Prove (by definition) that $\lim _{z\to 1+i}\left(\frac{1}{z^2+1}\right)=\frac{1}{2i+1}$

Question

How can I take this expression $$\left(\frac{1}{z^2+1}\right)-\frac{1}{2i+1}$$ to the form $z-(1+i)$?

I must use the definition of limit to prove this result. That is, use $\delta$ and $\epsilon$.

What I have done so far is to raise the definition: $$\lim _{z\to 1+i}\left(\frac{1}{z^2+1}\right)=\frac{1}{2i+1}\Leftrightarrow[\forall\epsilon>0,\exists\delta>0/\forall z\in\mathbb{C}:(0<|z-(1+i)|<\delta\Rightarrow|\left(\frac{1}{z^2+1}\right)-\frac{1}{2i+1}|<\epsilon)]$$ I must prove: $$\left|\left(\frac{1}{z^2+1}\right)-\frac{1}{2i+1}\right|<\epsilon$$ as long as $0<|z-(1+i)|<\delta$ We start from \begin{align} \left|\left(\frac{1}{z^2+1}\right)-\frac{1}{2i+1}\right|&=\left|\frac{1+2i-(z^2+1)}{(z^2+1)(2i+1)}\right|\\ &=\left|\frac{2i-z^2}{(z^2+1)(2i+1)}\right|\\ &=\left|\frac{(2i-z^2)(1-2i)}{(z^2+1)(2i+1)(1-2i)}\right|\\ &=\left|\frac{2i+4+i(2z^2+2)}{5(z^2+1)}\right|. \end{align} But I still don't get the expression I need.

Answer 1

The following response would be significantly shorter if you were allowed to use the Complex Analysis Theorem that
if $~\lim_{z \to z_0} a(z) = L \neq 0$
then $\lim_{z \to z_0} \frac{1}{a(z)} = \frac{1}{L}.$

I am assuming that the above theorem is out of bounds. Further, attempting to simply derive the above theorem (from scratch), would first require that some Real Analysis theorems be derived.

So, I will go for Kludginess.

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This problem requires a backward and offbeat approach. A straightforward conversion of $z$ to $x + iy$ will be very messy. Fortunately, there is a middle ground.

Preliminary Concepts

(1)
Given $~r \in \Bbb{R}, ~s \in \Bbb{R^+},$
the constraint that $|r| < s$
is equivalent to the constraint that
$-s < r < s.$

(2)
Suppose that $u,v \in \Bbb{C}$
and that you want to establish that
$|u - v| < \epsilon$.
This may be established by showing that
$|\text{Re}(u - v)| < \frac{\epsilon}{2}~~~$ and $~~~|\text{Im}(u - v)| < \frac{\epsilon}{2}.$
This is based on the idea that
$|\sqrt{\left[\frac{1}{2}\right]^2 + \left[\frac{1}{2}\right]^2} < 1.$

(3)
Suppose that $u,v \in \Bbb{C}$
and you are given that $0 < |u - v| < \delta.$
This implies that
$|\text{Re}(u - v)| < \delta~~~$ and $~~~|\text{Im}(u - v)| < \delta.$
The justification for this is very similar to that applied in (2) above.

(4)
In this problem, rather than identifying a relationship between $\delta$ and $\epsilon$ such that $\delta = g(\epsilon)$, I will make things easy on myself by also requiring that $\delta \leq \frac{1}{10}$.

Thus, the relationship will look like $\delta = \min\left[\frac{1}{10}, g(\epsilon)\right].$

(5)
As suggested at the start of my answer, I am going to start with the requirement that

$\displaystyle \left| ~\frac{1}{z^2 + 1} - \frac{1}{2i + 1} ~\right| < \epsilon.$
Then, I will see where this leads in terms of the necessary constraint on $\delta$, which will always be required to be $\leq \frac{1}{10}.$

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Assume that $z = (x + iy)$.
Let $w = (z^2 + 1)$ and let $\overline{w}$ represent the complex conjugate of $w$.

Let $f(z) = \displaystyle \left( ~\frac{\overline{w}}{|w|^2} - \frac{1 - 2i}{5} ~\right).$

It is desired that $|f(z)| < \epsilon.$
Based on (2) above, I will achieve this, by requiring that
$| ~\text{Re}[f(z)] ~| < \frac{\epsilon}{2}~~~$ and $~~~| ~\text{Im}[f(z)] ~| < \frac{\epsilon}{2}.$
Using (1) above, this will be done by requiring that

$$-\frac{\epsilon}{2} < \text{Re}[f(z)] < \frac{\epsilon}{2} ~~~\text{and} ~~~ -\frac{\epsilon}{2} < \text{Im}[f(z)] < \frac{\epsilon}{2}. \tag{A} $$

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The challenge is to specify $\delta$ so that

$$ 0 < \left| ~(x + iy) - (1 + i) ~\right| < \delta \leq \frac{1}{10} \tag{B} $$

implies that the constraints in (A) above are satisfied.

