A) \[-2\]
B) \[-1\]
C) \[2\]
D) \[1\]
Correct Answer: A
Solution :
Given \[{{z}^{\frac{1}{3}}}=p+iq\] \[\Rightarrow \,\,z={{p}^{3}}+{{(iq)}^{3}}+3p(iq)(p+iq)\] \[\Rightarrow \,\,x-iy={{p}^{3}}-3p{{q}^{2}}+i(3{{p}^{2}}q-{{q}^{3}})\] \[\therefore \,\,x={{p}^{3}}-3p{{q}^{2}}\Rightarrow \frac{x}{p}={{p}^{2}}-3{{q}^{2}}\] and \[y={{q}^{3}}-3{{p}^{2}}q\Rightarrow \frac{y}{q}={{q}^{2}}-3{{p}^{2}}\] \[\therefore \,\,\frac{x}{p}+\frac{y}{q}=-2{{p}^{2}}-2{{q}^{2}}\] \[\therefore \,{\,\left( \frac{x}{p}+\frac{y}{q} \right)}/{({{p}^{2}}+{{q}^{2}})=-2}\;\]You need to login to perform this action.
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