• # question_answer If $z=x+iy,\,{{z}^{1/3}}=a-ib$ and $\frac{x}{a}-\frac{y}{b}=k\,({{a}^{2}}-{{b}^{2}})$  then value of k equals [DCE 2005] A) 2 B) 4 C) 6 D) 1

${{(x+iy)}^{1/3}}=a-ib$ $x+iy={{(a-ib)}^{3}}=({{a}^{3}}-3a{{b}^{2}})+i({{b}^{3}}-3{{a}^{2}}b)$ Þ   $x={{a}^{3}}-3a{{b}^{2}},\,y={{b}^{3}}-3{{a}^{2}}b$ Þ  $\frac{x}{a}={{a}^{2}}-3{{b}^{2}},\,\frac{y}{b}={{b}^{2}}-3{{a}^{2}}$ $\therefore$  $\frac{x}{a}-\frac{y}{b}={{a}^{2}}-3{{b}^{2}}-{{b}^{2}}+3{{a}^{2}}$    $\frac{x}{a}-\frac{y}{b}=4({{a}^{2}}-{{b}^{2}})=k({{a}^{2}}-{{b}^{2}})$     $\therefore$  $k=4$.