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