Question

Find all solutions for the following systems with complex coefficients. (If there are an infinite number of solutions, use t as your parameter.) ix1 + x2 = 4 2x1 + (1 − i)x2 = 3i

Answer 1

The solutions are:

x1 = (-1/2)(5 + 3i)

and

x2 = (1/2)(11 + 5i)

Explanation:

Given the equations:

ix1 + x2 = 4 .................................(1)

2x1 + (1 − i)x2 = 3i ......................(2)

From (1), make x2 the subject, to have:

x2 = 4 - ix1 ...................................(3)

Using (3) in (2)

2x1 + (1 - i)(4 - ix1) = 3i

2x1 + 4 - ix1 - 4i + i²x1 = 3i

2x1 + 4 - ix1 - 4i - x1 = 3i

(because i² = -1)

(2 - i - 1)x1 + 4(1 - i) = 3i

(1 - i)x1 + 4(1 - i) = 3i

Divide both sides by (1 - i)

x1 + 4 = 3i/(1 - i)

Multiply both the numerator and denominator of the right hand side by the conjugate of (1 - i). The conjugate of (1 - i) is (1 + i), the aim of this multiplication is to makes the denominator a real number, rather than complex.

x1 + 4 = 3i(1 + i)/(1 - i)(1 + i)

x1 + 4 = (3i - 3)/2

x1 = (3/2)(i - 1) - 4

= [3(1 -i) - 8]/2

= (3 - 3i - 8)/2

= (-5 - 3i)/2

x1 = (-1/2)(5 + 3i)..............................(4)

Using this in (3)

x2 = 4 - ix1

x2 = 4 - i(-1/2)(5 + 3i)

= 4 + (1/2)(5i - 3)

= (8 + 5i + 3)/2

= (11 + 5i)/2

x2 = (1/2)(11 + 5i)