If z=x+iy is a complex number such that bar{z | vedclass

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(B) Given,$\bar{z}^{\frac{1}{3}} = a+ib$.Taking the cube on both sides,we get $\bar{z} = (a+ib)^3$.Since $\bar{z} = x-iy$,we have $x-iy = (a+ib)^3$.Ex

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-$2$

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(B) Given,$\bar{z}^{\frac{1}{3}} = a+ib$.Taking the cube on both sides,we get $\bar{z} = (a+ib)^3$.Since $\bar{z} = x-iy$,we have $x-iy = (a+ib)^3$.Expanding the right side: $x-iy = a^3 + 3a^2(ib) + 3a(ib)^2 + (ib)^3$.$x-iy = a^3 + 3a^2bi - 3ab^2 - ib^3$.Grouping real and imaginary parts: $x-iy = (a^3-3ab^2) + i(3a^2b-b^3)$.Comparing real and imaginary parts,we get $x = a^3-3ab^2$ and $-y = 3a^2b-b^3$,which implies $y = b^3-3a^2b$.Now,$\frac{x}{a} = a^2-3b^2$ and $\frac{y}{b} = b^2-3a^2$.Adding these,$\frac{x}{a} + \frac{y}{b} = (a^2-3b^2) + (b^2-3a^2) = -2a^2-2b^2 = -2(a^2+b^2)$.Therefore,$\frac{1}{a^2+b^2}\left(\frac{x}{a}+\frac{y}{b}\right) = \frac{-2(a^2+b^2)}{a^2+b^2} = -2$.

If $z=x+iy$ is a complex number such that $\bar{z}^{\frac{1}{3}}=a+ib$,then the value of $\frac{1}{a^2+b^2}\left(\frac{x}{a}+\frac{y}{b}\right)$ is equal to

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