+118654 You need to check each step of this answer as I likely made one or more carelsee errors.
Redone on a later post with the careless error removed.
https://web2.0calc.com/questions/imaginary-numbers-precalc/edit-2#r12
\(\frac{(1-i)^{10}(\sqrt 3 + i)^5}{(-1-i\sqrt 3)^{10}}\\~\\ =\frac{(1-i)^{10}(\sqrt 3 + i)^5}{(1+i\sqrt 3)^{10}}\times \frac{(1-i\sqrt 3)^{10}}{(1-i\sqrt 3)^{10}}\\~\\ =\frac{[(1-i)(1-i\sqrt 3)]^{10}(\sqrt 3 + i)^5}{[(1+i\sqrt 3)(1-i\sqrt 3)]^{10}}\\~\\ =\frac{[[(1-\sqrt3)-(1+\sqrt3)i]^{2}]^5(\sqrt 3 + i)^5}{4^{10}}\\~\\ =\frac{[4i]^5(\sqrt 3 + i)^5}{4^{10}}\\~\\ =\frac{[i]^5(\sqrt 3 + i)^5}{4^{5}}\\~\\ =\frac{i(\sqrt 3 + i)^5}{4^{5}}\\~\\ =\frac{i(16(\sqrt3+i)}{4^{5}}\\~\\ =\frac{i(\sqrt3+i)}{4^3}\\~\\ =\frac{-1+\sqrt3i}{64}\\~\\ \)
Latex:
\frac{(1-i)^{10}(\sqrt 3 + i)^5}{(-1-i\sqrt 3)^{10}}\\~\\
=\frac{(1-i)^{10}(\sqrt 3 + i)^5}{(1+i\sqrt 3)^{10}}\times \frac{(1-i\sqrt 3)^{10}}{(1-i\sqrt 3)^{10}}\\~\\
=\frac{[(1-i)(1-i\sqrt 3)]^{10}(\sqrt 3 + i)^5}{[(1+i\sqrt 3)(1-i\sqrt 3)]^{10}}\\~\\
=\frac{[[(1-\sqrt3)-(1+\sqrt3)i]^{2}]^5(\sqrt 3 + i)^5}{4^{10}}\\~\\
=\frac{[4i]^5(\sqrt 3 + i)^5}{4^{10}}\\~\\
=\frac{[i]^5(\sqrt 3 + i)^5}{4^{5}}\\~\\
=\frac{i(\sqrt 3 + i)^5}{4^{5}}\\~\\
=\frac{i(16(\sqrt3+i)}{4^{5}}\\~\\
=\frac{i(\sqrt3+i)}{4^3}\\~\\
=\frac{-1+\sqrt3i}{64}\\~\\
