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.

1. Evaluate the following expressions
   a) (3 + 2i) - (8 - 5i)
   b) (4 - 2i)*(1 - 5i)
   c) (- 2 - 4i) / i
   d) (- 3 + 2i) / (3 - 6i)
   

2. If (x + yi) / i = ( 7 + 9i ) , where x and y are real, what is the value of (x + yi)(x - yi)?


3. Determine all complex number z that satisfy the equation
   z + 3 z' = 5 - 6i
   where z' is the complex conjugate of z.
   

4. 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.
   

5. 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.
   

6. Find all complex numbers z such that z² = -1 + 2 sqrt(6) i.
   

7. Find all complex numbers z such that (4 + 2i)z + (8 - 2i)z' = -2 + 10i, where z' is the complex conjugate of z.
   

8. 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.
   

9. 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.
   

10. 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.
    

1. a) -5 + 7i
   b) -6 - 22i
   c) -4 + 2i
   d) -7/15 - 4i/15


2. (x + yi) / i = ( 7 + 9i )
   (x + yi) = i(7 + 9i) = -9 + 7i
   (x + yi)(x - yi) = (-9 + 7i)(-9 - 7i) = 81 + 49 = 130


3. 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


4. 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.


5. 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


6. 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


7. 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


8. 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.


9. 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.


10. 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 References and links