Complex Numbers And Quadratic Equations question 429
Question: If $ x+iy=\frac{3}{2+\cos \theta +i\sin \theta }, $ then $ x^{2}+y^{2} $ is equal to
Options:
A) $ 3x-4 $
B) $ 4x-3 $
C) $ 4x+3 $
D) None of these
Show Answer
Answer:
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 } $ $ =[ \frac{6+3\cos \theta }{5+4\cos \theta } ]+i[ \frac{-3\sin \theta }{5+4\cos \theta } ] $
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[ \frac{6+3\cos \theta }{5+4\cos \theta } ]-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.
BETA
