(a) 1+i= v2 = 2 cos 2n+ t +isin 2n+ co 2e log (1 +i) = log 2e4logv2 + i(2n+- h similarly (1- = 2e log(1-1)=log2e = log2 +i(2n)t 4

in above picture .....I don't understand why (2n+1/4 )pi is used ....and in the marked step √2e^-(2n+1/4)π ...... should be there as we know (1-i)=√2{1/√2-i/√2}=√2{cos(2n+1/4π)-sin(2n+1/4π)=e^-theta