Let a,b∈ℝ and a2+b2≠0 .suppose S=z∈C:z=1a+ibt,t∈ℝ,t≠0, where i=-1 . If z=x+iy and z∈S, then (x,y) lies on
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a
the circle with radius 12α and center 12a,0 and a>0,b≠0
b
the circle with radius -12a and center -12a,0a<0,b≠0
c
the x -axis for a≠0,b=0
d
the y -axis for a=0,b≠0
answer is Æ.
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Detailed Solution
z=1a+bit,t∈R,t≠0z=a−bita2+b2t2=x+iy xy=−abt⇒t=−aybx since x=aa2+b2t2⇒x=ax2+y2x2+y2=xa Circle with radius =12a, centre =12a,0 Also if b=0,a≠0⇒z=1a purely real lies on x -axis if b=0,a≠0⇒z=1bit purely imaginary lies on y -axis