Using the Theorem: exp^i(theta) = cos(theta) + i*sin(theta)

Show that:

log(-abs(x)) = log(abs(x)) + i*pi

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- Feb 22nd 2007, 02:51 AMsuperdaelComplex Numbers
Using the Theorem: exp^i(theta) = cos(theta) + i*sin(theta)

Show that:

log(-abs(x)) = log(abs(x)) + i*pi

- Feb 22nd 2007, 05:29 AMSoroban
Hello, superdael!

Quote:

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

- Feb 22nd 2007, 06:50 AMsuperdaelThanks
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?