cormglakes
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I got to the same point, but then ran into an issue.
There appears to be a sign error here: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?
(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 withThere 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.)
This first question is flawed. Puttinghow do i do question ii:
View attachment 30443
In my solution, theThis first question is flawed. Puttingdoes 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?
You are correct, of course, and your method is valid as those all satisfyIn my solution, thefor 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.
If
sorry yea the answer is -4,0Ifis a root of
where
then
.
Since, let
be the real number equal to
, and so
Substituting this value forinto the above equation yields:
Equating the real and imaginary parts gives two equations:
and
From (1), we see thatwhich can be put into (2) to yield
Makingthe subject of (2) yields
and thus:
So, there are two solutions,and
.
@cormglakes, can you check if youranswer is a typo?