+505 Ok! There are two important formulas! Euler's formula and de Moivre's theorem. Here are links to them:
Euler's Formula: https://www.math.columbia.edu/~woit/eulerformula.pdf
de Moivre's theorem: https://byjus.com/jee/de-moviers-theorem/
\(\frac{(1-i)^{10}(\sqrt 3 + i)^5}{(-1-i\sqrt 3)^{10}} \)
First, you find the modulus for each complex number. You get the modulus by taking the coefficients of the real and the imaginary parts. An example from this problem would be | 1-i | = \(\sqrt{(1^2)+(-1^2)} = \sqrt2 \) and \(\sqrt2\) would be the modulus.
I would think the people reading this question would be taking precalc, so I'll assume they know how to use these formulas correctly. If not, go read the articles posted above.
Solution:
\(\frac{(1-i)^{10}(\sqrt 3 + i)^5}{(-1-i\sqrt 3)^{10}} =\)
\(\frac{ (\sqrt{2} )^{10} \left( \cos (-\frac{\pi}4) + i \sin(-\frac{\pi}4) \right)^{10} \cdot 2^5 \left( \cos\frac\pi 6 + i \sin\frac\pi 6 \right)^5}{2^{10}\left( \cos\frac{4\pi}3 + i \sin \frac{4\pi}3 \right)^{10}} = \)
If you don't know how we got this \(\uparrow\), basically, we found the modulus, then took the coefficients, and found the angle that matches the cos and sin values.
\(\frac{ 2^{10} \left( \cos (-\frac{5\pi}2) + i \sin (-\frac{5\pi}2) \right)\left( \cos\frac{5\pi}6 + i \sin \frac{5\pi}6 \right)}{ 2^{10} \left( \cos\frac{40\pi}3 + i \sin\frac{40\pi}3 \right) } =\)
Here we used de Moivre's theorem. Go look at the article for an explanation.
\(\frac{ \cos (-\frac{5\pi}3) + i \sin (-\frac{5\pi}3) }{ \cos(\frac{40\pi}3) + i\sin(\frac{40\pi}3) } =\)
We simplified to get this.
\(\cos (-15\pi) + i\sin (-15\pi) = -1\)
Final Solution!!! :D
