1. Tardigrade
  2. Question
  3. Mathematics
  4. If z=x-iy and z1/3=p-iq, then ( (x/p)+(y/p) )/(p2+q2) is equal to

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. If $ z=x-iy $ and $ {{z}^{1/3}}=p-iq, $ then $ {\left( \frac{x}{p}+\frac{y}{p} \right)}/{({{p}^{2}}+{{q}^{2}})}\; $ is equal to

JamiaJamia 2007

Solution:

$ {{z}^{1/3}}=+iq $ $ \Rightarrow $ $ {{(x-iy)}^{1/3}}=(p+iq) $ $ (\because z=x-iy) $ $ \Rightarrow $ $ (x-iy)=p{{(p+iq)}^{3}} $ $ \Rightarrow $ $ (x-iy)={{p}^{3}}+{{(iq)}^{3}}+3{{p}^{2}}qi+3p{{q}^{2}}{{i}^{2}} $ $ \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}{p}={{p}^{2}}-3{{q}^{2}}+{{q}^{2}}-3{{p}^{2}} $ $ \Rightarrow $ $ \frac{x}{p}+\frac{y}{q}=-2{{p}^{2}}-2{{q}^{2}} $ $ \Rightarrow $ $ \frac{x}{p}+\frac{y}{q}=-2({{p}^{2}}+{{q}^{2}}) $ $ \Rightarrow $ $ \frac{x/p+y/q}{({{p}^{2}}+{{q}^{2}})}=-2 $