- #1

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## Homework Statement

Given that [itex]g(z) = ln(1-z^2)[/itex], defined on [itex]\mathbb{C}\backslash \left(-\infty, 1\right][/itex], i.e. the branch cut is from [itex]-\infty[/itex] to [itex]1[/itex] along the real axis. Find [itex]g(-i)[/itex] given [itex]g(i) = ln(2)[/itex].

## Homework Equations

## The Attempt at a Solution

I tried drawing it out but I'm having trouble making sense of this. I understand how [itex](-\infty, 0][/itex] works as a branch cut for [itex]ln(z)[/itex]. If [itex]ln(z) = a + bi[/itex] we restrict [itex]-\pi<b < \pi[/itex] so there is a single value for [itex]ln(z)[/itex]. This means every time [itex]z[/itex] passes through [itex](-\infty, 0][/itex] there is a discontinuous jump in the value of [itex]ln(z)[/itex].

However I've not been successful for explaining how [itex]-\infty[/itex] to [itex]1[/itex] works for [itex]g(z) = ln(1-z^2)[/itex].

From my drawing I think this means that the branch cut for [itex]ln(g)[/itex] is

*not*[itex](-\infty, 0][/itex]. But maybe that's not the right way to think about this.

It seems like if [itex]\left| z\right|^2 > 1[/itex], then making the argument of z go around a circle causes the argument of [itex]1 - z^2[/itex] to go around a circle

*twice*.