It's ok that z=0 is a singularity, but why , where k is a integer, are singularities too? sinh(z)=0 only when z=0 Regards
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Originally Posted by FernandoRevilla so, why the argument of the sinh must be an even multiple of ?
Originally Posted by hurz so, why the argument of the sinh must be an even multiple of ? You asked if were singularities of and the answer is yes. If you want to find all of them:
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