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Question

# lf |z−25i|≤15, then |max(arg(z))−min(arg(z))|=

A
2cos135
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B
2cos145
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C
π2+cos135
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D
sin135cos135
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Solution

## The correct option is A 2cos−135Let z=x+iyHence the above expression reduces tox2+(y−25)2=152Hencex=15cosθy=25+15sinθ.Hencearg(z)=tan−1(25+15sinθ15cosθ)=tan−1(5+3sinθ3cosθ).NowLet k=5+3sinθ3cosθHencedkdθ=3cosθ(3cosθ)+3sinθ(5+3sinθ)9cos2θ=0Hence9cos2θ+9sin2θ+15sinθ=0Or sinθ=−915=−35.Hencecosθ=±45.Hencearg(z)max=tan−1(5+3−353.45)=tan−1(25−912)=tan−1(43)Similarlyarg(z)min=tan−1(−43)=−tan−1(43).Hencearg(z)max−arg(z)min=2tan−1(43)=2cos−1(35)

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