If z is a complex number such that z−iz−1 is purely imaginary, then the minimum value of |z–(3+3i)| is
A
6√2
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B
2√2
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C
3√2
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D
2√2−1
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Solution
The correct option is B2√2 ∵z−iz−1 is purely imaginary number ∴z−iz−1+¯¯¯z+i¯¯¯z−1=0 ⇒(z−i)(¯z−1)+(z−1)(¯z+i)=0 ⇒2z¯z−(z+¯z)+i(z−¯z)=0
Let z=x+iy 2(x+iy)(x−iy)−(x+iy+x−iy)+i(x+iy−x+iy)=0 ∴x2+y2−x−y=0 ⇒(x−12)2+(y−12)2=12
Which is a circle of centre (12,12) and radius 1√2 ∴ Minimum value of |z−(3+3i)|=QR