MHB Calculating the Value of (1+i)^56

[Petrus]

Hello MHB,

I got no clue about that, well I start wounder but I don't think so but is that same as
$$(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}})^{56}$$
then I could go to polar form and get
$$r=1$$ that means $$cos\theta=\frac{1}{\sqrt{2}} \ sin\theta=\frac{1}{\sqrt{2}}$$ that means we got $$(e^{i\frac{\pi}{4}})^{56} <=>e^{i14\pi} $$ so we got $$1^{56}(1+0i) =1$$
Is this correct?

Regards,
$$|\rangle$$

 

[MarkFL]

Re: polar form

You have correctly found the last term of the summation.

I would try writing the summation in trigonometric form using de Moivre's theorem...

 

[I like Serena]

Re: polar form

Hello MHB,

I got no clue about that, well I start wounder but I don't think so but is that same as
$$(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}})^{56}$$
then I could go to polar form and get
$$r=1$$ that means $$cos\theta=\frac{1}{\sqrt{2}} \ sin\theta=\frac{1}{\sqrt{2}}$$ that means we got $$(e^{i\frac{\pi}{4}})^{56} <=>e^{i14\pi} $$ so we got $$1^{56}(1+0i) =1$$
Is this correct?

Regards,
$$|\rangle$$

Well... you've only got the last term.
Which term do you get for k=0? And k=1? k=2? k=3?

 

[Petrus]

Re: polar form

You have correctly found the last term of the summation.

I would try writing the summation in trigonometric form using de Moivre's theorem...

Is this correct?
$$\cos(\frac{\frac{\pi}{4}}{56}+\frac{k2\pi}{56})+ \sin(\frac{\frac{\pi}{4}}{56}+i \frac{k2\pi}{56})$$ k=0,1,2,3..,56

Regards
$$|\rangle$$

 

[Nono713]

Re: polar form

Plotting the first few partial sums in the complex plane may provide some insight..

(it draws regular octagons.. why? what happens at the last partial sum?)

 

[alyafey22]

Re: polar form

You can use

$$\sum^{n}_{k=0} e^{ix k}= \frac{1-e^{(n+1)ix}}{1-e^{ix}}$$

 

[I like Serena]

Re: polar form

Is this correct?
$$\cos(\frac{\frac{\pi}{4}}{56}+\frac{k2\pi}{56})+ \sin(\frac{\frac{\pi}{4}}{56}+i \frac{k2\pi}{56})$$ k=0,1,2,3..,56

Regards
$$|\rangle$$

Nope. It's not. Both the $i$ and the $k$ are misplaced.

 

[Plato2]

Re: polar form

You can use

$$\sum^{n}_{k=0} e^{ix k}= \frac{1-e^{(n+1)ix}}{1-e^{ix}}$$

Your number is really $$\alpha = \exp \left( \frac{i\pi }{4} \right)$$.

Show that $$\sum\limits_{k = 0}^7 {{\alpha ^k}} = 0$$

This repeats in blocks of eight.

So only $$\alpha^0$$ is left.

 

[Petrus]

Re: polar form

Nope. It's not. Both the $i$ and the $k$ are misplaced.

Hm... I get $$\cos(\frac{\frac{\pi}{4}}{56}+\frac{k2\pi}{56})+i \sin(\frac{\frac{\pi}{4}}{56}+ \frac{ik2\pi}{56})$$ k=0,1,2,3..,56
What I am doing wrong?

Regards,
$$|\rangle$$

 

[I like Serena]

Re: polar form

Hm... I get $$\cos(\frac{\frac{\pi}{4}}{56}+\frac{k2\pi}{56})+i \sin(\frac{\frac{\pi}{4}}{56}+ \frac{ik2\pi}{56})$$ k=0,1,2,3..,56
What I am doing wrong?

Regards,
$$|\rangle$$

For each term you've got $$(e^{i\frac \pi 4})^k = e^{i \frac {k\pi} 4} = \cos \frac {k\pi} 4 + i \sin \frac {k\pi} 4$$
Somehow it just doesn't look the same.

 

[alyafey22]

Re: polar form

Your number is really $$\alpha = \exp \left( \frac{i\pi }{4} \right)$$.

Show that $$\sum\limits_{k = 1}^7 {{\alpha ^k}} = 0$$

This repeats in blocks of seven.

So only $$\alpha^0$$ is left.

but you should start from k=0 , to get 0 .

 

[Petrus]

Re: polar form

Hello MHB,
Thanks for all responed but I totaly got confused. I need to read over my book and see if I missed something, I totaly got confused and don't understand...$$|\rangle$$

 

[I like Serena]

Re: polar form

Hello MHB,
Thanks for all responed but I totaly got confused. I need to read over my book and see if I missed something, I totaly got confused and don't understand...$$|\pi\rangle$$

Okay...
What is it you got confused over? And that you don't understand...?

 

[Petrus]

Re: polar form

Okay...
What is it you got confused over? And that you don't understand...?

I just got diffrent tips. I just lost myself, in facit it says use the trick zaid said. $$\sum^{n}_{k=0} e^{ix k}= \frac{1-e^{(n+1)ix}}{1-e^{ix}}$$[/QUOTE]
but I can't find that in my book, I don't understand where it comes from

Regards,
$$|\rangle$$

 

[Nono713]

Re: polar form

It's just the geometric series partial sum formula, with $e^{ix}$ as a base.

$$\sum_{k=0}^n a^k = \frac{1 - a^{n+1}}{1 - a}$$

Do you know this formula?

 

[Petrus]

Re: polar form

It's just the geometric series partial sum formula, with $e^{ix}$ as a base.

$$\sum_{k=0}^n a^k = \frac{1 - a^{n+1}}{1 - a}$$

Do you know this formula?

Nop, I have not seen it. is this something famous formula?

Regards,
$$|\rangle$$

 

[Nono713]

Re: polar form

Well it's pretty useful. You can derive it easily:

$$\sum_{k = 0}^n a^k = a^0 + a^1 + a^2 + \cdots + a^n = S$$

$$aS = a^1 + a^2 + a^3 + \cdots + a^{n + 1} = S + a^{n + 1} - a^0$$

$$(a - 1)S = a^{n + 1} - 1$$

$$S = \frac{a^{n + 1} - 1}{a - 1} = \frac{1 - a^{n + 1}}{1 - a}$$

Now let $a = e^{ix}$ and you have the formula you were confused about. It still works for complex numbers :)

The sum to infinity is similar except in that case there is no $a^{n + 1}$ obviously so you just get $\frac{1}{1 - a}$ :p (only if $|a| < 1$ of course otherwise the sum does not converge)

 

[Petrus]

Re: polar form

Hello,
Thanks evryone! Now I understand!

Regards,
$$|\rangle$$

 

[MarkFL]

Re: polar form

Hey Petrus,

As others hinted at, this is what I wanted you to write:

$$S=\sum_{k=1}^{56}\left(\cos\left(\frac{k\pi}{4} \right)+i\sin\left(\frac{k\pi}{4} \right) \right)$$

and then look at:

$$S=7\sum_{k=1}^{8}\left(\cos\left(\frac{k\pi}{4} \right)+i\sin\left(\frac{k\pi}{4} \right) \right)=7\cdot0=0$$