# Prove of complex numbers

1. May 20, 2008

### Theofilius

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

Hello!

a)Prove that $$(a+bi)^n$$ and $$(a-bi)^n$$, n $$\in \mathbb{N}$$ are conjugate complex numbers;

b)Prove that quotient of any two numbers from the set of $$\sqrt[n]{1}$$ is again number from the set of $$\sqrt[n]{1}$$

c)Prove that reciproca value of any number from the set of $$\sqrt[n]{1}$$, is again number from the set $$\sqrt[n]{1}$$

2. Relevant equations

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

$$\bar{z}=r(cos\alpha-isin\alpha)$$

$$w_k=\sqrt[n]{r}(cos\frac{\alpha + 2k\pi}{n}+isin{\alpha + 2k\pi}{n}) ; k=0,1,2,...,n-1$$

3. The attempt at a solution

a) $$(a+bi)=r(cos\alpha+isin\alpha)$$

$$(a+bi)^n=r^n(cosn\alpha+isinn\alpha)$$

$$(a-bi)=r(cos\alpha-sin\alpha)$$

$$(a-bi)^n=r^n(cosn\alpha-sinn\alpha)$$

Is this enough to prove that they are conjugate complex numbers?

b)$$E_k=\sqrt[n]{1}=cos\frac{2k\pi}{n}+isin{2k\pi}{n} , k=0,1,2,3,...,n-1$$

Should I make like this?

$$E_n_-_1=\sqrt[n]{1}=cos\frac{2(n-1)\pi}{n}+isin\frac{2(n-1)\pi}{n}$$

What should I do next?

c)If I know how to prove b) I will prove c)

In this case just $$(\sqrt[n]{1})^-^1=\sqrt[n]{1}$$, right?

2. May 20, 2008

### Defennder

Multiply eqns 2 and 4 and you're done.

For b,c you don't have to resort to any fanciful complex number manipulation. Suppose a complex number $$w = \sqrt[n]{1}$$ That implies $$w^n = 1$$ Similarly consider a different nth complex root of unity, say z. z^n = 1. Quotient = w/z. How can we tell if w/z is also an nth root of unity?

c)

Again suppose $$w = \sqrt[n]{1}$$. It's reciprocal is $$\frac{1}{w}$$. What is $$\left (\frac{1}{w}\right)^n$$

3. May 20, 2008

### Theofilius

But I need to prove like I wrote at b)

First, let me apologise, I have some typo error in b), it should look like:

$$E_k=\sqrt[n]{1}=cos\frac{2k\pi}{n}+isin\frac{2k\pi}{n} , k=0,1,2,3,...,n-1$$

In a) why I need to multiply them ?

4. May 20, 2008

### Defennder

For b,c, what kind of proof are you looking for? Is there something unacceptable about the suggestion I gave?

5. May 20, 2008

### Theofilius

I think your way of thinking is good, but probably I should get two general solutions from the set of solutions of $$\sqrt[n]{1}$$ and prove...

6. May 20, 2008

### Kurret

For b), switching to exponential form simplifies a lot.

7. May 20, 2008

### Theofilius

I need to prove with complex numbers like the $$E_k$$ formula above. Please help!

8. May 21, 2008

### Theofilius

Any help?

9. May 21, 2008

### HallsofIvy

Staff Emeritus
Suppose a and b are both "from the set $^n\sqrt{1}[/tex]. What is (ab)n? 10. May 22, 2008 ### Theofilius Maybe again $$(ab)^n$$ is from the set of $$\sqrt[n]{1}$$. I am not sure. 11. May 22, 2008 ### tiny-tim Hint: [itex](ab)^n$ = ababab … ababab = … ?

12. May 22, 2008

### Theofilius

$$a^nb^n$$ ?

13. May 22, 2008

### tiny-tim

Yes … but … why the question-mark?

Theofilius , you keep answering questions with a question … even when, as in this case, it's actually the complete answer.

ok … and $$a^nb^n$$ = … ?

… and don't answer with a question … !

14. May 22, 2008

### Theofilius

$$a^nb^n=$$? I honesly don't know.. I don't even know what are you giving me that question for?

15. May 22, 2008

### tiny-tim

Because of:

16. May 22, 2008

### Physicsissuef

$$(ab)^n$$ is complex or real number.

17. May 23, 2008

### Theofilius

I don't know, we are just going in circle.

18. May 23, 2008

### tiny-tim

So $a^n\,=\,b^n\,=\,1$

19. May 24, 2008

### Physicsissuef

Is that the proof?

20. May 24, 2008

### tiny-tim

Well, it needs to be put together into one continuous piece, and the actual question was about a/b rather than ab, but apart from that … yes!