Complex Numbers

I think it's n, not (n+1)

[MATH] B e^{-?\phi}= \frac{1-a^n \mathrm{e}^{? n\phi}}{1-a \mathrm{e}^{? \phi}}[/MATH]
Next step, can you make the denominator real?
Yes. What I thought. Since I figured in place of n, it will be n-1 as we took out one term common.
And how do I do the denominator real? And how come only denominator? Shouldn't I make numerator real as well? Should I like, substitute e^niΦ as cosnΦ + isinnΦ ?
 
I'm not sure what route @Subhotosh Khan was going to take with post#20, perhaps he has seen a quick way that I've missed.

I think it's n, not (n+1)

[MATH] B e^{-?\phi}= \frac{1-a^n \mathrm{e}^{? n\phi}}{1-a \mathrm{e}^{? \phi}}[/MATH]
Next step, can you make the denominator real?

Actually, before making the above denominator real, I would divide through by the \( e^{-?\phi} \) in order to isolate B on the LHS. This will change the denominator. Then you'll need to make the "new denominator" real.

And how do I do the denominator real? And how come only denominator? Shouldn't I make numerator real as well? Should I like, substitute e^niΦ as cosnΦ + isinnΦ ?

Do you know how to find the real part of the following if a,x,y are real?

[MATH]\Re{\left(\frac{a}{x+iy}\right)}[/MATH]
The word "conjugate" is a big clue!

You don't seem very familiar with complex numbers since you needed to click this spoiler. Please review your previous work on this topic. You won't learn anything if we need to suggest every step.
 
I'm not sure what route @Subhotosh Khan was going to take with post#20, perhaps he has seen a quick way that I've missed.

I was taking the same route (in different words) - you did not miss anything.
B * e^(-iΦ) = 1 + a*e^(iΦ) + {a*e^(iΦ)}^2 + {a*e^(iΦ)}^3 ......

B * e^(-iΦ) * [ 1 - a*e^(iΦ)] = 1 + a^n *e^(inΦ)

B * [e^(-iΦ) - 1] = [1 + a^n*cos(nΦ)] + i * a^n*sin(nΦ)

Equating real parts:

B * [cos(Φ) - 1] = [1 + a^n*cos(nΦ)] ......... and so on ..................................
Incorrect - follow Cubist's suggestion of "conjugate"
 
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I was taking the same route (in different words) - you did not miss anything.


B * e^(-iΦ) * [ 1 - a*e^(iΦ)] = 1 + a^n *e^(inΦ)

B * [e^(-iΦ) - 1] = [1 + a^n*cos(nΦ)] + i * a^n*sin(nΦ)

Equating real parts:

B * [cos(Φ) - 1] = [1 + a^n*cos(nΦ)] ......... and so on
I'm not sure what route @Subhotosh Khan was going to take with post#20, perhaps he has seen a quick way that I've missed.



Actually, before making the above denominator real, I would divide through by the \( e^{-?\phi} \) in order to isolate B on the LHS. This will change the denominator. Then you'll need to make the "new denominator" real.



Do you know how to find the real part of the following if a,x,y are real?

[MATH]\Re{\left(\frac{a}{x+iy}\right)}[/MATH]
The word "conjugate" is a big clue!

You don't seem very familiar with complex numbers since you needed to click this spoiler. Please review your previous work on this topic. You won't learn anything if we need to suggest every step.


Yea. That would be by multiplying numerator and denominator by (x-iy) . We then will get real part as ax / (x^2 + y^2)
And I'm asking steps just in case cause I don't want to get the answer wrong. It's already been days since I posted the question. If I had been taught this in college, I wouldn't be here asking questions now, would I? Thankyou for being patient with me anyway. I don't mean to come off rude, but this topic, I am learning it on my own. And this is the first time I'm asking for steps since I'm still getting hang of how to simplify the complex equations. That's why I came here in hopes of getting to know the steps so that I can study them and understand them. Thanks for your help.
 
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I was taking the same route (in different words) - you did not miss anything.


B * e^(-iΦ) * [ 1 - a*e^(iΦ)] = 1 + a^n *e^(inΦ)

B * [e^(-iΦ) - 1] = [1 + a^n*cos(nΦ)] + i * a^n*sin(nΦ)

Equating real parts:

B * [cos(Φ) - 1] = [1 + a^n*cos(nΦ)] ......... and so on ..................................
Incorrect - follow Cubist's suggestion of "conjugate"

Yea. Thanks.
 
Yea. That would be by multiplying numerator and denominator by (x-iy) . We then will get real part as ax / (x^2 + y^2)
And I'm asking steps just in case cause I don't want to get the answer wrong. It's already been days since I posted the question. If I had been taught this in college, I wouldn't be here asking questions now, would I? Thankyou for being patient with me anyway. I don't mean to come off rude, but this topic, I am learning it on my own. And this is the first time I'm asking for steps since I'm still getting hang of how to simplify the complex equations. That's why I came here in hopes of getting to know the steps so that I can study them and understand them. Thanks for your help.

