You can continue by noting that $\displaystyle 1+ i= \sqrt{2}e^{i\pi/2+ 2ik\pi}$ so that $\displaystyle ln(1+ i)= ln(\sqrt{2})+ \pi/2+ 2k\pi$ for k any integer.
You can continue by noting that $\displaystyle 1+ i= \sqrt{2}e^{i\pi/2+ 2ik\pi}$ so that $\displaystyle ln(1+ i)= ln(\sqrt{2})+ \pi/2+ 2k\pi$ for k any integer.
For some reason it had not occured to me that the i was the imaginary unit!?