Question

write (1-root3i)^8 in the standard form a+bi

Answer 1

There's a couple ways to do this - have you learned binomial expansion?

Answer 2

well, if not, you can muscle your way through it
\[(1- \sqrt{3}i)^8 = \big ((1- \sqrt{3}i)^2 \big)^4\]
\[ = (1-2 \sqrt{3}i-3)^4
\\ \
\\ =(-2-2\sqrt{3}i)^4
\\ \
\\ =(-2)^4(1+ \sqrt{3}i)^4
\\ \
\\ =16\big( (1+ \sqrt{3}i)^2\big)^2
\\ \
\\ =16\big( (1+2 \sqrt{3}i-3)\big)^2
\\ \
\\ = .......\]

Can you finish it from there?

Answer 3

remember that
\[\sqrt{-1} =i
\\ \
\\ i^2=-1\]