Sorry, it's difficult to provide the right advice when I don't know your full situation. This question seems advanced and I get the impression that you have not (yet) mastered the knowledge that you need to solve it. This might not be your fault (missed time at school, the course that you're on perhaps assumes too much knowledge, etc). The thing is, I honestly don't think that it will help you if we just (mostly) solve this for you.

I would recommend that you spend some time doing some simpler work with complex numbers if you have the time. Try to master the less difficult questions first - IF time allows. Actually do some simpler questions rather than just reading about it.

I'm sure you can catch up. Please post back with your attempts and I, or others, will try to check it for you. You'll learn MUCH MORE by attempting it yourself, failing, showing us what you've tried, and then we can (hopefully) point you towards the solution and tell you why your attempt didn't work.

EDIT: Sorry that it's been days. I log in here fairly regularly because I enjoy helping out. It's volunteer work that I do during my "hobby time". The helpers here are much more inclined to help if students show what they have tried, and DON'T ask for a step by step solution (because this doesn't help you). If there's a question that you need to complete by a certain date then try to start it as early as you can, which obviously allows more time if you get stuck.
 
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Sorry, it's difficult to provide the right advice when I don't know your full situation. This question seems advanced and I get the impression that you have not (yet) mastered the knowledge that you need to solve it. This might not be your fault (missed time at school, the course that you're on perhaps assumes too much knowledge, etc). The thing is, I honestly don't think that it will help you if we just (mostly) solve this for you.

I would recommend that you spend some time doing some simpler work with complex numbers if you have the time. Try to master the less difficult questions first - IF time allows. Actually do some simpler questions rather than just reading about it.

I'm sure you can catch up. Please post back with your attempts and I, or others, will try to check it for you. You'll learn MUCH MORE by attempting it yourself, failing, showing us what you've tried, and then we can (hopefully) point you towards the solution and tell you why your attempt didn't work.

EDIT: Sorry that it's been days. I log in here fairly regularly because I enjoy helping out. It's volunteer work that I do during my "hobby time". The helpers here are much more inclined to help if students show what they have tried, and DON'T ask for a step by step solution (because this doesn't help you). If there's a question that you need to complete by a certain date then try to start it as early as you can, which obviously allows more time if you get stuck.


Yea. Thanks for all the help. I will try to find answer on my own. I'm gonna submit the same even if that's wrong. Thanks.
 
Just one last piece of advice then... you could ask your college if you can get free access to "Matlab". If they don't have this, then you could download "gnu Octave" for free (it's similar to Matlab). Then you can check your own work by putting some random values in to see if your work obtains the expected result (I find this easier than using a calculator)...

Rich (BB code):
phi=0.4
a=4
n=2

% Calculate the sum
s=0;
for k=1:n
  s += a^(k-1)*cos(phi*k);
end
s
s = 3.7079

% Try my own expression
(1 - a^n * e^(i*n*phi)) * e^(i*phi) / (1 - a*e^(i*phi))
ans =  3.7079 + 3.2588i
...so far so good :)
 
In fact I just found https://octave-online.net/ (click) so you don't need to install anything. Just copy and paste the commands above into that site! (Try with different values of n, a, etc, to be more sure of your final answer)
 
In fact I just found https://octave-online.net/ (click) so you don't need to install anything. Just copy and paste the commands above into that site! (Try with different values of n, a, etc, to be more sure of your final answer)
I don't need that value. I just need the expression in terms of cosϕ. Like, there was a similar question in which I had to find the sum of cosϕ + acos 2ϕ/(1!) + a ^ 2 cos 3ϕ/(2!) +… ∞

For that, I used the same method by considering sin series and adding them together to bring them in cosϕ + isinϕ format. And after simplifying and separating the real and imaginary part, I got e^acosϕ * cos[ϕ + asinϕ] as the answer. For that I substituted e^iϕ as cosϕ + isinϕ and got that answer as real part. That's why I'm asking if I should do it in similar way? And that simplification was easy but this one, I find it bit complicated that's why I was asking for step wise answer since I don't know if it's correct or not. Sorry if that bothered you or you think that it won't be helping me anyway. I won't bother anyone here anymore. Thanks for the help.
 
I think you have got as far as, the answer:

the real part of [MATH]\dfrac{e^{i \phi} (1 - a^n e^{i n \phi})}{1 - a e^{i \phi}}[/MATH]
What form do you want the final answer? I think if you write down an expression for the real part of this, it will be hideous.
 
...I find it bit complicated that's why I was asking for step wise answer since I don't know if it's correct or not. Sorry if that bothered you or you think that it won't be helping me anyway. I won't bother anyone here anymore. Thanks for the help.


Next time you should "bother" to read the posting guidelines. This is a math help forum not a provide the step wise answer forum.
 
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