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Question

log(1+3i) can be expressed in cartesian form as

A
ln2(2π3)i
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B
ln2+(2π3)i
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C
ln2(π3)i
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D
ln2+(π3)i
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Solution

The correct option is B ln2+(2π3)i
log(z)=ln(|z|)+iarg(z)

For z=1+3i,|z|=2 and
tanα=31
α=π3
arg(z)=θ=ππ3=2π3

Hence, log(1+3i)=ln2+(2π3)i

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