A) 1
B) 2
C) 3
D) 4
Correct Answer: D
Solution :
\[{{z}^{1/3}}=a-ib\Rightarrow z={{(a-ib)}^{3}}\] \[\therefore \,\,\,x+iy={{a}^{3}}+i{{b}^{3}}-3i{{a}^{2}}b-3a{{b}^{2}}.\]Then \[x={{a}^{3}}-3a{{b}^{2}}\Rightarrow \frac{x}{a}={{a}^{2}}-3{{b}^{2}}\] \[y={{b}^{3}}-3{{a}^{2}}b\Rightarrow \frac{y}{b}={{b}^{2}}-3{{a}^{2}}\] So, \[\frac{x}{a}-\frac{y}{b}=4({{a}^{2}}-{{b}^{2}})\]You need to login to perform this action.
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