(1-i)ln(1+i) - Math Central

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I am stuck on the expansion of (1-i)ln(1+i) =(1-i)[ln(square root of 2)+i(3.14/4 = 2n3.14)]

Kim,

1 + i = √2e^{(π/4 + 2nπ)i}and hence ln(1 + i) = ln√2 + i(π/4 + 2nπ) which is of the form a + ib. Hence (1 - i) ln(1+ i) is of the form

(1 - i) (a + ib)

Expand and express as the real part plus i times the imaginary part.

Harley

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