Math Question...(Epsilon-Delta definition of a limit)

[eLiu]

Note: I posted this in OT too...but then it dawned on me that I might get more responses here, lol

Hey guys...I know there are some math-majors on here, so hopefully someone can give me a push in the right direction. Note: this is not for homework--I plan on taking this class (Complex Analysis) in the fall, and I'm trying to learn some of the material ahead of time to make it easier.

So...the problem... (E denotes epsilon, D delta)
For anyone that may need a reminder...what we're trying to show is the limit of f(z) = w0 as z->z0 by demonstrating that for any E>0 there exists a positive D such that:
|f(z) - w0| < E whenver 0 < |z-z0| < D.

The limit i'm trying to prove: limit of 1/z = i as z->-i.
So that gives me: |1/z - i| < E whenever |z+i|<D.

Then I know I need to try and get the |1/z-i| in terms of |z+i|...but that's where my hang-up is.

Through some manipulations...I've gotten that |1/z-i| = |z+i|/|z|...but there's still that annoying z term.

If I change it to |z+i|/|z+i-i|, I can do the triangle inequality in reverse and get:

|z+i|/|z+i-i| <= |z+i|/(|z+i|-1).

And there...I'm stuck. Trying to say D=E/k and substitute that into the above doesn't work out too good. Dividing top/bottom by |z+i| gets me something in the form of 1/1-x...but the power-expansion doesn't converge >.<

So with that, I've just about run out of ideas... Do I even have the right approach?? I don't want the whole proof...just a hint.

Thanks,
-Eric

 

[bonkers325]

this looks very much like the Calculus I definition of limits. not much i can say about it, since i hate limits and haven't taken a complex analysis course.

 

[eLiu]

Well, it's more of a generalized form of the Calculus-I definition. Those functions work in single variables, so you can think of it as converging on a line--like, when x->0, it approaches from the left and the right of the real line.

In this situation, you have to think about it as 2 planes--one for the domain and one for the range. That is, complex functions are written w=f(z). Just as z decomposes into real/imaginary parts (z=x+iy), w does also: w=u(x,y)+i*v(x,y). Thus, a complex function is essentially 2 real-valued functions of 2 variables.

So...2 real functions, 2 variables results in a 4D graph. Not very happy eh? So forget thinking about what it looks like...but you can imagine some of the properties by making 2 graphs--one for domain, one for range. The domain graph would be the traditional x-y graph showing the region over which f(z) exists. The range graph is u-v showing a map of the domain onto f(z).

What that all means for the limit is that you're now convering from all sides (a circle) instead of just 2.

 

[Matthew Daws]

Well, I nearly have a PHD in Analysis, so I should be able to help I guess...

You want, for E>0, to find D>0 so that |z+i|<D means |1/z-i|<E.

So, you have |1/z-i| = |1-iz|/|z| = |z+i|/|z| < D/|z| but as you say, how to get rid of the |z| term?

Well, I think a good hint might be the triangle-inequality. Intuition says that if |z+i| is very small, then z is roughly -i, so that |z| is roughly 1. The triangle inequality is |a+b| <= |a|+|b| for any complex numbers a and b.

So, another hint: you need to show that D/|z| < E, which is if and only if D < |z|E. But |z| can vary.

So, use the triangle-inequality to show that if |z+i|<D (where D is small) then |z| cannot be too small. Then |z|E is large(i.e. never gets much smaller than E) so you can pick a suitable D which will work.

I can explain more if you want more of a hint!

--Matt

 

[eLiu]

Hey Matthew Daws,
Uhh I'm not sure I'm understanding you 100%. (Sorry I've never done this stuff before...doh.) Anyway uh I included a link to my thread in Off-Topic: Link. There I posted a proof for lim z->i of x^2 = -1. And so far I've basically been trying to re-apply that...so if you want to look there and see if I have a logic flaw with the current problem...*shrug*

But yeah, I'm not quite sure I got this part fully:

So, use the triangle-inequality to show that if |z+i|<D (where D is small) then |z| cannot be too small. Then |z|E is large(i.e. never gets much smaller than E) so you can pick a suitable D which will work.

 

[Matthew Daws]

ELiu,

Right, well, your proof that z^2->-1 as z->-i seems correct, if not 100% rigourous (sorry, I'm a bit of a maths stickler ).

So, fix E>0. Then |1/z-i| < D/|z| and we want |1/z-i|<E. So, we basically want D/|z|<E, remembering that |z+i|<D.

Suppose we can prove that 1/|z| < 2 say. Then D/|z| = D*(1/|z|) < D*2 so D=E/2 will certainly work. How can we show that 1/|z|<2? Well, we need |z| > 1/2. You should be able to use the triangle-inequality and the fact that |z+i|<D to show this (as long as D is sufficiently small).

I'm not sure if I can hint any harder without giving the answer away. Maybe some other people here can give a better hint?

Cheers, --Matt

 

[eLiu]

Oh lol forgive me for the lack of rigor--i'm actually a senior in high school...math is more of a hobby that I'll be turnin into a major soon. Either way I've never had any formal analysis classes so all the conventions of rigorousness are kind of foreign

But yes...I dunno, I'm feeling dumb now b/c I'm still not seeing the key to this problem >.< Ah well, I'll come back to it tomorrow in school or something...hehe

Thanks for the help so far,
-Eric

 

[eLiu]

well, I did come up with this...
starting from |(z+i)/(z+i-i)|

Then...|z+i|*|1/(-i+(z+i))|...we'll just look at the 2nd part of that for now:

multiplying top/bottom by i: |i/(1+i(z+i))| = |1/(1+i(z+i))|

Power series expansion of 1/(1+x): 1+x+x^2+x^3+...

So: |z+i||1/(1+i(z+i))|= |z+i||1-i(z+i)+i^2(z+i)^2....|

Then, since the i^2 terms make this not-exactly an alternating series, use the general triangle inequality:
|z+i||1-i(z+i)+i^2(z+i)^2....| <= |z+i|(1+|i(z+i)|+|i(z+i)|^2....) = (|z+i|+|i(z+i)|^2+|i(z+i)|^3+...)

So yes, the ^2, ^3, etc terms vanish when compared to |z+i|, so the above "reduces" to |z+i|

So now, setting D=E, we have:
|1/z - i| <= |z+i| < E

I guess that's not fully rigourous and all...but I think it makes sense? What was your way of doing it?

 

[Matthew Daws]

Yep, seems to be the right idea.

My way was as follows. Given E>0 we want to find D>0 so that if |z+i|<D then |1/z-i| < E.

Now, |1/z-i| = |z-iz|/|z| = |z+i|/|z|.

The idea is that if z is close to -i then z cannot be close to 0, so that |z| is large. We can prove this by doing something a bit weird with the triangle-inequality, namely note that i=(z+i)-z so that 1=|i| = |(z+i)-z| <= |z+i| + |z| <= D+|z| so that |z| >= 1-D. So assume D<1/2, then 1-D>1/2 so |z|>1/2, so 1/|z|<2.

Thus |1/z-i| = |z+i|/|z| < 2|z+i| = 2D so let D=E/2 or 1/2, whichever is smaller.

A bit like black-magic maybe, but I hope you get the idea of what I've done.

Cheers, --Matt

Edit: Spelling...

 

[eLiu]

Oh that's cool! Yep it makes sense, and thanks for the help Matthew Daws.