If `z=x+i y, z^(1/3)=a-i b " and " x/a-y/b=lambda(a^2-b^2),` then `lambda` is equal to

Options

A. 2
B. 4
C. 6
D. 1

Answer

B

Explanation

We have,
`z=x-iy` and `z^(1//3) = p+iq`
`rArr z=x-iy, z^(1//3)=a-ib`
`rArr x+iy=(a-ib)^(3)`
`rArr x+iy=(a^(3)-3ab^(2))+i(-3a^(2)b+b^(3))`
`rArr x/a-y/b=a^(2)-3b^(2)+3a^(2)-b^(2)`
`rArr x/a-y/b=4(a^(2)-b^(2))`
`rArr k(a^(2)-b^(2))=4(a^(2)-b^(2))` `[therefore x/a-y/b=k(a^(2)-b^(2))]`
`rArr k=4`