Advertisement

Advertisement

Advertisement

Sum

**Solve the following equation for x, y ∈ R: **

2x + i^{9} y (2 + i) = x i^{7} + 10 i^{16}

Advertisement

#### Solution

2x + i^{9} y (2 + i) = x i^{7} + 10 i^{16}

∴ 2x + (i^{4})^{2}.i.y (2 + i) = x (i^{2})^{3}.i + 10.(i^{4})^{4}

∴ 2x + (1)^{2}.iy (2 + i) = x (– 1)^{3}.i +10 (1)^{4} ...[∵ i^{2} = – 1, i^{4} = 1]

∴ 2x + 2yi + yi^{2} = – 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

Is there an error in this question or solution?