complex number question help!!! (1 Viewer)

cormglakes

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how do i do question ii:
1616673702900.png
and this question
1616710324655.png

ANS k = -4, 0
 
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CM_Tutor

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NOTE: The results presented here contain a mistake corrected below and an approach predicated on the question not being flawed... see later posts for details.

I got to the same point, but then ran into an issue.



Now, as must be real, the LHS must be real, and so we can conclude that:



This equation has no solution as the LHS is even and the RHS is odd.

However, this reasoning requires to be defined, which is untrue if .

In this case, .

Returning the original equation



it is clear that LHS is undefined for and leads to LHS = 0.

Thus, there appear to be no solutions.

However, if start from the original equation:



Let and
let .

We have now reduced the problem to two equations, and



This gives solutions and



This gives four more solutions and

However, none of these solutions is valid as all are real and no real can have an eighth power that is negative.


Am I making some silly mistake, or is this really an elaborate "there are no solutions" problem... and where does the second method shown here break down, as it is giving 8 real answers but all are invalid, and I don't see the flaw in the method that produced them?
 
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fan96

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Am I making some silly mistake, or is this really an elaborate "there are no solutions" problem... and where does the second method shown here break down, as it is giving 8 real answers but all are invalid, and I don't see the flaw in the method that produced them?
There appears to be a sign error here:
(The first attempt at a solution fails because it assumes all solutions to part ii) lie on the unit circle.)
 
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fan96

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One possible solution:

Define





Note that these functions are inverses when .

Let be the solutions to .

Now if



then we know that any solution satisfies for some .

Apply to both sides, so we get

Because we have an explicit formula for we're now done, but we can use part i) to get a nicer form for the solutions.

From part i), we know that for any on the unit circle.
(Geometrically, this means that sends the unit circle to the imaginary axis. In other words, all our solutions will be purely imaginary.)

From here, we just need to evaluate

The co-tangent function has periodicity , so take to obtain the full set of solutions.

Pedantic note: We show here that every solution corresponds to a for some .
Technically, we should also explain why every is a valid solution .
This is because there are exactly eight possible values each for and .
None of these values are or , so the function is a one-to-one mapping between these sets.

---

Fun fact: and are examples of Möbius transformations - transformations of the type



Such transformations in the complex plane send a line or circle to another line or circle.
 
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CM_Tutor

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There appears to be a sign error here:


(The first attempt at a solution fails because it assumes all solutions to part ii) lie on the unit circle.)
Thanks for pointing out the sign error, I will come back and post the corrected solution... we now have all the cases with leading to only non-real complex solutions, as was expected. :)
 

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Returning the original equation and correcting my earlier mistake, to find the actual 8 solutions of the equation:



Let and
let .

With both and being negative, we can be certain that are non-real.

Further, we can rewrite as:



and rewrite as:



We have now reduced the problem to two equations, and



Our eight solutions thus occur when



and so are:







 

CM_Tutor

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how do i do question ii:
View attachment 30443
This first question is flawed. Putting does yield the result . Applying it to solve the second equation, however, is only valid if the second part of the equation has restricted to the unit circle and thus having a modulus of 1. Looking at the eight solutions that I have found algebraically, none of them has and so none can be found using the result in part (i).

So, there is no "hence" solution possible here and this is a textbook mistake. Where does this come from, @cormglakes?
 

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Adapting from @fan96 's approach, we can demonstrate that my 8 solutions correspond to solutions in the form of and so can be used to prove results like:

 

fan96

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This first question is flawed. Putting does yield the result . Applying it to solve the second equation, however, is only valid if the second part of the equation has restricted to the unit circle and thus having a modulus of 1. Looking at the eight solutions that I have found algebraically, none of them has and so none can be found using the result in part (i).

So, there is no "hence" solution possible here and this is a textbook mistake. Where does this come from, @cormglakes?
In my solution, the for which the substitution occurs are the eighth roots of , not the actual solutions of ii). This is a valid substitution, just not the typical one that would be expected here.
 

CM_Tutor

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In my solution, the for which the substitution occurs are the eighth roots of , not the actual solutions of ii). This is a valid substitution, just not the typical one that would be expected here.
You are correct, of course, and your method is valid as those all satisfy .

I was explaining why the approach implied by the question that leads to



is an invalid approach because it supposes that the roots of



have the property that and they don't.
 

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If is a root of where then .

Since , let be the real number equal to , and so



Substituting this value for into the above equation yields:



Equating the real and imaginary parts gives two equations:



and



From (1), we see that which can be put into (2) to yield

Making the subject of (2) yields and thus:





So, there are two solutions, and .

@cormglakes, can you check if your answer is a typo?
 

cormglakes

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If is a root of where then .

Since , let be the real number equal to , and so



Substituting this value for into the above equation yields:



Equating the real and imaginary parts gives two equations:



and



From (1), we see that which can be put into (2) to yield

Making the subject of (2) yields and thus:





So, there are two solutions, and .

@cormglakes, can you check if your answer is a typo?
sorry yea the answer is -4,0
 

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