Anonymous
Anonymous asked in Science & MathematicsMathematics · 7 years ago

# Calculate the following: I. log(1+i√3) II. Log((e^2)i)?

Complex analysis

Relevance

I.

1+i√3 in polar form is 2 cis(π/3), which can be written 2e^(iπ/3)

So ln(1+i√3) = ln(2e^(iπ/3))

ln(2) + ln(e^(iπ/3))

ln(2) + iπ/3

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If you account for rotations about the polar axis then there are in fact infinitely many solutions of the form ln(2) + iπ(2n + 1/3).

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• log of complex numbers do not exist, so I'd say

I log(1 + 0) = log 1 = 0

II log e^2 + log i

2loge + 0 = 2loge

Edit

• Anonymous
7 years ago

I. Log(1+i(3)^(1/2))

=Log(2) + Log(1/2 + i(3)^(1/2))

=Log(2) + Log(cos(pi/3) + isin(pi/3))

=Log(2) + Log(e^(ipi/3))

=Log(2) + ipi/3

II. Log(e^2*i)

=Log(e^2) + Log(i)

=2 + Log(e^(ipi/2))

=2 + ipi/2