I think you're on the right track. z2 + (1 + i)z + 2 - i factors into (z - i)(z + 1 + 2i). I came up with this by long division.MelissaJL said:Homework Statement
The lim(z->i) of [z^2+(1+i)z+2] using the epsilon-delta proof.
Homework Equations
z=x+iy
Triangle Inequality: |z+w|<or=|z|+|w|
The Attempt at a Solution
For every epsilon>0, there exists a delta>0 such that
|(z^2+(1+i)z+2)-(i)|<epsilon whenever 0<|z-i|<delta
I'm not sure how to change z^2+(1+i)z+2 into z-i. Or am I approaching the situation wrong?
Are you dividing ##z^2+(1+i)z+2## by ##z-i##? Mark divided ##z^2+(1+i)z+2-i## by ##z-i##.MelissaJL said:When I try the long division I get a remainder of i. Is that okay? I haven't ever done long division with complex numbers.
MelissaJL said:Homework Statement
The lim(z->i) of [z^2+(1+i)z+2] using the epsilon-delta proof.
Homework Equations
z=x+iy
Triangle Inequality: |z+w|<or=|z|+|w|
The Attempt at a Solution
For every epsilon>0, there exists a delta>0 such that
|(z^2+(1+i)z+2)-(i)|<epsilon whenever 0<|z-i|<delta
I'm not sure how to change z^2+(1+i)z+2 into z-i. Or am I approaching the situation wrong?
No. You should have (z - i) as one of the factors, not (z + i). For the other factor, you'll need to establish some bounds on it.MelissaJL said:Ahh I see what I did wrong with the long division. I definitely forgot that -i in there. So now I have:
|(z+i)(z+1+2i)|<epsilon for 0<|z-i|<delta
|z+i||z+1+2i|<epsilon but what can I do with z+1+2i to isolate z+1?
You want to establish the largest and smallest values it (|z + 1 + 2i|) can have.MelissaJL said:What do you mean by bounds on it?
That's no help because you would be trading one unknown, z, for two unknowns, a and b.MelissaJL said:trying the modulus wouldn't do anything for me right now since I still have that pesky z there in z+1+2i, can I expand z out as z=a+ib?
Something like that. vela's recommendation to review some similar limits with real-valued function is good advice.MelissaJL said:Oh like making an upper or lower bound on it by taking one of the factors to be less than 1 then plugging it back into the other inequality??
No, it doesn't work this way with complex numbers, since you can't compare complex numbers using < or >. For instance, if |z| < 1, all that means is that z is a complex number inside the unit circle with center at the origin.MelissaJL said:Oh okay, thank you. So if I want to make an upper bound I first want to take 0<|z+1+2i|<1 but by definition of absolute value it turns into -1<z+1+2i<1
MelissaJL said:and if I want to take this and make it z-i I have to add -i-1-2i to each side which gives -i-2-2i<z-i<-i-2i. So |z-i|<-i-2i
MelissaJL said:When I try the long division I get a remainder of i. Is that okay? I haven't ever done long division with complex numbers.
Ray, I don't see how this is helpful. Melissa already has the limit (which is i), and now she needs to prove that this is the limit using a delta-epsilon proof.Ray Vickson said:Why don't you try out the suggestion I made in Post #5?
Mark44 said:Ray, I don't see how this is helpful. Melissa already has the limit (which is i), and now she needs to prove that this is the limit using a delta-epsilon proof.
I think here (https://www.physicsforums.com/forums/math-learning-materials.178/ ) would be a good choice.MelissaJL said:If I were to make a new thread later on finding resources, what category should I put it under?
You can put a bound on the distance between complex numbers. You can't say z>w, but it's perfectly fine to say |z|>|w| because the modulus of a complex number is a real number.MelissaJL said:How do you put bounds on complex numbers then?
The definition of a limit is the value that a function approaches as the input approaches a given point. It is denoted by the notation lim(x->a) f(x) and is calculated by evaluating the function at values closer and closer to the given point.
To use the definition of a limit to find a complex limit, we first rewrite the given function in the form f(z) = u(x,y) + iv(x,y), where u and v are real-valued functions of x and y. Then, we evaluate the limit separately for the real and imaginary parts, and if both limits exist, the complex limit exists as well and is equal to the combination of the two real limits.
A real limit refers to the limit of a real-valued function as the input approaches a real number, while a complex limit refers to the limit of a complex-valued function as the input approaches a complex number. In other words, a real limit is a limit in one dimension, while a complex limit is a limit in two dimensions.
Yes, there are two special cases to consider when using the definition of a limit to find a complex limit. The first is the case where the real and imaginary parts of the function do not exist separately, but the limit of the function itself exists. The second is the case where the real and imaginary limits exist, but their combination does not exist.
Yes, the definition of a limit can be extended to find limits at infinity. In this case, the input approaches a large value instead of a specific point, and the limit is evaluated by seeing how the function behaves as the input gets larger and larger. This is known as a one-sided limit, where the input only approaches the given point from one direction.