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4-1.Complex numbers
medium
यदि $z = x + iy,\,{z^{1/3}} = a - ib$ तथा $\frac{x}{a} - \frac{y}{b} = k\,({a^2} - {b^2})$, तब $k$ का मान है
A
$2$
B
$4$
C
$6$
D
$1$
Solution
(b) ${(x + iy)^{1/3}} = a – ib$
$x + iy = {(a – ib)^3} = ({a^3} – 3a{b^2}) + i({b^3} – 3{a^2}b)$
$⇒$ $x = {a^3} – 3a{b^2},\,y = {b^3} – 3{a^2}b$
$⇒$ $\frac{x}{a} = {a^2} – 3{b^2},\,\frac{y}{b} = {b^2} – 3{a^2}$
$\therefore $ $\frac{x}{a} – \frac{y}{b} = {a^2} – 3{b^2} – {b^2} + 3{a^2}$
$\frac{x}{a} – \frac{y}{b} = 4({a^2} – {b^2}) = k({a^2} – {b^2})$
$\therefore $ $k = 4$.
Standard 11
Mathematics
