# Simplifying complex numbers

1. May 20, 2008

### Physicsissuef

1. The problem statement, all variables and given/known data

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

2. Relevant equations

$$z=a+bi$$

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

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

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

3. The attempt at a solution

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

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

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

2. May 20, 2008

### Defennder

3. May 20, 2008

### Physicsissuef

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

4. May 20, 2008

### BrendanH

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

5. May 20, 2008

### 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 text book results: $-22i\sqrt{2}$, we both get same result, but the question is how they get integer numbers?

6. May 21, 2008

### 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.

7. May 21, 2008

### 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...

8. May 21, 2008

### Hootenanny

Staff Emeritus
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. May 22, 2008

### tiny-tim

Oh come on, guys!

(a + b)^5 - (a - b)^5 = … ?

10. May 22, 2008

### Theofilius

By binom formula?

11. May 22, 2008

### HallsofIvy

Staff Emeritus
Have you tried it?

12. May 22, 2008

### 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

13. May 22, 2008

### BrendanH

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. May 22, 2008

### Theofilius

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

15. May 22, 2008

### tiny-tim

… and don't answer with a question … !

16. May 22, 2008

### Theofilius

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

17. May 22, 2008

### tiny-tim

erm … no.
eh? … but this is Physicsissuef's question, not yours!

What makes you think he has to use de Moivre?

18. May 22, 2008

### Theofilius

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

19. May 23, 2008

### Physicsissuef

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

20. May 23, 2008

### 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)