Grade 12 Problems on Complex Numbers with Solutions and Answers
Grade 12 problems on complex numbers with detailed solutions are presented.
Evaluate the following expressions
a) (3 + 2i) - (8 - 5i)
b) (4 - 2i)*(1 - 5i)
c) (- 2 - 4i) / i
d) (- 3 + 2i) / (3 - 6i)
If (x + yi) / i = ( 7 + 9i ) , where x and y are real, what is the value of (x + yi)(x - yi)?
Determine all complex number z that satisfy the equation
z + 3 z' = 5 - 6i
where z' is the complex conjugate of z.
Find all complex numbers of the form z = a + bi , where a and b are real numbers such that z z' = 25 and a + b = 7
where z' is the complex conjugate of z.
The complex number 2 + 4i is one of the root to the quadratic equation x2 + bx + c = 0, where b and c are real numbers.
a) Find b and c
b) Write down the second root and check it.
Find all complex numbers z such that z2 = -1 + 2 sqrt(6) i.
Find all complex numbers z such that (4 + 2i)z + (8 - 2i)z' = -2 + 10i, where z' is the complex conjugate of z.
Given that the complex number z = -2 + 7i is a root to the equation:
z3 + 6 z2 + 61 z + 106 = 0
find the real root to the equation.
a) Show that the complex number 2i is a root of the equation
z4 + z3 + 2 z2 + 4 z - 8 = 0
b) Find all the roots root of this equation.
P(z) = z4 + a z3 + b z2 + c z + d is a polynomial where a, b, c and d are real numbers. Find a, b, c and d if two zeros of polynomial P are the following complex numbers: 2 - i and 1 - i.
Solutions to the Above Questions
a) -5 + 7i
b) -6 - 22i
c) -4 + 2i
d) -7/15 - 4i/15
(x + yi) / i = ( 7 + 9i )
(x + yi) = i(7 + 9i) = -9 + 7i
(x + yi)(x - yi) = (-9 + 7i)(-9 - 7i) = 81 + 49 = 130
Let z = a + bi , z' = a - bi ; a and b real numbers.
Substituting z and z' in the given equation obtain
a + bi + 3*(a - bi) = 5 - 6i
a + 3a + (b - 3b) i = 5 - 6i
4a = 5 and -2b = -6
a = 5/4 and b = 3
z = 5/4 + 3i
z z' = (a + bi)(a - bi)
= a2 + b2 = 25
a + b = 7 gives b = 7 - a
Substitute above in the equation a2 + b2 = 25
a2 + (7 - a)2 = 25
Solve the above quadratic function for a and use b = 7 - a to find b.
a = 4 and b = 3 or a = 3 and b = 4
z = 4 + 3i and z = 3 + 4i have the property z z' = 25.
a) Substitute solution in equation: (2 + 4i)2 + b(2 + 4i) + c = 0
Expand terms in equation and rewrite as: (-12 + 2b + c) + (16 + 4b)i = 0
Real part and imaginary part equal zero.
-12 + 2b + c = 0 and 16 + 4b = 0
Solve for b: b = -4 , substitute and solve for c: c = 20
b) Since the given equation has real numbers, the second root is the complex conjugate of the given root: 2 - 4i is the second solution.
Check: (2 - 4i)2 - 4 (2 - 4i) + 20
(Expand) = 4 - 16 - 16i - 8 + 16i + 20
= (4 - 16 - 8 + 20) + (-16 + 16)i = 0
Let z = a + bi
Substitute into given equation: (a + bi)2 = -1 + 2 sqrt(6) i
Expand: a2 - b2 + 2 ab i = - 1 + 2 sqrt(6) i
Real part and imaginary parts must be equal.
a2 - b2 = - 1 and 2 ab = 2 sqrt(6)
Equation 2 ab = 2 sqrt(6) gives: b = sqrt(6) / a
Substitute: a2 - ( sqrt(6) / a )2) = - 1
a4 - 6 = - a2
Solve above equation and select only real roots: a = sqrt(2) and a = - sqrt(2)
Substitute to find b and write the two complex numbers that satisfies the given equation.
z1 = sqrt(2) + sqrt(3) i , z2 = - sqrt(2) - sqrt(3) i
Let z = a + bi where a and b are real numbers. The complex conjugate z' is written in terms of a and b as follows: z'= a - bi. Substitute z and z' in the given equation
(4 + 2i)(a + bi) + (8 - 2i)(a - bi) = -2 + 10i
Expand and separate real and imaginary parts.
(4a - 2b + 8a - 2b) + (4b + 2a - 8b - 2a )i = -2 + 10i
Two complex numbers are equal if their real parts and imaginary parts are equal. Group like terms.
12a - 4b = -2 and - 4b = 10
Solve the system of the unknown a and b to find:
b = -5/2 and a = -1
z = -1 - (5/2)i
Since z = -2 + 7i is a root to the equation and all the coefficients in the terms of the equation are real numbers, then z' the complex conjugate of z is also a solution. Hence
z3 + 6 z2 + 61 z + 106 = (z - (-2 + 7i))(z - (-2 - 7i)) q(z)
= (z2 + 4z + 53) q(z)
q(z) = [ z3 + 6 z2 + 61 z + 106 ] / [ z2 + 4z + 53 ]
= z + 2
Z + 2 is a factor of z3 + 6 z2 + 61 z + 106 and therefore z = -2 is the real root of the given equation.
a) (2i)4 + (2i)3 + 2 (2i)2 + 4 (2i) - 8
= 16 - 8i - 8 + 8i - 8 = 0
b) 2i is a root -2i is also a root (complex conjugate because all coefficients are real).
z4 + z3 + 2 z2 + 4 z - 8 = (z - 2i)(z + 2i) q(z)
= (z2 + 4)q(z)
q(z) = z2 + z - 2
The other two roots of the equation are the roots of q(z): z = 1 and z = -2.
Since all coefficients of polynomial P are real, the complex conjugate to the given zeros are also zeros of P. Hence
P(z) = (z - (2 - i))(z - (2 + i))(z - (1 - i))(z - (1 + i)) =
= z4 - 6 z3 + 15 z2 - 18 z + 10
Hence: a = -6, b = 15, c = -18 and d = 10.
More High School Math (Grades 10, 11 and 12) - Free Questions and Problems With Answers
More Middle School Math (Grades 6, 7, 8, 9) - Free Questions and Problems With Answers
More Primary Math (Grades 4 and 5) with Free Questions and Problems With Answers
Free Trigonometry Questions with Answers
Interactive HTML5 Math Web Apps for Mobile LearningNew !
Free Online Graph Plotter for All Devices
Home Page --
HTML5 Math Applets for Mobile Learning --
Math Formulas for Mobile Learning --
Algebra Questions -- Math Worksheets
Free Compass Math tests Practice
Free Practice for SAT, ACT Math tests
Precalculus Tutorials --
Precalculus Questions and Problems
Precalculus Applets --
Equations, Systems and Inequalities
Online Calculators --
Geometry Tutorials --
Geometry Calculators --
Calculus Tutorials --
Calculus Questions --
Applied Math --
Math Software --
High School Math --
Middle School Math --
Math Videos From Analyzemath
Updated: 2 April 2013
Copyright © 2003 - 2014 - All rights reserved