Question

If \(z_1, z_2, z_3\) are three distinct complex numbers and \(a,b,c\) are three positive real numbers such that \[ \frac{a}{|z_2-z_3|}=\frac{b}{|z_3-z_1|}=\frac{c}{|z_1-z_2|} \] then \[ \frac{a^2}{z_2-z_3}+\frac{b^2}{z_3-z_1}+\frac{c^2}{z_1-z_2}= \]

• A. \(1\)
• B. \(0\)
• C. \(z_1+z_2+z_3\)
• D. \(z_1^2+z_2^2+z_3^2\)

Hint:

Useful identity in complex numbers: \[ |z|^2=z\bar z \quad \Rightarrow \quad \frac{|z|^2}{z}=\bar z \] Cyclic sums of differences often cancel out to zero.

Answer 1

The Correct Option is B

Step 1: Given \[ \frac{a}{|z_2-z_3|}=\frac{b}{|z_3-z_1|}=\frac{c}{|z_1-z_2|}=k \] for some positive real constant \(k\).
Step 2: Hence, \[ a=k|z_2-z_3|,\quad b=k|z_3-z_1|,\quad c=k|z_1-z_2| \]
Step 3: Squaring, \[ a^2=k^2|z_2-z_3|^2,\quad b^2=k^2|z_3-z_1|^2,\quad c^2=k^2|z_1-z_2|^2 \]
Step 4: Substitute into the given expression: \[ k^2\left(\frac{|z_2-z_3|^2}{z_2-z_3} +\frac{|z_3-z_1|^2}{z_3-z_1} +\frac{|z_1-z_2|^2}{z_1-z_2}\right) \]
Step 5: Using the identity \[ \frac{|z|^2}{z}=\bar z \quad (z\neq 0) \] the expression becomes: \[ k^2\big[(\overline{z_2-z_3})+(\overline{z_3-z_1})+(\overline{z_1-z_2})\big] \]
Step 6: Simplifying: \[ \overline{z_2-z_3+z_3-z_1+z_1-z_2}= \overline{0}=0 \]
Step 7: Therefore, the required value is: \[ \boxed{0} \]