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A potatoe of a residu!


What should be spelled C-I-N-C-H is the result of inputting

Residue[1/Sinh[z],{z,I Pi}] .


This should be the same as inputting

Residue[1/(I Sin[-I z]),{z,I Pi}] .

Both inputs result in the output

Residue[Csch[z],{z, I Pi}].

To feed Mathematica the dictionary necessary to get over the hurdle,  
one must input

Residue[2/(Exp[z]-Exp[-z]),{z,I Pi}].
This results in the correct output, -1.

Eeeeeeeeeeeeeekk!






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