probably shouldnt put qed unless u are sure mate.I fixed the wording for my answer to part (i)
(ii)
Establish the fact that there will always exist some complex u |u|=1 , such that:
There are an infinitely many u such that this condition is true, for ANY z.
Now, if we look at the complex number:
For That complex number
is a 'region' of the WHOLE unit circle, that is, every and any complex number will satisfy if and only if r is irrational, this is the converse from the result proven in part (i)
Hence there must exist infinitely many cis nrpi such that the inequality holds true for any z.
Q.E.D
EDIT: Fixed the proof.
Well I was quite sure so I put it there.probably shouldnt put qed unless u are sure mate.
"cis(nr*pi) is a 'region' of the WHOLE unit circle, that is, every and any complex number will satisfy if and only if r is irrational, this is the converse from the result proven in part (i)"
its not true that every complex number on the circle will be cis(nr*pi) for some n whenever r is irrational if that is what you are trying to say. part (i) says nothing of this sort.
"Of these infinitely many values of u, cis npi r covers infinitely many of them for infinitely many n"Well I was quite sure so I put it there.
It is the converse argument from the first part.
The first part says that there are finitely many values if and only if r is rational
Conversely, there are infinitely man values if and only if r is irrational. This is the condition given in the question.
If there are infinitely many values for cis nr pi for all n=1,2,3.. and so on, for infinitely many values of n, cis n pi r will satisfy the every complex number of modulus 1.
There exists infinitely many values of u such that |u-z| < c for any c.
Of these infinitely many values of u, cis npi r covers infinitely many of them for infinitely many n. That should finish it shouldn't it?
cis npi r can cover the whole unit circle, since it satisfies infinitely many values for integers n."Of these infinitely many values of u, cis npi r covers infinitely many of them for infinitely many n"
no
infinitely many integers are even
infinitely many integers are odd
therefore there must be infinitely many integers that are both even and odd at the same time
qed
did you not read my example of why that logic doesn't work? you are doing exactly the same thing as in my obviously erroneous argument. think about it.cis npi r can cover the whole unit circle, since it satisfies infinitely many values for integers n.
That is, u= cis npi r for some irrational r and number n.
Now there are infinitely many u, such that |u-z| < c
So it must follow that there is an infinite number of cis n pi r which satisfy |cis npi r - z| < c
wrong. let theta be zero for example and it doesnt work.Here is something I have been working on:
If you put in theta=0 and a=1wrong. let theta be zero for example and it doesnt work.
heres a better way of rotating parabolas.
if we identify complex numbers and the points in the argand plane that represent them, the standard form parabola is given parametrically by:
z(t)=2at+iat^2.
to rotate by @ counterclockwise, multiply by cis(@).
gives (2atcos(@)-at^2sin(@))+i(2atsin(@)+at^2cos(@))
so x=2atcos(@)-at^2sin(@)
y=2atsin(@)+at^2cos(@)
for our rotated parabola.
so ycos(@)-xsin(@)=at^2=(2at)^2/4a=(xcos(@)+ysin(@))^2/4a.
finally leading to x^2cos^2(@)+y^2sin^2(@)+2xysin(@)cos(@)+4a(xsin(@)-ycos(@))=0
the correct cartesian equation.
lol convenient that you only checked situations with a=1, this is the only time your equation is correct.If you put in theta=0 and a=1
y^2=-4x
Which is correct because the focus is at (-1,0) and directrix at x=1....
It is also correct for theta = pi/2, pi, 3pi/2
Looking over my working for the third time, I found that when I should of multiplied by 1/a I instead multiplied by 1/a^2 which is why it only works for a=1lol convenient that you only checked situations with a=1, this is the only time your equation is correct.
no, that's still wrong.Looking over my working for the third time, I found that when I should of multiplied by 1/a I instead multiplied by 1/a^2 which is why it only works for a=1
The result of:
Should be the answer.
With the conditions above, create the variable point P(X,Y)no, that's still wrong.
1997 4U HSC Sydney boys? I remmeber doing this!!
Correct, it was a really good question1997 4U HSC Sydney boys? I remmeber doing this!!
Yeh man!Correct, it was a really good question