Complex number very confused

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Loki123

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I seem to be confused when it comes to understanding how they solved the following problem. I don't understand what happenes after e comes in, and how it comes in. I would prefer an explanation for this way, however, if you have easier or better methods tell me that too. Thanks in advance.
IMG_20220312_175219.jpg
I don't have the original problem. These are notes I was given.
 
I simply do not follow your workings. This file may help you.
 

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Loki123 said:
I seem to be confused when it comes to understanding how they solved the following problem. I don't understand what happenes after e comes in, and how it comes in. I would prefer an explanation for this way, however, if you have easier or better methods tell me that too. Thanks in advance.
View attachment 31633
I don't have the original problem. These are notes I was given.
Click to expand...
Do you not recognize that w is any of the fourth roots of 1, in exponential form? (I don't know why they omit k=0.)

I think they have an extra i on the bottom line; when I solve (z+1)/(z-1) = w, I get z = (w+1)/(w-1), without the i.

On the rest of the bottom line, they are using some identities related to [imath]e^{ix}=\cos(x)+i\sin(x)[/imath].

I don't know whether I've covered the points you are unsure of. Please be specific about what you need help with.
 
Dr.Peterson said:
Do you not recognize that w is any of the fourth roots of 1, in exponential form? (I don't know why they omit k=0.)

I think they have an extra i on the bottom line; when I solve (z+1)/(z-1) = w, I get z = (w+1)/(w-1), without the i.

On the rest of the bottom line, they are using some identities related to [imath]e^{ix}=\cos(x)+i\sin(x)[/imath].

I don't know whether I've covered the points you are unsure of. Please be specific about what you need help with.
Click to expand...
I don't understand where e comes from. I have not been given a formula with it. It was not in any lecture.
 
Loki123 said:
I don't understand where e comes from. I have not been given a formula with it. It was not in any lecture.
Click to expand...
Loki123 said:
Is e here any number, or 2,7...?
what is e^{ix}=\cos(x)+i\sin(x)eix=cos(x)+isin(x)?
Click to expand...
Clearly you have not learned about the exponential form of complex numbers, so you can't be expected to understand this solution at all. I assume this solution is not from any class you have taken.

Yes, e is the natural exponential base, 2.71828... .

Here is a lesson about exponential form, which may help depending on what you already know:

That doesn't justify Euler's formula at all, but just offers it, in effect, as a new notation. For a deeper explanation, see section 6 here, which doesn't give an actual proof, but reasons to accept it:
 
@Loki123
Assuming you have read the notes recommended and know that [imath]e^{i\theta}=\cos \theta + i \sin \theta[/imath]
then here is an expanded version of your solution.
(Note there were a couple of errors (1 should be i) in the second-last line of your original notes).
z.jpg
 
An alternative proof:
[imath]\left(\dfrac{z+i}{z-i}\right)^4=1[/imath]
[imath]\Rightarrow (z+i)^4=(z-i)^4[/imath]
Either do difference of 2 squares twice or expand brackets using Binomial Theorem, and simplify to:
[imath]z^3-z=0[/imath]
[imath]z(z^2-1)=0[/imath]
[imath]z=0, z=1, z=-1[/imath]
 
lex said:
@Loki123
Assuming you have read the notes recommended and know that [imath]e^{i\theta}=\cos \theta + i \sin \theta[/imath]
then here is an expanded version of your solution.
(Note there were a couple of errors (1 should be i) in the second-last line of your original notes).
View attachment 31637
Click to expand...
Thank you for this, it helped A LOT. The only part I am confused about is why k = 1,2,3
 
@Loki123
I'm glad.
If you follow as far as [imath]w=e^{k\pi i/2}[/imath] for all integer values k, i.e. 0, 1, -1, 2, -2...
Substituting these values in, we find that the values of w just repeat after k=0, 1, 2, 3
So we simply say [imath]w=e^{k\pi i/2}[/imath] for k = 0, 1, 2, 3

Now in this particular question we know that w≠1, so k=0 does not give a solution to this problem.
So [imath]w=e^{k\pi i/2}[/imath] for k = 1, 2, 3

(I'm not sure if this was your problem).
 
lex said:
@Loki123
I'm glad.
If you follow as far as [imath]w=e^{k\pi i/2}[/imath] for all integer values k, i.e. 0, 1, -1, 2, -2...
Substituting these values in, we find that the values of w just repeat after k=0, 1, 2, 3
So we simply say [imath]w=e^{k\pi i/2}[/imath] for k = 0, 1, 2, 3

Now in this particular question we know that w≠1, so k=0 does not give a solution to this problem.
So [imath]w=e^{k\pi i/2}[/imath] for k = 1, 2, 3

(I'm not sure if this was your problem).
Click to expand...
Okay now I am little more confused. I get why z can't be 1. But why can't w be one? that is why can't z+1/z-2 be one??
 
z can be 1. It is z-i on the bottom line (not z-1 a little error in your original notes).
w cannot be 1 because then (z+i)/(z-i)=1 which you can see is not possible.
 
lex said:
z can be 1. It is z-i on the bottom line (not z-1 a little error in your original notes).
w cannot be 1 because then (z+i)/(z-i)=1 which you can see is not possible.
Click to expand...
If w=1, then z+i=z-i which would be 0=2i. I get that's not possible. But how did you determine it was not possible without doing this? Like If I got this on a test, how should my process of thinking go so I realize that w can't be 1.
For example If I see 2/x-1 on a test, my mind goes, x can't be 1 because then we get 2/0 which is not possible. How do I do that with this? How do I know that it exists for other numbers, 2,3,4...??
 
Loki123 said:
If w=1, then z+i=z-i which would be 0=2i. I get that's not possible. But how did you determine it was not possible without doing this? Like If I got this on a test, how should my process of thinking go so I realize that w can't be 1.
For example If I see 2/x-1 on a test, my mind goes, x can't be 1 because then we get 2/0 which is not possible. How do I do that with this? How do I know that it exists for other numbers, 2,3,4...??
Click to expand...
You don't really have to worry about it! If you don't eliminate k=0 at the start, then at the end you should observe that k=0 leads to an undefined cotangent, so you'll eliminate it then.

The benefit of seeing it at the start (as I didn't!) is that you're safe if you forget to check at the end. The same is true if an equation contains 2/(x-1); you can often ignore the implied restriction in your work, as long as you check your final answer if it turns out to be x=1.
 
Dr.Peterson said:
You don't really have to worry about it! If you don't eliminate k=0 at the start, then at the end you should observe that k=0 leads to an undefined cotangent, so you'll eliminate it then.

The benefit of seeing it at the start (as I didn't!) is that you're safe if you forget to check at the end. The same is true if an equation contains 2/(x-1); you can often ignore the implied restriction in your work, as long as you check your final answer if it turns out to be x=1.
Click to expand...
so should i check for k=0, k=1,k=2 and k=3?
 
Loki123 said:
so should i check for k=0, k=1,k=2 and k=3?
Click to expand...
At least check that each answer makes sense (i.e. is defined), if you don't fully check that they satisfy the equation, which would be wise if you have time on a test.
 
I echo what Dr.Peterson says. You can plough on and at the end check your candidate solutions.
In this question, you would probably have spotted that w≠1 before reaching the end.
At the point where we got:

[imath]z(w-1)=i(w+1)[/imath]

[imath]w=1[/imath], leads to the contradiction [imath]0=2i[/imath]

and so [imath]z=\dfrac{i(w+1)}{w-1}[/imath] and [imath]w≠1[/imath]
(which is probably where you would have spotted the problem, in the normal fashion).
 
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