Answer
$2(cos(ln(2))-isin(ln(2)))$
Work Step by Step
$2^{1-i} = 2^1 2^{-i}$
Applying the identity $a^{ib} = cos(ln(a)b) + isin(ln(a)b)$ yields:
$cos(-ln(2)) + isin(-ln(2))$
Cosine and sine are even and odd functions, respectively, so
$cos(-ln(2)) + isin(-ln(2)) = cos(ln(2)) -isin(ln(2))$
In total:
$2(cos(ln(2))-isin(ln(2)))$