$(z+1)^2=1-i$ which is $(1, -1)$ on the axis with angle $-\pi /4$ and length $2^{1/2}$ so $1-i = 2^{1/2}e^{i(-\pi /4+2k\pi )}$
$$z+1=(1-i)^{1/2} =\bigl(2^{1/2}e^{i(-\pi /4+2k\pi )}\bigr)^{1/2}$$ $$z=2^{1/4}e^{i(-\pi /8 + k\pi )}-1$$
when $k=0$ $z=2^{1/4}e^{-i\pi /8}-1=2^{1/4}(\cos \pi /8 - i\sin \pi /8)-1$
when $k=1$ $z=2^{1/4}e^{i7\pi /8}-1=-2^{1/4}(\cos \pi /8 - i\sin \pi /8)-1$
Do not use pitiful html options to display mathematical snippets! I fixed it
