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# Imaginary Numbers Precalc

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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?

Dec 17, 2021

#1
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The complex number works out to 4.

Dec 17, 2021
#2
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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}\\~\\

Dec 17, 2021
edited by Melody  Dec 18, 2021
#3
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Oh I forgot to say that it should work out to be a real number. So I think there were a few errors.

MobiusLoops  Dec 17, 2021
edited by MobiusLoops  Dec 17, 2021
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I expect there were a number of trivial errors.

However, there is nothing wrong with my logic.

I am quite sure you are able to use my logic and find the errors.

You asked for a little help.  You did not ask to be given a full copiable answer!

Melody  Dec 17, 2021
edited by Melody  Dec 17, 2021
#5
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You're right. I'm gonna try to do it lol :D

MobiusLoops  Dec 18, 2021
#6
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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

MobiusLoops  Dec 18, 2021
#7
+117443
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That would be an alternate way of doing it.  My method (minus careless errors) also works.

Melody  Dec 18, 2021
#8
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Well thanks Melody for pointing me in the right direction :)

MobiusLoops  Dec 18, 2021
#9
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Why don't you display your solution.

I expect it is an easier method.

Melody  Dec 18, 2021
#10
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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

MobiusLoops  Dec 18, 2021
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Thanks MobiusLoops

Melody  Dec 18, 2021
#12
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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

Melody  Dec 18, 2021
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Cool!

MobiusLoops  Dec 18, 2021