= Solution
Squaring first is simpler than expanding the tenth powers. Since $(1-i)^2=-2i$ and $(1+i)^2=2i$, while $i^4=1$, the requested Cartesian forms are
$$
\boxed{\bigl((1-i)^2\bigr)^5=-32i=0-32i,\qquad
\bigl((1+i)^2\bigr)^5=32i=0+32i.}
$$
Thus the <real part> is zero in each case, and the <imaginary parts> are $-32$ and $32$. The results are <complex conjugates>, as also follows because the two inputs are <complex conjugates> and the fifth-power polynomial has real coefficients.
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