Question

If \( z_1 = 2 - 3i \) and the roots of the equation \( z^3 + bz^2 + cz + d = 0 \) are \( i, z_1 \) and \( \overline{z_1} \), then \( b + c + d = \)

• A. \( 13 \)
• B. \( 7 \)
• C. \( 9 - 10i \)
• D. \( 10 - 10i \)

Hint:

If a polynomial equation with real coefficients has a complex root \( a + bi \), then its conjugate \( a - bi \) is also a root. However, in this problem, the coefficients \( b, c, d \) are not necessarily real. To solve, construct the polynomial from its roots and then compare the coefficients. Remember to handle complex number arithmetic carefully.

Answer 1

The Correct Option is C

The roots of the cubic equation \( z^3 + bz^2 + cz + d = 0 \) are given as \( i, z_1 = 2 - 3i \), and \( \overline{z_1} = 2 + 3i \).
For a cubic equation with roots \( \alpha, \beta, \gamma \), the equation can be written as \( (z - \alpha)(z - \beta)(z - \gamma) = 0 \).
In our case, the roots are \( i, 2 - 3i, 2 + 3i \).
So the equation is: $$ (z - i)(z - (2 - 3i))(z - (2 + 3i)) = 0 $$ $$ (z - i)((z - 2) + 3i)((z - 2) - 3i) = 0 $$ Using the identity \( (x + y)(x - y) = x^2 - y^2 \) with \( x = z - 2 \) and \( y = 3i \): $$ (z - i)((z - 2)^2 - (3i)^2) = 0 $$ $$ (z - i)((z^2 - 4z + 4) - (9i^2)) = 0 $$ Since \( i^2 = -1 \): $$ (z - i)(z^2 - 4z + 4 - (-9)) = 0 $$ $$ (z - i)(z^2 - 4z + 13) = 0 $$ Expanding this product: $$ z(z^2 - 4z + 13) - i(z^2 - 4z + 13) = 0 $$ $$ z^3 - 4z^2 + 13z - iz^2 + 4iz - 13i = 0 $$ $$ z^3 + (-4 - i)z^2 + (13 + 4i)z - 13i = 0 $$ Comparing this with the given equation \( z^3 + bz^2 + cz + d = 0 \), we have: $$ b = -4 - i $$ $$ c = 13 + 4i $$ $$ d = -13i $$ We need to find \( b + c + d \): $$ b + c + d = (-4 - i) + (13 + 4i) + (-13i) $$ $$ b + c + d = -4 - i + 13 + 4i - 13i $$ $$ b + c + d = (-4 + 13) + (-1 + 4 - 13)i $$ $$ b + c + d = 9 + (-10)i $$ $$ b + c + d = 9 - 10i $$