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# Analysis questions watch

1. https://www8.imperial.ac.uk/content/...als/m1p1s5.pdf

I need help with 4.

For the "if" part this is what I did (which seems to be different to what is suggested):

Assume xn -> X and yn -> Y. Fix e > 0.

There exists N1 such that n > N1 implies |xn - X| < e/2
There exists N2 such that n > N1 implies |yn - Y| < e/2

Set N = max{N1, N2)

for n > N

|zn - Z| = |xn - X + i(yn - Y)| which is less than or equal to (by the triangle inequality)
|xn - X| + |i(yn - Y)| = |xn - X| + |(yn - Y)| = e/2 + e/2 = e
So we have |zn - Z| < e and we're done.

Is this proof ok?

How about the "only if" part? I can't prove the suggested inequality (trying the triangle inequality doesn't seem to help) and I don't see how it would help us.

On a side note, is this exactly the same as the vector sequence (xn, yn) converging to (X, Y)? In that case wouldn't it just follow by definition?
2. The above is fine for the if direction. For the only if, your best bet is probably to assume that but and derive a contradiction. Then you can either do the same with or use a symmetry argument of some sort.
3. (Original post by nuodai)
The above is fine for the if direction. For the only if, your best bet is probably to assume that but and derive a contradiction. Then you can either do the same with or use a symmetry argument of some sort.
Is that the method that's being hinted at?
4. Looking at the hints, you're not supposed to use epsilon-type proofs at all.
5. Any other ideas?
6. The hint pretty much gives you the answer for the first part; by the looks of it you don't even need to derive it, you can just quote that any linear combination of convergent sequences is convergent, and bam. For the other direction, you're given that and . Directly from this we have and and, trivially, . Why, then, must and converge if converges?
7. (Original post by nuodai)
The hint pretty much gives you the answer for the first part; by the looks of it you don't even need to derive it, you can just quote that any linear combination of convergent sequences is convergent, and bam. For the other direction, you're given that and . Directly from this we have and and, trivially, . Why, then, must and converge if converges?
How would I go about proving those inequalities (|z| greater than or equal to |x| and |z| greater than or equal to |y|)? I tried to use the triangle inequality but it didn't seem to help. Given those inequalities, I think that we could do something like

epsilon > |zn - Z| >= |xn - X|

and the same for yn.
8. (Original post by gangsta316)
How would I go about proving those inequalities (|z| greater than or equal to |x| and |z| greater than or equal to |y|)? I tried to use the triangle inequality but it didn't seem to help. Given those inequalities, I think that we could do something like

epsilon < |zn - Z| <= |xn - X|

and the same for yn.
So do that It seems alright to me (except you've got your inequality signs the wrong way round), since and so on.
9. (Original post by nuodai)
So do that It seems alright to me (except you've got your inequality signs the wrong way round), since and so on.
But how do I prove those inequalities?
10. (Original post by gangsta316)
But how do I prove those inequalities?
Well is a good starting point. If you can prove that and (where ) for general , then it must also be true for .

Spoiler:
Show
Remember that for
11. (Original post by nuodai)
Well is a good starting point. If you can prove that and (where ) for general , then it must also be true for .

Spoiler:
Show
Remember that for
I think that I managed to prove them.

I proved that |z|^2 > |x|^2 and |z|^2 > |y|^2. I'm pretty sure that we're allowed for those to imply |z| > |x| and |z| > |y|.

How would I do question 6? The logic is confusing me a little bit. I start with the fact that convergent <=> Cauchy for real sequences. Then from that do I need to prove both (separately) convergent => Cauchy for complex sequences and Cauchy => convergent for complex sequences?
12. (Original post by gangsta316)
How would I do question 6? The logic is confusing me a little bit. I start with the fact that convergent <=> Cauchy for real sequences. Then from that do I need to prove both (separately) convergent => Cauchy for complex sequences and Cauchy => convergent for complex sequences?
The logical statement being tested is "if convergent Cauchy for real sequences, then convergent Cauchy for complex sequences". You can proceed with this one in a number of ways. You already know that converges if and only if and converge... in fact just me saying that has almost done the question for you
13. I finished the sheet. The last question wasn't that hard.

https://www8.imperial.ac.uk/content/...1P1/m1p1s6.pdf

I need help on 5. This is what I have done

Let zn be a complex sequence. zn = xn + i yn where xn and yn are real sequences. zn is bounded implies (from that inequality earlier) that xn and yn are bounded. Therefore xn and yn both have convergent subsequences. Now the first thought I had was to just add these to subsequences with an i in front of yn's so we have a sequence of complex numbers. But the x and y subscripts have to match for each term or it's not a subsequence of zn. I mean, if the convergent subsequence of xn is x1, x3, x5 etc. and for yn it's y2, y4, y6 etc. then if we add them as above we get x1 + iy2, x3 + iy4, x5 + iy6 etc. This sequence should converge but it's not a subsequence of zn. So how do we get one that is a subsequence of zn? I mean, how can we be sure that xn and yn's subsequences have subscripts in common?
14. Did you read the whole question? There's a very strong hint there.
15. I still can't figure out this question (question 5 on sheet 6). I've talked to some of my peers but they just say that, because xn and yn have convergent subsequences, zn has one (where zn = xn + iyn is our complex sequence). But, as I said if the convergent subsequence of xn is x1, x3, x5 etc. and the convergent subsequence of yn is y2, y4, y6 etc. then the only supposed subsequence of zn is x1+iy2, x3+iy4, x5+iy6 etc. which is of course not a subsequence of zn. So we need to be sure that something like that never happens and that the subsequences of xn and yn have subscripts in common. But why would they in general?
16. By considering x_n, we see there's a subsequence (s_n) of (z_n) such that Re(s_n) converges.
But now consider Im(s_n): it's bounded, so we can find a subsequence (t_n) of (s_n) such that Im(t_n) converges.
Now since Re(t_n) is just a subsequence of Re(s_n), Re(t_n) converges as well.
And of course, as t_n is a subseq of s_n and s_n is a subseq of z_n, t_n is a subseq of z_n so we're done.

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