A) \[3x-4\]
B) \[4x-3\]
C) \[4x+3\]
D) None of these
Correct Answer: B
Solution :
If \[x+iy=\frac{3}{2+\cos \theta +i\sin \theta }\] \[=\frac{3(2+\cos \theta -i\sin \theta )}{{{(2+\cos \theta )}^{2}}+{{\sin }^{2}}\theta }=\frac{6+3\cos \theta -3i\sin \theta }{4+{{\cos }^{2}}\theta +4\cos \theta +{{\sin }^{2}}\theta }\] \[=\left[ \frac{6+3\cos \theta }{5+4\cos \theta } \right]+i\,\left[ \frac{-3\sin \theta }{5+4\cos \theta } \right]\] Þ \[x=\frac{3(2+\cos \theta )}{5+4\cos \theta },y=\frac{-3\sin \theta }{5+4\cos \theta }\] \ \[{{x}^{2}}+{{y}^{2}}=\frac{9}{{{(5+4\cos \theta )}^{2}}}\]\[[4+{{\cos }^{2}}\theta +4\cos \theta +{{\sin }^{2}}\theta ]\] \[=\frac{9}{5+4\cos \theta }=4\left[ \frac{6+3\cos \theta }{5+4\cos \theta } \right]-3=4x-3\] Trick: \[x+iy=\frac{3(2+\cos \theta -i\sin \theta )}{(2+\cos \theta +i\sin \theta )(2+\cos \theta -i\sin \theta )}\] Let\[y=0\], then\[\sin \theta =0\]i.e.,\[\theta =0\] Now put \[x=1\] then\[{{x}^{2}}+{{y}^{2}}={{1}^{2}}+0=1\]. Also option b gives 4(1) - 3=1.
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