Question

Find the value of m for which ${{\rm{z}}^3} + \left( {3 + {\rm{i}}} \right){{\rm{z}}^2} - 3{\rm{z}} - \left( {{\rm{m}} + {\rm{i}}} \right) = 0$ where ${\rm{m}} \in {\rm{R}}$ has at least one real root.
A) 1
B) 2
C) 3
D) 5

Answer 1

Hint:
We suppose let a be real root then replace z with a, then take conjugate of the equation in terms of a. Solve both equations to find a and by using value of a, find m.

Complete step by step solution:
 ${{\rm{z}}^3} + \left( {3 + {\rm{i}}} \right){{\rm{z}}^2} - 3{\rm{z}} - \left( {{\rm{m}} + {\rm{i}}} \right) = 0$---------(1)
Let ‘a’ be the real root.
 $ \Rightarrow {\rm{z}} = {\rm{a}}$
 ${{\rm{a}}^3} + \left( {3 + {\rm{i}}} \right){{\rm{a}}^2} - 3{\rm{a}} - \left( {{\rm{m}} + {\rm{i}}} \right) = 0$---------(2)
(since, a is the root of the equation 1)
Conjugating eq (2) we get.
 ${{\rm{\bar a}}^3} + \left( {3 - {\rm{i}}} \right){{\rm{\bar a}}^2} - 3{\rm{\bar a}} - \left( {{\rm{m}} - {\rm{i}}} \right) = 0$
 $ \Rightarrow {{\rm{a}}^3} + \left( {3 - {\rm{i}}} \right){{\rm{a}}^2} - 3{\rm{a}} - \left( {{\rm{m}} - {\rm{i}}} \right) = 0$----------(3)
Subtracting eq (2) and eq .3 we get
 ${{\rm{a}}^2}\left( {2{\rm{i}}} \right) = 2{\rm{i}}$
 ${\rm{a}} = \pm 1$ (4)
Multiplying eq 2 by (3 - i) and eq 3 by (3 + i) and then subtracting we get
 $\left( {{{\rm{a}}^3} - 3{\rm{a}}} \right)\left( {3 - {\rm{i}} - 3 - {\rm{i}}} \right) = \left( {3 - {\rm{i}}} \right)\left( {{\rm{m}} + {\rm{i}}} \right) - \left( {3 + {\rm{i}}} \right)\left( {{\rm{m}} - {\rm{i}}} \right)$
 $ \Rightarrow {{\rm{a}}^3} - 3{\rm{a}} = {\rm{m}} - 3$
 $ \Rightarrow 1 - 3 = {\rm{m}} - 3$ (when a =1)
 $ \Rightarrow {\rm{m}} = 1$
 $ \Rightarrow - 1 + 3 = {\rm{m}} - 3$ (when a = -1)
 $ \Rightarrow {\rm{m}} = 5$
So, m = 1, 5

So, from above option both A and D is correct.

Note:
Don’t miss any equation which we have taken. To remove i multiply the equation in terms of a with (3 - i) and its conjugate equation with (3 + i). then subtract both equations. Find all the value of m by putting a = 1 and a = -1.