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Question

The argument of 1i1+i is


A

π2

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B

π2

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C

3π2

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D

5π2

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Solution

The correct option is A

π2


Let z = 1i1+i z=1i1+i×1i1i z=1+i22i1i2 z=112i1+1 z=2i2 z=i

Since, z lies on negative direction of imaginary axis.

Therefore, arg (z) = π2


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