Question

Let $z_1 = \dfrac{5+7i}{7-5i}, \, z_2 = \dfrac{3+2i}{3-2i}$ and $z_3 = \dfrac{1+11i}{11-i}$. Then $z_1\overline{z_1} + z_2\overline{z_2} + z_3\overline{z_3}$ is equal to:

• A. $2$
• B. $1+2i$
• C. $1$
• D. $3$
• E. $1-2i$

Hint:

Always use $z\overline{z} = |z|^2$ to simplify complex number expressions quickly.

Answer 1

The Correct Option is D

Concept:
• $z\overline{z} = |z|^2$
• For $\dfrac{a+bi}{c+di}$, modulus squared is: \[ \left|\frac{a+bi}{c+di}\right|^2 = \frac{a^2 + b^2}{c^2 + d^2} \]

Step 1: Compute $z_1\overline{z_1}$
\[ |z_1|^2 = \frac{5^2 + 7^2}{7^2 + (-5)^2} = \frac{25 + 49}{49 + 25} = \frac{74}{74} = 1 \]

Step 2: Compute $z_2\overline{z_2}$
\[ |z_2|^2 = \frac{3^2 + 2^2}{3^2 + (-2)^2} = \frac{9 + 4}{9 + 4} = 1 \]

Step 3: Compute $z_3\overline{z_3}$
\[ |z_3|^2 = \frac{1^2 + 11^2}{11^2 + (-1)^2} = \frac{1 + 121}{121 + 1} = 1 \]

Step 4: Add all values
\[ z_1\overline{z_1} + z_2\overline{z_2} + z_3\overline{z_3} = 1 + 1 + 1 = 3 \] Final Conclusion:
\[ = 3 \]