Complex Numbers Problems with Solutions - Grade 12

Explore a variety of complex number problems with step-by-step solutions. Learn how to solve complex numbers, including operations, polar form, and applications. Complex numbers play a crucial role in applied mathematics, physics, electrical engineering, and other technical fields.

In what follows, i is the imaginary unit.

Question 1

Evaluate the following expressions:

  1. ) (3+2i)(85i)
  2. ) (42i)(15i)
  3. ) 24ii
  4. ) 3+2i36i

Solution:

  1. ) 5+7i
  2. ) 622i
  3. ) 4+2i
  4. ) 715415i

Question 2

If x+yii=7+9i, where x and y are real, what is the value of (x+yi)(xyi)?

Solution:

x+yii=7+9i x+yi=i(7+9i)=9+7i (x+yi)(xyi)=(9+7i)(97i)=81+49=130

Question 3

Determine all complex numbers z that satisfy the equation: z+3z=56i where z denotes the complex conjugate of z.

Solution:

Let z=a+bi, and its conjugate z=abi; a and b real numbers

Substituting z and z in the given equation obtain

a+bi+3(abi)=56i a+3a+(b3b)i=56i Simplify by grouping terms on theleft side: 4a2bi=56i Two complex numbers are equal if both their real and imaginary parts are equal.. Hence 4a=5and2b=6 a=54andb=3 z=54+3i

Question 4

Find all complex numbers of the form z=a+bi, where a and b are real numbers, such that:

zz=25anda+b=7 where, z represents the complex conjugate of z.

Solution:

Let z=a+bi

Hence its conjugate z=abi therefore zz=(a+bi)(abi) =a2+b2=25 a+b=7givesb=7a Substitute above in the equation a2+b2=25 a2+(7a)2=25

Solve the above quadratic function for a and use b=7a to find b. a=4andb=3 or a=3andb=4 The complex numbers z=4+3i and z=3+4i satisfy zz=25.

Check that: z=4+3i and z=3+4i have the property zz=25

Question 5

The complex number 2+4i is one of the roots of 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.

Solution:

a) Substitute the root in the 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 are both equal to zero. 12+2b+c=0and16+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:

24i is the second solution.

Check: (24i)24(24i)+20 Expand: =416i+16i16+820+20 =(4168+20)+(16+16)i=0

Question 6

Find all complex numbers z such that: z2=1+26i

Solution:

Let z=a+bi

Substitute into given equation: (a+bi)2=1+26i

Expand: a2b2+2abi=1+26i

Real part and imaginary parts must be equal. a2b2=1and2ab=26 Equation 2ab=26 gives: b=6a

Substitute: a2(6a)2=1 a46=a2 Solve above equation and select only real roots: a=2 and a=2

Substitute to find b and write the two complex numbers that satisfy the given equation. z1=2+3i,z2=23i

Question 7

Find all complex numbers z such that (4+2i)z+(82i)z=2+10i, where z is the complex conjugate of z.

Solution:

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=abi.

Substitute z and z in the given equation (4+2i)(a+bi)+(82i)(abi)=2+10i Expand and separate real and imaginary parts. (4a2b+8a2b)+(4b+2a8b2a)i=2+10i Two complex numbers are equal if their real parts and imaginary parts are equal. Group like terms. 12a4b=2and4b=10 Solve the system of the unknown a and b to find: b=52anda=1 z=152i

Question 8

Given that the complex number z=2+7i is a root to the equation: z3+6z2+61z+106=0 find the real root to the equation.

Solution:

Since z=2+7i is a root of 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 we may factor the left side as follows: z3+6z2+61z+106=(z(2+7i))(z(27i))q(z) =(z2+4z+53)q(z) q(z)=z3+6z2+61z+106z2+4z+53=z+2 z+2 is a factor of z3+6z2+61z+106 and therefore z=2 is the real root of the given equation.

Question 9

a) Show that the complex number 2i is a root of the equation

z4+z3+2z2+4z8=0 b) Find all the roots of this equation.

Solution:

a) Substitute z by 2i in the left side of the expression:

z4+z3+2z2+4z8 (2i)4+(2i)3+2(2i)2+4(2i)8 =168i8+8i8=0

which shows that 2i is a root of the given equation.

b) Since 2i is a root and all coefficients are real, 2i is also a root (complex conjugate). Hence we may factor the left side of the given equation as follows: z4+z3+2z2+4z8=(z2i)(z+2i)q(z) =(z2+4)q(z) q(z)=z4+z3+2z2+4z8z2+4=z2+z2

The other two roots of the equation are the roots of q(z)=z2+z2 and are given by: z=1andz=2.

Question 10

P(z)=z4+az3+bz2+cz+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: 2i and 1i.

Solution:

Since all coefficients of polynomial P are real, the complex conjugate 2+iand1+i to the given zeros are also zeros of the polynomial P. Hence P(z) in factored form: P(z)=(z(2i))(z(2+i))(z(1i))(z(1+i))= =z46z3+15z218z+10

Identify a, b, c, and d to obtain: a=6,b=15,c=18,d=10.