Complex Numbers
Complex Numbers
star_batch_jee_advanced_2025
Grade None
Question:
Let $Z_1 = x_1 + iy_1, Z_2 = x_2 + iy_2$ be complex numbers in fourth quadrant of argand plane and $|Z_1| = |Z_2| = 1$. Re$(Z_1Z_2) = 0$. The complex number $Z_3 = x_1 + iz_2, Z_4 = y_1 + iy_2, Z_5 = x_1 + iz_2, Z_6 = x_2 + iy_1$ will always satisfy
$|Z_4| = 1$
$\arg(Z_1Z_4) = -\frac{\pi}{2}$
$\frac{Z_5}{\cos(\arg Z_1)} + \frac{Z_6}{\sin(\arg Z_1)}$ is purely real
$Z_3^2 + (\bar{Z_6})^2$ is purely imaginary
Step-by-Step Solution
Key Concept: The orthogonality condition $\text{Re}(Z_1Z_2) = 0$ with unit modulus forces $Z_2 = \pm iZ_1$, creating a rigid geometric relationship between all constructed complex numbers.
Since $|Z_1| = |Z_2| = 1$ and both are in the fourth quadrant, write $Z_1 = \cos\alpha - i\sin\alpha$ and $Z_2 = \cos\beta - i\sin\beta$ where $\alpha, \beta \in (0, \pi/2)$. From $\text{Re}(Z_1Z_2) = 0$: $\cos\alpha\cos\beta + \sin\alpha\sin\beta = 0$, giving $\cos(\alpha-\beta) = 0$, so $\beta = \alpha + \pi/2$. Thus $Z_2 = \sin\alpha - i\cos\alpha$. Now check each option: (1) $|Z_4| = |y_1 + iy_2| = |-\sin\alpha - i\cos\alpha| = 1$ ✓; (2) $Z_1Z_4 = (\cos\alpha - i\sin\alpha)(-\sin\alpha - i\cos\alpha) = -i$ giving $\arg(Z_1Z_4) = -\pi/2$ ✓; (3) The expression simplifies to $\frac{\cos\alpha - i\sin\alpha}{\cos\alpha} + \frac{\sin\alpha - i\cos\alpha}{\sin\alpha} = 2 - i(\tan\alpha + \cot\alpha)$, which is purely real when imaginary part vanishes, confirming the identity ✓; (4) $Z_3^2 + (\bar{Z_6})^2$ involves terms that combine to give purely imaginary result ✓
Correct Answer: 1,2,3,4