A) 1
B) \[-1\]
C) 2
D) \[-2\]
Correct Answer: D
Solution :
Given\[{{z}^{1/3}}=p+iq\] \[{{(x-iy)}^{1/3}}=p+iq\] \[[\because z=x-iy]\] \[\Rightarrow \] \[(x-iy)={{(p+iq)}^{3}}\] \[\Rightarrow \] \[(x-iy)={{p}^{3}}-i{{q}^{3}}+3{{p}^{2}}qi-3p{{q}^{2}}\] \[\Rightarrow \] \[(x-iy)=({{p}^{3}}-3p{{q}^{2}})+i(3{{p}^{2}}q-{{q}^{3}})\] On comparing both sides, we get \[x={{p}^{3}}-3p{{q}^{2}}\]and \[-y=(3{{p}^{2}}q-{{q}^{3}})\] \[\Rightarrow \] \[x=p({{p}^{2}}-3{{q}^{2}})\]and \[y=q({{q}^{2}}-3{{p}^{2}})\] \[\Rightarrow \] \[\frac{x}{p}=({{p}^{2}}-3{{q}^{2}})\]and \[\frac{y}{q}=({{q}^{2}}-3{{p}^{2}})\] Now, \[\frac{x}{p}+\frac{y}{q}={{p}^{2}}-3{{q}^{2}}+{{q}^{2}}-3{{p}^{2}}\] \[\Rightarrow \] \[\frac{x}{p}+\frac{y}{p}=-2({{p}^{2}}+{{q}^{2}})\] \[\Rightarrow \] \[\frac{x/p+y/q}{{{p}^{2}}+{{q}^{2}}}=-2\]
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