Grade 12 Problems on Complex Numbers with Solutions and Answers
Grade 12 problems on complex numbers with detailed solutions are presented.
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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.
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)
= 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.

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