- Home
- Standard 11
- Mathematics
4-1.Complex numbers
hard
Let $z$ be a complex number with non-zero imaginary part. If $\frac{2+3 z+4 z^2}{2-3 z+4 z^2}$ is a real number, then the value of $|z|^2$ is. . . . .
A
$0.20$
B
$0.50$
C
$0.55$
D
$0.60$
(IIT-2022)
Solution
Given that
$z \neq \overline{ z }$
Let $\alpha=\frac{2+3 z+4 z^2}{2-3 z+4 z^2}=\frac{\left(2-3 z+4 z^2\right)+6 z}{2-3 z+4 z^2}$
$\therefore \alpha=1+\frac{6 z}{2-3 z+4 z^2}$
If $\alpha$ is a real number, then
$\alpha=\bar{\alpha}$
$\Rightarrow \frac{z}{2-3 z+4 z^2}=\frac{\bar{z}}{2-3 \bar{z}+4 \bar{z}^2}$
$\therefore 2(z-\bar{z})=4 z \bar{z}(z-\bar{z})$
$\Rightarrow(z-\bar{z})(2-4 z \bar{z})=0$
As $z \neq \bar{Z}$ (Given)
$\Rightarrow z \bar{z}=\frac{2}{4}=\frac{1}{2}$
$\Rightarrow|z|^2=0.50$
Standard 11
Mathematics
