# Simplifying complex numbers

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

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

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

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

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?

Homework Helper
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...

Staff Emeritus
Gold Member
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.

Homework Helper
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?

Homework Helper
Have you tried it?

Homework Helper
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?

Homework Helper
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.

Homework Helper
$$-22i\sqrt{2}$$. .

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.