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AP EAMCET Engineering 21 Apr 2019 Shift 1 Paper

Section: Mathematics

Question : 7 of 160

Marks: +1, -0

z^{\frac{1}{3}}

\frac{ \left( \frac{x}{a} + \frac{y}{b} \right) }{ a^2+b^2 }

Solution:

Consider the given equation.

z=x+iy \; \text{ and } \; z^{\frac{1}{3}}=a+ib

z^{\frac{1}{3}} \; = (a+ib)

z \; = (a+ib)^3

z \; = a^3 + (i b)^3 + 3 a^2 i b + 3 a (i b)^2

x - i y \; = (a^3 - 3a b^2) - i (b^3 - 3 a^2 b)

The values are

x = a^3 - 3 a b^2

y = b^3 - 3 a^2 b

Find the value f function.

\; \frac{ \left( \frac{x}{a} + \frac{y}{b} \right) }{ a^2+b^2 } \; = \; \frac{ \left( \frac{a^3 - 3a b^2}{a} + \frac{b^3 - 3a^2 b}{b} \right) }{ a^2+b^2 }

\; = - \frac{ a^2 - 3b^2 + b^2 - 3a^2 }{ a^2+b^2 }

\; = \frac{ -2a - 2b^2 }{ a^2+b^2 }

\; = -2

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