If z=x+iy,x,yϵR then locus of equation b¯z+z¯b=c where b,cϵR,b≠0 are fixed,is
a straight line
an ellipse
a circle
a hyperbola
A
a straight line
B
a circle
C
a hyperbola
D
an ellipse
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Solution
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let b=m+ni... fixed point. z=x+iy .... variable. Hence ¯¯¯zb+¯¯bz=c (x−iy)(m+ni)+(m−ni)(x+iy)=c mx+inx−imy+ny+mx+imy−inx+ny=c 2mx+2ny=c x(2m)+y(2n)=c This is nothing but equation of a straight line. Hence z lies on a straight line.
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