A) 0
B) 1
C) \[-1\]
D) 2
Correct Answer: B
Solution :
Given,\[\frac{3}{2+\cos \theta +i\,\sin \theta }=a+ib\] \[\Rightarrow \]\[\frac{3[(2+\cos \theta )-i\,\sin \theta ]}{{{(2+\cos \theta )}^{2}}+{{\sin }^{2}}\theta }=a+ib\] \[\Rightarrow \]\[\frac{3[2+cos\theta -i\,sin\theta ]}{5+4\,\cos \theta }=a+ib\] \[\Rightarrow \]\[a=\frac{3(2+cos\theta )}{5+4\cos \theta },b=-\frac{3\sin \theta }{5+4\cos \theta }\] \[\therefore \]\[{{(a-2)}^{2}}+{{b}^{2}}={{\left( \frac{6+3\cos a}{5+4\cos \theta }-2 \right)}^{2}}\] \[+\frac{9\,{{\sin }^{2}}\theta }{{{(5+4\cos \theta )}^{2}}}\] \[=\frac{{{\left( -4-5-\cos \theta \right)}^{2}}+9{{\sin }^{2}}\theta }{{{(5+4cos\theta )}^{2}}}\] \[=\frac{16+25{{\cos }^{2}}\theta +40\cos \theta +9{{\sin }^{2}}\theta }{{{(5+4cos\theta )}^{2}}}\] \[=\frac{16+16{{\cos }^{2}}\theta +40\cos \theta +9}{{{(5+4cos\theta )}^{2}}}\] \[=\frac{{{(5+4\cos \theta )}^{2}}}{{{[5+4\cos \theta ]}^{2}}}=1\]
You need to login to perform this action.
You will be redirected in
3 sec
