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Question

# If z lies on the curve |z−3−4i|=3, then least value of |z| is

A
2
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B
3
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C
8
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D
None of these
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Solution

## The correct option is A 2Let z=x+iyHence, |(x−3)+(y−4)i|=3Therefore, (x−3)2+(y−4)2=9Hence z lies on the above circle.Let,x−3=3cosθx=3+3costhetay−4=3sinθy=4+3sinθHence ,z=3(1+cosθ)+i(4+3sinθ)|z|2=9(1+cosθ)2+(4+3sinθ)2 ...(i)=9[1+cos2θ+2cosθ]+[16+24sinθ+9sin2θ]=25+9+18cosθ+24sinθ=34+18cosθ+24sinθDifferentiating with respect to θ and equating with 0, we get⇒−18sinθ+24cosθ=0⇒3sinθ=4cosθ⇒tanθ=43Hence,⇒sinθ=±45⇒cosθ=±35Now Considering ⇒sinθ=−45⇒cosθ=−35⇒z=3(1−35)+i(4−125)⇒65+i85Hence ⇒|z|min=105=2

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