If z is a complex number satisfying |z|=1, then the range of arg(11−z) is
A
(0,π)
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B
[0,π]
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C
[0,π2]
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D
(−π2,π2)
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Solution
The correct option is D(−π2,π2) Let u=11−z ⇒z=1−1u⇒|z|=∣∣∣1−1u∣∣∣⇒∣∣∣1−1u∣∣∣=1⇒|u−1|=|u|
Locus of u is the perpendicular bisector of the line segment joining points (0,0) and (1,0) Re(u)=12
Therefore, arg(u)∈(−π2,π2)