Complex Numbers
Statements on Complex Inequalities
nta_pyq_2024_apr
Grade 11
Question:
Consider the following two statements:
Statement I: For any two non-zero complex numbers $z_1,z_2$, $(|z_1|+|z_2|)\left|\dfrac{z_1}{|z_1|}+\dfrac{z_2}{|z_2|}\right|\leq2(|z_1|+|z_2|)$.
Statement II: If $x,y,z$ are three distinct complex numbers and $a,b,c$ are three positive real numbers such that $\dfrac{a}{|y-z|}=\dfrac{b}{|z-x|}=\dfrac{c}{|x-y|}$, then $\dfrac{a^2}{y-z}+\dfrac{b^2}{z-x}+\dfrac{c^2}{x-y}=1$.
Between the above two statements,
Statement I is correct but Statement II is incorrect.
both Statement I and Statement II are correct.
both Statement I and Statement II are incorrect.
Statement I is incorrect but Statement II is correct.
Step-by-Step Solution
Key Concept: Statement I: $|z_1/|z_1|+z_2/|z_2||\leq2$ (unit vectors sum $\leq2$), so LHS $\leq2(|z_1|+|z_2|)$ ✓. Statement II: $\sum a^2/(y-z)=\lambda\sum(y-z)(\bar{y}-\bar{z})/(y-z)=\lambda\sum(\bar{y}-\bar{z})$... $=\lambda\cdot0=0\neq1$. False.
Statement I ✓. Statement II: $\sum a^2/(y-z)=\lambda(\bar{y}-\bar{z}+\bar{z}-\bar{x}+\bar{x}-\bar{y})=0\neq1$. ✗.
Correct Answer: 1