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If `z=z_0+A( z -( z )_0), w h e r eA`is a constant, then prove that locus of `z`is a straight line. |
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Answer» `bar(z)=bar(z)_(0)+A(z-z_(0))` `impliesAz-bar(z)-Az_(0)+bar(z)_(0)=0" "(1)` `impliesbar(A)bar(z)-z-bar(A)bar(z)_(0)+z_(0)=0" "(2)` Adding (1) and (2), we get `(A-1)z+(bar(A)-1)bar(z)-(Az_(0)+bar(A)bar(z)_(0))+z_(0)+bar(z)_(0)=0` This is of the form `bar(a)z+abar(z)+b=0," where "a=bar(A)-1" and "b=-1(Az_(0)+bar(A)bar(z)_(0))+z_(0)+bar(z)_(0)inR.` Hence, locus of z is a straight line. |
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