A) \[p=x,q=y\]
B) \[p={{x}^{2}},\,\,q={{y}^{2}}\]
C) \[x=q,y=p\]
D) None of these
Correct Answer: C
Solution :
\[(x+iy)(p+iq)=({{x}^{2}}+{{y}^{2}})i\] Þ \[(xp-yq)+i(xq+yp)=({{x}^{2}}+{{y}^{2}})i\] Þ \[xp-yq=0,xq+yp={{x}^{2}}+{{y}^{2}}\] Þ \[\frac{x}{q}=\frac{y}{p}\]and\[xq+yp={{x}^{2}}+{{y}^{2}}\] Let \[\frac{x}{q}=\frac{y}{p}=\lambda \]. then \[x=\lambda q,y=\lambda p\] \\[xq+yp={{x}^{2}}+{{y}^{2}}\]Þ \[\lambda ={{\lambda }^{2}}\]Þ \[\lambda =1\] \ \[x=q,y=p\].
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