If z=x-iy and z^{1/3}=a+ib,then frac{(x/a+y/b | vedclass

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(A) Given,$z=x-iy$ and $z^{1/3}=a+ib$. Taking the cube on both sides,we get: $(z^{1/3})^3 = (a+ib)^3$ $z = a^3 + (ib)^3 + 3a^2(ib) + 3a(ib)^2$ $z = a^

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-$2$

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(A) Given,$z=x-iy$ and $z^{1/3}=a+ib$. Taking the cube on both sides,we get: $(z^{1/3})^3 = (a+ib)^3$ $z = a^3 + (ib)^3 + 3a^2(ib) + 3a(ib)^2$ $z = a^3 - ib^3 + 3a^2bi - 3ab^2$ $z = (a^3 - 3ab^2) - i(b^3 - 3a^2b)$ Since $z = x - iy$,we compare the real and imaginary parts: $x = a^3 - 3ab^2$ and $y = b^3 - 3a^2b$. Now,substitute these into the expression: $\frac{(x/a + y/b)}{a^2+b^2} = \frac{((a^3 - 3ab^2)/a + (b^3 - 3a^2b)/b)}{a^2+b^2}$ $= \frac{(a^2 - 3b^2 + b^2 - 3a^2)}{a^2+b^2}$ $= \frac{-2a^2 - 2b^2}{a^2+b^2} = \frac{-2(a^2+b^2)}{a^2+b^2} = -2$.

If $z=x-iy$ and $z^{1/3}=a+ib$,then $\frac{(x/a+y/b)}{a^2+b^2}=$

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