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Question

The principal argument of i−3i−1 is

A
tan112
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B
tan132
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C
tan152
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D
tan172
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Solution

The correct option is A tan112
We have,
arg(i3)=tan1(13) &
arg(i1)=tan1(1)

arg(i3)arg (i1)=tan1⎜ ⎜ ⎜13+11+13⎟ ⎟ ⎟=tan1(12)

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