+505 Compute \(\frac{(1-i)^{10}(\sqrt 3 + i)^5}{(-1-i\sqrt 3)^{10}} \).
Here's what I tried to do:
For the first imaginary number, (1-i)^10, I simplified it down to -2^5 * i. The next two imaginary numbers I tried to simplify using DeMoivre's Theorem. But I got stuck on simplifying them down to make it easy to divide. A little help?
+118608 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}\\~\\
+505 Oh I forgot to say that it should work out to be a real number. So I think there were a few errors.
+505 Ok I found out how to do it!
We have to convert the complex numbers into cos + i * sin!
It makes so much sense now lol
+118608 That would be an alternate way of doing it. My method (minus careless errors) also works.
+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
+118608 Just to prove my point, I have redone it without any errors.
\(\frac{(1-i)^{10}(\sqrt 3 + i)^5}{(-1-i\sqrt 3)^{10}}\\~\\ =\frac{(1-i)^{10}(\sqrt 3 + i)^5}{(1+i\sqrt 3)^{10}}\times \frac{(1-i\sqrt 3)^{10}}{(1-i\sqrt 3)^{10}}\\~\\ =\frac{[(1-i)(1-i\sqrt 3)]^{10}(\sqrt 3 + i)^5}{[(1+i\sqrt 3)(1-i\sqrt 3)]^{10}}\\~\\ =\frac{[[(1-\sqrt3)-(1+\sqrt3)i]^{2}]^5(\sqrt 3 + i)^5}{4^{10}}\\~\\ \text{no errors to here}\\~\\ =\frac{[-4\sqrt3+4i]^5(\sqrt 3 + i)^5}{4^{10}}\\~\\ =\frac{(-4)^5[\sqrt3-i]^5(\sqrt 3 + i)^5}{4^{10}}\\~\\ =\frac{-(4)^5[(\sqrt3-i)(\sqrt 3 + i)]^5}{4^{10}}\\~\\ =\frac{-[3+1]^5}{4^{5}}\\~\\ =-1 \)
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}}\\~\\
\text{no errors to here}\\~\\
=\frac{[-4\sqrt3+4i]^5(\sqrt 3 + i)^5}{4^{10}}\\~\\
=\frac{(-4)^5[\sqrt3-i]^5(\sqrt 3 + i)^5}{4^{10}}\\~\\
=\frac{-(4)^5[(\sqrt3-i)(\sqrt 3 + i)]^5}{4^{10}}\\~\\
=\frac{-[3+1]^5}{4^{5}}\\~\\
=-1
