Complex Numbers
Algebra of Complex Numbers
Grade Class 11
Question:
<p>If \(\dfrac{2+3z+4z^2}{2-3z+4z^2}\) is real (with \(\text{Im}(z)\neq 0\)), then \(|z|^2\) is:</p>
Step-by-Step Solution
Key Concept: For A/B to be real: Im(A \cdot B̄) = 0. Here A = 2+3z+4z^2, B = 2-3z+4z^2. A-B = 6z, A+B = 4+8z^2. A/B real iff Im(A \cdot B̄)=0 iff A/B = Ā/B̄ iff cross-multiply and simplify.
<p>Let $f=\dfrac{2+3z+4z^2}{2-3z+4z^2}$. $f$ real $\Rightarrow f=\bar{f}\Rightarrow\dfrac{2+3z+4z^2}{2-3z+4z^2}=\dfrac{2+3\bar{z}+4\bar{z}^2}{2-3\bar{z}+4\bar{z}^2}$. Cross-multiply and use $\text{Im}(z)\neq 0$ to get $|z|^2=3/4$. ✓</p>
Correct Answer: C