Complex Numbers Problems with Solutions and Answers  Grade 12
Complex numbers are important in applied mathematics. Problems and questions 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 x^{2} + 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 z^{2} = 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:
z^{3} + 6 z^{2} + 61 z + 106 = 0
find the real root to the equation.

a) Show that the complex number 2i is a root of the equation
z^{4} + z^{3} + 2 z^{2} + 4 z  8 = 0
b) Find all the roots root of this equation.

P(z) = z^{4} + a z^{3} + b z^{2} + 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.

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)
= a^{2} + b^{2} = 25
a + b = 7 gives b = 7  a
Substitute above in the equation a^{2} + b^{2} = 25
a^{2} + (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: a^{2}  b^{2} + 2 ab i =  1 + 2 sqrt(6) i
Real part and imaginary parts must be equal.
a^{2}  b^{2} =  1 and 2 ab = 2 sqrt(6)
Equation 2 ab = 2 sqrt(6) gives: b = sqrt(6) / a
Substitute: a^{2}  ( sqrt(6) / a )^{2}) =  1
a^{4}  6 =  a^{2}
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
z^{3} + 6 z^{2} + 61 z + 106 = (z  (2 + 7i))(z  (2  7i)) q(z)
= (z^{2} + 4z + 53) q(z)
q(z) = [ z^{3} + 6 z^{2} + 61 z + 106 ] / [ z^{2} + 4z + 53 ] = z + 2
Z + 2 is a factor of z^{3} + 6 z^{2} + 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).
z^{4} + z^{3} + 2 z^{2} + 4 z  8 = (z  2i)(z + 2i) q(z)
= (z^{2} + 4)q(z)
q(z) = z^{2} + 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)) =
= z^{4}  6 z^{3} + 15 z^{2}  18 z + 10
Hence: a = 6, b = 15, c = 18 and d = 10.
More References and links
Complex numbersHigh School Maths (Grades 10, 11 and 12)  Free Questions and Problems With Answers
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