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Sum
Solve the following equation for x, y ∈ R:
2x + i9 y (2 + i) = x i7 + 10 i16
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Solution
2x + i9 y (2 + i) = x i7 + 10 i16
∴ 2x + (i4)2.i.y (2 + i) = x (i2)3.i + 10.(i4)4
∴ 2x + (1)2.iy (2 + i) = x (– 1)3.i +10 (1)4 ...[∵ i2 = – 1, i4 = 1]
∴ 2x + 2yi + yi2 = – xi + 10
∴ 2x + 2yi – y + xi = 10
∴ (2x – y) + (x + 2y)i = 10 + 0.i
Equating real and imaginary parts, we get
2x – y = 10 ...(i)
and x + 2y = 0 ...(ii)
Equation (i) x 2 + equation (ii) gives
5x = 20
∴ x = 4
Putting x = 4 in (i), we get
2(4) – y = 10
∴ y = 8 – 10
∴ y = – 2
∴ x = 4 and y = – 2
Concept: Solution of a Quadratic Equation in Complex Number System
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