Complex Numbers
Tangent intersection points on unit circle — polynomial equation
MJAT_TS4_P2
Grade 12
Question:
Let $z_1,z_2,z_3$ be three unimodular complex numbers which are also roots of the equation $z^3+az^2+bz+1=0$ (where $a,b$ are complex numbers). Tangents are drawn to $|z|=1$ at points $z_1,z_2,z_3$ which intersect pairwise at points $\omega_1,\omega_2,\omega_3$. Then $\omega_1,\omega_2,\omega_3$ are roots of the equation:
A) $(ab-1)z^3+2(b^2-a)z^2+8bz+8=0$
B) $(ab-1)z^3+2(b^2+a)z^2-8bz+8=0$
C) $(ab-1)z^3-2(b^2+a)z^2+8bz+8=0$
D) $(ab-1)z^3+2(b^2+a)z^2+8bz+8=0$
Step-by-Step Solution
Key Concept: The tangent to $|z|=1$ at the point $z_k$ is $z\bar{z}_k+\bar{z}z_k=2$, i.e., $z/\bar{z}_k+\bar{z}/z_k=2$. The intersection $\omega$ of tangents at $z_i$ and $z_j$ satisfies $\omega=\frac{2}{z_i+z_j}\cdot z_iz_j$ (for the unit circle, the intersection of tangents at $z_i,z_j$ is $\omega=\frac{2z_iz_j}{z_i+z_j}$).
After computing the elementary symmetric polynomials of $\omega_1,\omega_2,\omega_3$ in terms of $a,b$: answer is option **D**.
Correct Answer: D