# Limits of Complex Functions ... Zill & Shanahan, Theorem 3.1.1/ A1

#### Peter

##### Well-known member
MHB Site Helper
I am reading the book: Complex Analysis: A First Course with Applications (Third Edition) by Dennis G. Zill and Patrick D. Shanahan ...

I need some help with an aspect of the proof of Theorem 3.1.1 (also named Theorem A1 and proved in Appendix 1) ...

The statement of Theorem 3.1.1 (A1) reads as follows: In the proof of Theorem 3.1.1 (A1) [see below] we read the following:

" ... ... On the other hand, with the identifications $$\displaystyle f(z) = u(x,y) + i v(x,y)$$ and $$\displaystyle L = u_0 + i v_0$$, the triangle inequality gives

$$\displaystyle \mid f(z) - L \mid \ \leq \ \mid u(x,y) - u_0 \mid + \mid v(x,y) - v_0 \mid$$ ... ... "

My question is as follows:

How exactly do we apply the triangle inequality to get $$\displaystyle \mid f(z) - L \mid \ \leq \ \mid u(x,y) - u_0 \mid + \mid v(x,y) - v_0 \mid$$ ... ... ?

Note: Zill and Shanahan give the triangle inequality as

$$\displaystyle \mid z_1 + z_2 \mid \ \leq \ \mid z_1 \mid + \mid z_2 \mid$$

My thoughts are as follows:

$$\displaystyle \mid f(z) - L \mid \ = \ \mid u(x,y) + i v(x,y) - (u_0 + i v_0 ) \mid \ = \ \mid ( u(x,y) - u_0 ) + i ( v(x,y) - v_0 ) \mid$$

so in triangle inequality put

$$\displaystyle z_1 = u(x,y) - u_0 + i.0$$

and

$$\displaystyle z_2 = 0 + i ( v(x,y) - v_0$$)

and apply triangle inequality ...

Is that correct?

Hope someone can help ...

Peter

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The statement and proof of Theorem 3.1.1 (given in Appendix 1 where the theorem is called Theorem A.1) reads as follows:  Hope that helps ...

Peter

Last edited:

#### steep

##### Member
I'm going to suggest 2 things to streamline this. The main question is: what if, instead of $\mathbb C$ this was $\mathbb R^2$?

First, in general this is a good questions to ask as $\mathbb C$ in some sense is $\mathbb R^2$ with some 'tiny refinements' (i.e. geometrically pleasant field operations). One of the main points of complex analysis is that these 'tiny refinements' add a huge amount of structure and while definitions formally look about the same e.g. for differentiability on an open set in $\mathbb R^2$ vs $\mathbb C$, all kinds of things can be differentiable for the former that aren't for the latter -- and as a result only 'nice' functions are called holomorphic. So it's always a good question to have in the back of your mind and to try to make these comparisons.

Second, check on the definition of the limit in $\mathbb C$ vs that in $\mathbb R^2$. Consider $\mathbf z \in \mathbb R^2$ and
$f: \mathbb R^2 \to \mathbb R^2$. You'll see that $u(x,y) = z_1$ and $i v(x,y) = z_2$, i.e. they are isolating real and imaginary parts just like in the vector space interpretation of $\mathbb C$. Now apply results from your most recent thread about convergence in $\mathbb R^d$ iff there is component-wise convergence, selecting $d = 2$.

#### frapps11

##### New member
Yeah this shall be helpful, as one of the main points of complex analysis is that these 'tiny refinements' add a huge amount of structure and while definitions formally look about the same e.g. for differentiability on an open set in ℝ2 vs ℂ, all kinds of things can be differentiable for the former that aren't for the latter -- and as a result only 'nice' functions are called holomorphic. So it's always a good question to have in the back of your mind and to try to make these comparisons.

#### Peter

##### Well-known member
MHB Site Helper
Yeah this shall be helpful, as one of the main points of complex analysis is that these 'tiny refinements' add a huge amount of structure and while definitions formally look about the same e.g. for differentiability on an open set in ℝ2 vs ℂ, all kinds of things can be differentiable for the former that aren't for the latter -- and as a result only 'nice' functions are called holomorphic. So it's always a good question to have in the back of your mind and to try to make these comparisons.

Thanks to steep and frapps11 for most helpful posts ...

I am still reflecting on what you have written ...

Thanks again!

Peter