EQUALITY OF TWO COMPLEX NUMBERS EXAMPLES

Problem 1 :

Find real numbers x and y such that :

a) 2x + 3yi = -x - 6i 

b) x² + xi = 4 - 2i

c) (x + yi) (2 - i) = 8 + i 

d)  (3 + 2i) (x + yi) = -i

Solution

Problem 2 :

Find x and y if x, y ϵ ℝ and :

a)  2(x + yi) = x - yi 

b)  (x + 2i) (y - i) = -4 - 7i

c)  (x + i) (3 - iy) = 1 + 13i

d)  (x + yi) (2 + i) = 2x - (y + 1)i

Solution

Problem 3 :

Write z in the form a + bi where a, b ϵ ℝ and i = √-1, if the complex number z satisfies the equation 3z + 17i = iz + 11

Solution

Problem 4 :

The complex number z is a solution of the equation 

√z = 4/(1 + i) + 7 - 2i.

Express z in the form a + bi where a and b ϵ ℤ.

Solution

Problem 5 :

Find the real values of m and n for which

3(m + ni) = n - 2mi - (1 - 2i).

Solution

Problem 6 :

Express z = 3i/(√2 - i) + 1 in the form a + bi where a, b ϵ ℝ, giving the exact values of the real and imaginary parts of z.

Solution

Problem 1 :

Find real numbers a and b such that :

a)  a + ib = 4 

b) (1 – 2i) (a + bi) = -5 – 10i

c)  (a + 2i) (1 + bi) = 17 – 19i

Solution

Problem 2 :

Find a complex number z such that 2z -1 = iz – i. Write your answer in the form z = a + bi where a, b ϵ ℝ.

Solution

Problem 3 :

if z = 4 + i and w = 3 – 2i find 2w* - iz.

Solution

Problem 4 :

Find rationals a and b such that

(2 – 3i)/(2a + bi) = 3 + 2i.

Solution

Problem 5 :

a + ai is a root of x² – 6x + b = 0 where a, b ϵ ℝ. Explain why b has two possible values. Find a in each case.

Solution

Problem 6 :

Find real numbers x and y such that

(3x + 2yi) (1 - i) = (3y + 1)i – x.

Solution

Problem 7 :

Find real x and y such that :

a) x + iy = 0

b) (3 – 2i) (x + i) = 17 + yi 

c) (x + iy)² = x – iy

Solution

Problem 8 :

Find z if √z = 2/(3 – 2i) + 2 + 5i

Solution