# Simple calculus proof

1. Mar 13, 2006

### stunner5000pt

I was supposed to have learnt this in my first year but i seem to have forgotten it because i havent kept in touch with it

Definition:A sequence of complex numbers $\left{z_{n}\right}_{1}^{\infty}$ is said to have the limit Z0 or to converge to Zo and we write $$\lim_{n \rightarrow \infty} z_{n} = z_{0}$$ if for any epsilon>0 there exists an integer N such taht |Zn-Zo|< epsilon for all n>N
Using the given definition prove that the sequence of complex number
Zn = Xn + iYn converges to Zo = Xo + iYo iff Xn converges to Xo and Yn converges to Yo.
[Hint: |Xn - Xo|<=|Zn - Zo|
|Yn - Yo|<=|Zn-Zo|
|Zn - Zo|<=|Xn - Xo|+|Yn - Yo|

so we suppose the first part that
Zn = Xn + iYn converges to Zo = Xo + iYo then Xn converges to Xo and Yn converges to Yo.

well suppose it was triue then
$$|Z_{n} - Z_{0}| = |X_{n} - X_{0} + iY_{n} - iY_{0}| \leq |X_{n} - X_{0}| + i|Y_{n} - Y_{0}| < \delta_{1} + \delta_{2} < \epsilon$$

not sure how to choose the pepsilon... do i make it the min of delta1 and delta 2 or the max?

for the other way around i get that easily

2. Mar 14, 2006

### benorin

You should not have an "i" in your inequality

3. Mar 14, 2006

### benorin

The iff will mandate a two-fold proof: the "if" part, and the "only if" part.

Proof:

The "if" part: If Zn = Xn + iYn converges to Zo = Xo + iYo, then Xn converges to Xo and Yn converges to Yo.

Since Zn = Xn + iYn converges to Zo = Xo + iYo, we have

$$\mbox{For every } \epsilon >0,\mbox{ there exists a }N\in\mathbb{N}\mbox{ such that }n>N\Rightarrow |z_n-z_0|<\epsilon$$

4. Mar 14, 2006

### HallsofIvy

You do not "choose" epsilon. You have to show how to choose delta for any given epsilon.

5. Mar 14, 2006