How can I prove properties of complex numbers in my homework?

[Theofilius]

Homework Statement

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}

Homework 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

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?

 

[Defennder]

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?

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

 

[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 ?

 

[Defennder]

Well, okay I misread your answer for a). It's fine.

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

 

[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...

 

[Kurret]

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

 

[Theofilius]

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

 

[Theofilius]

Any help?

 

[HallsofIvy]

Suppose a and b are both "from the set ^n\sqrt{1}[/tex]. What is (ab)ⁿ?

 

[Theofilius]

Maybe again (ab)^n is from the set of \sqrt[n]{1}. I am not sure.

 

[tiny-tim]

Maybe again (ab)^n is from the set of \sqrt[n]{1}. I am not sure.

Hint: (ab)^n = ababab … ababab = … ?

 

[Theofilius]

a^nb^n ?

 

[tiny-tim]

a^nb^n ?

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 … ! ​

 

[Theofilius]

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

 

[tiny-tim]

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

Because of:

Suppose a and b are both "from the set ^n\sqrt{1}[/tex]. What is (ab)ⁿ?

 

[Physicsissuef]

(ab)^n is complex or real number.

 

[Theofilius]

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

 

[tiny-tim]

Suppose a and b are both "from the set ^n\sqrt{1}

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

 

[Physicsissuef]

Is that the proof?

 

[tiny-tim]

Is that the proof?

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!