Calculate the following: I. log(1+i√3) II. Log((e^2)i)?
- 7 years agoBest Answer
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
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).
@Admire, above, I gave you an up thumb. Why? Because you gave a valid albeit naïve answer. And I respect your humble edit.
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- AdmireLv 77 years ago
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
Seems others know a lot about this, so ignore my answer.
- Anonymous7 years ago
=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
=Log(e^2) + Log(i)
=2 + Log(e^(ipi/2))
=2 + ipi/2