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Question
Find the value of x and y which satisfy the following equation (x, y∈R).
If x(1 + 3i) + y(2 − i) − 5 + i3 = 0, find x + y
Solution
x(1 + 3i) + y(2 − i) − 5 + i3 = 0
∴ x + 3xi + 2y − yi − 5 − i = 0 ...[∵ i3 = − i]
∴ (x + 2y − 5) + (3x − y − 1)i = 0 + 0i
Equating real and imaginary parts, we get
x + 2y − 5 = 0 ...(i)
and 3x − y − 1 = 0 ...(ii)
Equation (i) + equation (ii) × 2 gives
7x − 7 = 0
∴ 7x = 7
∴ x = 1
Putting x = 1 in (i), we get
1 + 2y − 5 = 0
∴ 2y = 4
∴ y = 2
∴ x = 1 and y = 2
∴ x + y = 1 + 2 = 3
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