Question

Why argument of mod z1/z2=arg z1-arg z2+2kpi why it is added 2kpi?

Answer 1

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Similarly, log (-(1+i)/sqrt(2)) = -i 3pi/4 + 2kpi i = (2n - 3/4)pi i where n is an integer. ... Therefore, what we need to show is that Arg( z1 z2) = Arg z1 + Arg z2 + 2Npi where N is 0 ...

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Answer 2

Answer:

Similarly, log (-(1+i)/sqrt(2)) = -i 3pi/4 + 2kpi i = (2n - 3/4)pi i where n is an integer. ... Therefore, what we need to show is that Arg( z1 z2) = Arg z1 + Arg z2 + 2Npi where N is 0 ...