. Since $|z|=1$, we can use the property that if $z$ is a root of $a{z}^2+bz+c=0$ and $|z|=1$, then $z$ is also a root of $c{z}^2+\overline{b}z+\overline{a}=0$. As $k\in \mathbb{R}$, this gives a second equation: $2z^2-(k^2-k)z-k=0$. Solving these two equations simultaneously for a common root $z$ (by eliminating $z^2$) leads to $(k+2)[(k-2)z+k(k-1)]=0$. This implies either $k=-2$ or $(k-2)z+k(k-1)=0$.

If $k=-2$, the original equation becomes $z^2-3z+1=0$, whose roots are $z=\frac{3\pm \sqrt{5}}{2}$. Neither of these roots has modulus 1, so $k=-2$ is not a valid solution. Thus, we must have $(k-2)z+k(k-1)=0$. If $k=2$, this simplifies to $2=0$, which is impossible. So $k\ne 2$. For $k\ne 2$, we have $z=\frac{k(1-k)}{k-2}$. Applying the condition $|z|=1$, we get $|k(1-k)|=|k-2|$. This leads to two cases: $k(1-k)=k-2⟹k^2=2⟹k=\pm \sqrt{2}$, or $k(1-k)=-(k-2)⟹k^2-2k+2=0$, which has no real solutions for $k$. Therefore, the only possible values for $k$ are $k=\pm \sqrt{2}$. Verifying these values in the original equation shows that for both $k=\sqrt{2}$ and $k=-\sqrt{2}$, $z=1$ is the solution, and $|z|=1$ is satisfied.