Hello, i have problem with this, can someone help me?

Question

z=((1-i)^10*(sqrt(3)+i)^5)/(-1-i*sqrt(3))^10

Answer 1

In polar form, (1-i) = √2(cos315° + isin315°)

So, by DeMoivre's Theorem, (1-i)¹⁰ = (√2)¹⁰(cos3150° + isin3150°) = 32(cos3150° + isin3150°)

In polar form, (√3 + i) = 2(cos30° + isin30°)

So, (√3 + i)⁵ = 2⁵(cos150° + isin150°) = 32(cos150° + isin150°)

In polar form, (-1-√3i) = 2(cos240° + isin240°)

So, (-1-√3i)¹⁰ = 2¹⁰(cos2400° + isin2400°) = 1024(cos2400° + isin2400°)

Therefore, using DeMoivre's Theorem, the original expression is equivalent to :

(32)(32)[cos3300° + isin3300°] / [1024(cos2400° + isin2400°)]

= cos(3300°-2400°) + isin(3300°-2400°) = cos900° + isin900° = cos180° + isin180° = -1