Branch Cut for Log(1 - 1/z^2)
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When you're given Log(f(z)) of any kind of f(z), is there a particular method to find the domain of analyticity of Log(f(z)), how can you find its branch cuts?
I'm stuck with the following Log(1-1/z^2), i don't understand how to prove that the branch cut is [-1,1]
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Solution Summary
We explain how to find the branch cut of the multivalued complex analytic function f(z) = Log(1 + 1/z^2).
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Since Log(z) is analytic everywhere except at z=0, the domain of analyticity of Log(f(z)) is the same as the domain of analyticity of f(z) - f(0). To identify the branch cuts of Log(f(z)), we first need to find the ...
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