If |z−1|=1, then the value of tan(arg(z−1)2)−2iz is
A
1
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B
−1
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C
i
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D
−i
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Solution
The correct option is D−i Given that |z−1|=1
Here z−1=cosθ+isinθ
Where θ=arg(z−1) ⇒z=1+cosθ+isinθ⇒z=2cos2θ2+i2sinθ2cosθ2⇒z=2cosθ2(cosθ2+isinθ2)
From above equation, we get |z|=2cosθ2,argz=θ2
Hence tan(arg(z−1)2)−2iz=tanθ2−icosθ2(cosθ2+isinθ2)=tanθ2−i(cosθ2−isinθ2)cosθ2=−i