If the multiplicative inverse of x2-iy2x-iy is purely imaginary- where x- y 0- then the value of x y is

Let z be the multiplicative inverse, then z×x^2 iy^2x+iy=1 ⇒ z = x+iyx^2 iy^2 ⇒ z = x+iyx^2 iy^2×x^2+iy^2x^2+iy^2 ⇒ z = x^3+ixy^2+ix^2y y^3x^4+y^4 ⇒ z = (x^3 y^ ...

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Standard XII

Mathematics

Conjugate of a Complex Number

If the multip...

Question

If the multiplicative inverse of $\frac{x^{2} - i y^{2}}{x + i y}$ is purely imaginary, where $x, y \neq 0$ , then the value of $\frac{x}{y}$ is

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Solution

The correct option is B $1$
Let $z$ be the multiplicative inverse, then
$z \times \frac{x^{2} - i y^{2}}{x + i y} = 1 \Rightarrow z = \frac{x + i y}{x^{2} - i y^{2}} \Rightarrow z = \frac{x + i y}{x^{2} - i y^{2}} \times \frac{x^{2} + i y^{2}}{x^{2} + i y^{2}} \Rightarrow z = \frac{x^{3} + i x y^{2} + i x^{2} y - y^{3}}{x^{4} + y^{4}} \Rightarrow z = \frac{(x^{3} - y^{3}) + i (x y^{2} + x^{2} y)}{x^{4} + y^{4}}$
As the multiplicative inverse is purely imaginary, so
$x^{3} - y^{3} = 0 \Rightarrow (x - y) (x^{2} + x y + y^{2}) = 0 \Rightarrow x = y ∴ \frac{x}{y} = 1$

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Conjugate of a Complex Number

Standard XII Mathematics

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If the multiplicative inverse of x2−iy2x+iy is purely imaginary, where x,y≠0, then the value of xy is

The correct option is B 1Let z be the multiplicative inverse, then z×x2−iy2x+iy=1⇒z=x+iyx2−iy2⇒z=x+iyx2−iy2×x2+iy2x2+iy2⇒z=x3+ixy2+ix2y−y3x4+y4⇒z=(x3−y3)+i(xy2+x2y)x4+y4 As the multiplicative inverse is purely imaginary, so x3−y3=0⇒(x−y)(x2+xy+y2)=0⇒x=y∴xy=1

If the multiplicative inverse of $\frac{x^{2} - i y^{2}}{x + i y}$ is purely imaginary, where $x, y \neq 0$ , then the value of $\frac{x}{y}$ is