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Simplifying complex numbers

  • #1

Homework Statement



Simplify [itex](1+i\sqrt{2})^5-(1-i\sqrt{2})^5[/itex]

Homework Equations



[tex]z=a+bi[/tex]

[tex]z=r(cos\varphi+isin\varphi)[/tex]

[tex]tg\varphi=\frac{b}{a}[/tex]

[tex]r=\sqrt{a^2+b^2}[/tex]

The Attempt at a Solution



[tex](\sqrt{3}(arccos\frac{\sqrt{3}}{3}+iarcsin\frac{\sqrt{6}}{3}))^5-(\sqrt{3}(arccos\frac{\sqrt{3}}{3}+iarcsin\frac{-\sqrt{6}}{3}))^5
[/tex]
How will I get integer angle out of here?

[tex]arccos\frac{\sqrt{3}}{3} \approx 54.74^\circ[/tex]

[tex]arcsin\frac{\sqrt{-6}}{3} \approx -54.74^\circ [/tex]
 

Answers and Replies

  • #3
Yes, I know de Moivre's theorem, but I don't know how will I get integer at the end...
 
  • #4
63
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Why do you need an integer at the end? Is this a part of the question that hasn't been specified? ???
 
  • #5
Ok.

[tex]\sqrt{3^5}(cos5*54.74^\circ+isin5*54.74^\circ)-\sqrt{3^5}(cos5*54.74^\circ-isin5*54.74^\circ)[/tex]

[tex]9\sqrt{3}(0.065-0.997i)-9\sqrt{3}(0.065+0.997i)=9\sqrt{3}(0.065-0.997i-0.065-0.997i)=9\sqrt{3}(-2*0.997i)=-i17.946\sqrt{3} \approx -31.08i[/tex]

And in my text book results: [itex]-22i\sqrt{2}[/itex], we both get same result, but the question is how they get integer numbers?
 
  • #6
Defennder
Homework Helper
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I can think of one possible way to get that: Expand the complex numbers out by binomial theorem and then simplify the expression. This would be very tedious, no doubt.
 
  • #7
Defennder, I know that I can solve it like that, but it is far more complicated, and the possibility that you may miss some value is very big...
 
  • #8
Hootenanny
Staff Emeritus
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I think it was be best to put the two complex numbers into exponential form for the powers and then convert back to Cartesian to perform the subtraction.
 
  • #9
tiny-tim
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Simplify [itex](1+i\sqrt{2})^5-(1-i\sqrt{2})^5[/itex]
Oh come on, guys! :rolleyes:

:smile: (a + b)^5 - (a - b)^5 = … ? :smile:
 
  • #10
86
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By binom formula?
 
  • #11
HallsofIvy
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Have you tried it?
 
  • #12
Defennder
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Well actually he's asking for a quicker way to get the answer in terms of radicals apart from the binomial theorem or de Moivre's theorem
 
  • #13
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Oh come on, guys! :rolleyes:

:smile: (a + b)^5 - (a - b)^5 = … ? :smile:
Tiny Tim is right to point this out. But the explanation for a general rule of a difference of this sort is in line, for learning purposes of course ;)

If you have an equation of the above kind e.g. (a+b)^n - (a-b)^n , an expansion shows that there will be n+1 terms for (a+b)^n and (a-b)^n. The difference of the two, however, eliminates all but the even terms: for these terms, the coefficients are doubled.
Let's look at an easier example, (a+b)^3 - (a-b)^3 .
Using the binomial theorem (by writing the coefficients as they would appear in Pascale's triangle and by ordering terms by increasing exponents of b and decreasing of a) we get

[a^3 + 3(a^2)(b)+3(a)(b^2) + b^2] - [a^3 - 3(a^2)(b)+3(a)(b^2) -b^3]
= 6 (a^2)(b)+ 2 b^3

As you can see, this is just twice the even-numbered terms in the expansion of (a+b)^3.
The case for (a+b)^5 is analogous.

This is a succinct way of arriving at the result.
 
  • #14
86
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Ahh... I understand now. So I should also use the binom formula, right?
 
  • #15
tiny-tim
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Ahh... I understand now. So I should also use the binom formula, right?
Theofilius , you keep answering questions with a question …
:smile: (a + b)^5 - (a - b)^5 = … ? :smile:
:smile: … and don't answer with a question … ! :smile:
 
  • #16
86
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[tex]-22i\sqrt{2}[/tex]. :smile: I know that, but I should have do that with De Moivre's formula.
 
  • #17
tiny-tim
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[tex]-22i\sqrt{2}[/tex]. :smile:.
erm … no.
… I know that, but I should have do that with De Moivre's formula
eh? … but this is Physicsissuef's question, not yours! :confused:

What makes you think he has to use de Moivre?
 
  • #18
86
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Since I have same problem in my book. And I solve it correctly, why you say no?
 
  • #19
Yes, since logically we need to solve this problem as simple as possible, but no problem.
 
  • #20
tiny-tim
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oops! it is -22i√2. Sorry! :redface:

If you must do it by de Moivre, just put (1 + i√2) = r(cosθ + i sinθ), but leave putting the numbers in until the end.

Then you want r^5[(cos5θ + i sin5θ) - (cos5θ - i sin5θ)], = 2 i r^5 sin5θ.

You know r = √3, and tanθ = √2, so it's fairly easy to work out from that what sin5θ is. :smile:

(But the binomial method is probably a more straightforward way of calculating sin5θ, sin 7θ, etc)
 

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