If [tex]z=1-i[/tex], then [tex]|z|=\sqrt{1^2+(-1)^2}=\sqrt2[/tex] and [tex]\arg z=-\dfrac\pi4[/tex] because [tex]z[/tex] lies in the fourth quadrant.
By DeMoivre's theorem,
[tex]z=|z|e^{i\arg z}\implies z^{10}=(\sqrt2)^{10}e^{-i10\pi/4}=32e^{-i5\pi/2}=32e^{-i\pi/2}=-32i[/tex]
so the answer is B.
