Using the Theorem: exp^i(theta) = cos(theta) + i*sin(theta)
Show that:
log(-abs(x)) = log(abs(x)) + i*pi
Hello, superdael!
Using the theorem: . e^{iθ} .= .cosθ + i·sinθ
show that: .ln(-|x|). = .ln(|x| + i·π
In the theorem, let θ = π
. . and we have: .e^{i·π} .= .cos(π) + i·sin(π) .= .-1
Hence: .ln(-1) = i·π .[1]
We have: .ln(-|x|) .= .ln[|x|·(-1)] .= .ln(|x|) + ln(-1)
From [1], we have: .ln(-|x|) .= .ln(|x|) + i·π
Thanks a lot.
I got a sort of backward variation on your answer but had one sticky point that I couldn't figure out. Your solution helped simplify it.
The next part of the question asks:
"What can you conclude about the the relationship betweein complex numbers and the ln function? Explain"
Not to sure of the reasoning...
I've just started working with Complex Numbers and wonder if anyone knows a good web-based explanation?