How can complex numbers be simplified?

[Physicsissuef]

Homework Statement

Simplify (1+i\sqrt{2})^5-(1-i\sqrt{2})^5

Homework Equations

z=a+bi

z=r(cos\varphi+isin\varphi)

tg\varphi=\frac{b}{a}

r=\sqrt{a^2+b^2}

The Attempt at a Solution

(\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<br />
How will I get integer angle out of here?

arccos\frac{\sqrt{3}}{3} \approx 54.74^\circ

arcsin\frac{\sqrt{-6}}{3} \approx -54.74^\circ

 

[Defennder]

Use de Moivre's theorem:

http://en.wikipedia.org/wiki/De_Moivre's_formula

This makes it a lot easier.

 

[Physicsissuef]

Yes, I know de Moivre's theorem, but I don't know how will I get integer at the end...

 

[BrendanH]

Why do you need an integer at the end? Is this a part of the question that hasn't been specified? ?

 

[Physicsissuef]

Ok.

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

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

And in my textbook results: -22i\sqrt{2}, we both get same result, but the question is how they get integer numbers?

 

[Defennder]

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.

 

[Physicsissuef]

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...

 

[Hootenanny]

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.

 

[tiny-tim]

Simplify (1+i\sqrt{2})^5-(1-i\sqrt{2})^5

Oh come on, guys!

(a + b)^5 - (a - b)^5 = … ? ​

 

[Theofilius]

By binom formula?

 

[HallsofIvy]

Have you tried it?

 

[Defennder]

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

 

[BrendanH]

Oh come on, guys!

(a + b)^5 - (a - b)^5 = … ? ​

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.

 

[Theofilius]

Ahh... I understand now. So I should also use the binom formula, right?

 

[tiny-tim]

Ahh... I understand now. So I should also use the binom formula, right?

Theofilius , you keep answering questions with a question …

(a + b)^5 - (a - b)^5 = … ? ​

… and don't answer with a question … ! ​

 

[Theofilius]

-22i\sqrt{2}. I know that, but I should have do that with De Moivre's formula.

 

[tiny-tim]

-22i\sqrt{2}. .

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!

What makes you think he has to use de Moivre?

 

[Theofilius]

Since I have same problem in my book. And I solve it correctly, why you say no?

 

[Physicsissuef]

Yes, since logically we need to solve this problem as simple as possible, but no problem.

 

[tiny-tim]

oops! it is -22i√2. Sorry!

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.

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