UPSC NDA I 2024 Mathematics Paper

Given the condition that the modulus of the complex fraction $\left| \frac{z_1+z_2}{z_1-z_2} \right|$ is equal to 1 , we can infer that this fraction lies on the unit circle in the complex plane. This implies that the magnitude of the numerator is equal to the magnitude of the denominator.So, we have:$\left| z_1+z_2 \right| = \left| z_1-z_2 \right|$Let us write $z_1$ and $z_2$ in their general form as complex numbers: $z_1 = x_1 + i y_1$ and $z_2 = x_2 + i y_2$, where $x_1, x_2, y_1$, and $y_2$ are real numbers.Thus, the equation becomes:$\left| (x_1+x_2) + i (y_1+y_2) \right| = \left| (x_1-x_2) + i (y_1-y_2) \right|$The modulus of a complex number $a + b i$ is given by $\sqrt{a^2+b^2}$. Applying this to our equation, we get:$\sqrt{(x_1+x_2)^2 + (y_1+y_2)^2} = \sqrt{(x_1-x_2)^2 + (y_1-y_2)^2}$Squaring both sides, we obtain:$(x_1+x_2)^2 + (y_1+y_2)^2 = (x_1-x_2)^2 + (y_1-y_2)^2$Expanding both sides gives:$x_1^2 + 2 x_1 x_2 + x_2^2 + y_1^2 + 2 y_1 y_2 + y_2^2 = x_1^2 - 2 x_1 x_2 + x_2^2 + y_1^2 - 2 y_1 y_2 + y_2^2$By canceling out the common terms from both sides, we get:$2 x_1 x_2 + 2 y_1 y_2 = -2 x_1 x_2 - 2 y_1 y_2$ This further simplifies to:$4 x_1 x_2 + 4 y_1 y_2 = 0$Dividing both sides by 4 , we obtain:$x_1 x_2 + y_1 y_2 = 0$The expression for the real part of a complex fraction $\frac{z_1}{z_2}$ is given by:$\operatorname{Re} \left( \frac{z_1}{z_2} \right) = \frac{x_1 x_2 + y_1 y_2}{x_2^2 + y_2^2}$Given that $x_1 x_2 + y_1 y_2 = 0$, we substitute this into the expression above:$\operatorname{Re} \left( \frac{z_1}{z_2} \right) = \frac{0}{x_2^2 + y_2^2} = 0$So, we have:$\operatorname{Re} \left( \frac{z_1}{z_2} \right) + 1 = 0 + 1 = 1$