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Say z = 2(pi)i * n where n can be anything.e^(2pi*n*i) = cos(2pi*n) + i sin(2pi*n)Hence this can only be 1 when imaginory part of this equation is 0. Hence,sin(2pi*n) = 0This is valid only when n is an integer.Hence Proved.Thanks & RegardsBharat BajajaskIITians FacultyQualification.IIT Delhi
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