Sum

If (x + iy)^{3} = y + vi then show that `(y/x + "v"/y)` = 4(x^{2} – y^{2})

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#### Solution

(x + yi)^{3} = y + vi

∴ x^{3} + 3x^{2}yi + 3xy^{2}i^{2} + y^{3}i^{3} = y + vi

∴ x^{3} + 3x^{2}yi + 3xy^{2} (–1) – y^{3}i = y + vi ...[∵ i^{2} = – 1, i^{3} = – 1]

∴ (x^{3} – 3xy^{2}) + (3x^{2}y – y^{3})i = y + vi

Equating real and imaginary parts, we get

y = x^{3} – 3xy^{2} and v = 3x^{2}y – y

∴ `y/x` = x^{2} – 3y^{2} and `"v"/y` = 3x^{2} – y^{2}

∴ `y/x + "v"/y` = x^{2} – 3y^{2} + 3x^{2} – y^{2} = 4x^{2} – 4y^{2}

∴ `y/x + "v"/y` = 4(x^{2} – y^{2})

Concept: Algebra of Complex Numbers

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