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