JEE Main & Advanced Mathematics Complex Numbers and Quadratic Equations Question Bank Integral power of iota, Algebraic operations and Equality of complex numbers

  • question_answer If \[x+iy=\frac{3}{2+\cos \theta +i\sin \theta },\]then \[{{x}^{2}}+{{y}^{2}}\] is equal to

    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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