Write z in exponential form:
[tex]z=1-i=\sqrt2 e^{-i\frac\pi4}[/tex]
Then taking the logarithm, we get
[tex]\mathrm{Ln}(z)=\ln(\sqrt2) + \ln e^{-i\frac\pi4} = \boxed{\ln(\sqrt2)-\dfrac\pi4i}[/tex]
so a is the correct answer.