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$w = (z^2 + 1) = (x^2 - y^2 + 1) + i(2xy).$
This implies that
$\overline{w} = (x^2 - y^2 + 1) - i(2xy)~~~$ and $~~~|w|^2 = (x^2 - y^2 + 1)^2 + (4x^2y^2).$

Using (1) and (3) above, (B) above implies that

$$1 - \delta < x < 1 + \delta ~~~\text{and} ~~~ 1 - \delta < y < 1 + \delta. \tag{C} $$

Since $~\delta \leq \frac{1}{10},~$ you have that
$(1 - \delta) > 0~~~$ and $~~~\delta^2 < \delta.$

Therefore,

$$ 1 - 2\delta < x^2 < 1 + 3\delta ~~~\text{and} ~~~ 1 - 2\delta < y^2 < 1 + 3\delta. \tag{D} $$

Consequently lower and upper bounds for $(x^2 - y^2 + 1)$ are

$$(1 - 5\delta) = (1 - 2\delta) - (1 + 3\delta) + 1 \leq (x^2 - y^2 + 1) \\ \leq [\text{similarly}] ~(1 + 5\delta). \tag{E} $$

Similarly, lower and upper bounds for $(-2xy)$ are

$$(-2 - 6\delta) < (-2)(1 + \delta)^2 < (-2xy) \\ < (-2)(1 - \delta)^2 < (-2 + 4\delta). \tag{F} $$

Using (E) and (F), you have that

$$(1 - 5\delta) < \text{Re}(\overline{w}) < (1 + 5\delta) ~~~\text{and} \\ (-2 - 6\delta) < \text{Im}(\overline{w}) < (-2 + 4\delta). \tag{G}$$

Using (G) above, lower and upper bounds for $|w|^2$ are

$$(5 - 26\delta) < (1 - 5\delta)^2 + (-2 + 4\delta)^2 \leq |w|^2 \\ \leq (1 + 5\delta)^2 + (-2 - 6\delta)^2 < (5 + 95\delta). \tag{H}$$

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Now, using (G), (H) above, lower and upper bounds for $~$ Re$\left(\frac{\overline{w}}{|w|^2}\right)~$ and $~$ Im$\left(\frac{\overline{w}}{|w|^2}\right)~$ may be identified.

A lower bound for $~$ Re$\left(\frac{\overline{w}}{|w|^2}\right)~$ is

$$\frac{1 - 5\delta}{5 + 95\delta} = \frac{1 + 19\delta}{5 + 95\delta} - \frac{24\delta}{5 + 95\delta} > \frac{1}{5} - \frac{24}{5}\delta > \frac{1}{5} - 5\delta.\tag{I} $$

An upper bound for $~$ Re$\left(\frac{\overline{w}}{|w|^2}\right)~$ is

$$\frac{1 + 5\delta}{5 - 26\delta} = \frac{1 - \frac{26}{5}\delta}{5 - 26\delta} + \frac{\frac{51}{5}\delta}{5 - 26\delta} < \frac{1}{5} + \frac{51}{10}\delta < \frac{1}{5} + 6\delta.\tag{J} $$

A lower bound for $~$ Im$\left(\frac{\overline{w}}{|w|^2}\right)~$ is

$$\frac{-2 - 6\delta}{5 - 26\delta} = \frac{-2 + \frac{52}{5}\delta}{5 - 26\delta} - \frac{\frac{82}{5}\delta}{5 - 26\delta} > -\frac{2}{5} - \frac{82}{10}\delta > -\frac{2}{5} - 9\delta.\tag{K} $$

An upper bound for $~$ Im$\left(\frac{\overline{w}}{|w|^2}\right)~$ is

$$\frac{-2 + 4\delta}{5 + 95\delta} = \frac{-2 - 38\delta}{5 + 95\delta} + \frac{42\delta}{5 + 95\delta} < -\frac{2}{5} + \frac{42}{5}\delta < -\frac{2}{5} + 9\delta.\tag{L} $$

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Using (I) and (J)

$$-5\delta = \frac{1}{5} - 5\delta - \frac{1}{5} < \text{Re}[f(z)] < \frac{1}{5} + 6\delta - \frac{1}{5} = 6\delta.$$

Using (K) and (L)

$$-9\delta = -\frac{2}{5} - 9\delta + \frac{2}{5} < \text{Im}[f(z)] < -\frac{2}{5} + 9\delta + \frac{2}{5} = 9\delta.$$

$\delta$ must be chosen so that

$$-\frac{\epsilon}{2} < -5\delta, ~6\delta < \frac{\epsilon}{2}, ~-\frac{\epsilon}{2} < -9\delta, ~9\delta < \frac{\epsilon}{2}.$$

Coupled with the constraint that $\delta \leq \frac{1}{10}$
a workable final specification is

$$\delta = \min\left(\frac{1}{10}, \frac{\epsilon}{20}\right).$$

Answer 2

It is unfortunate that most textbooks and teachers do not teach how to use asymptotic analysis. This is an intuitive and powerful technique that can be used even if you want to obtain an ε-δ proof. $ \def\lfrac#1#2{{\large\frac{#1}{#2}}} \def\abs#1{\left|#1\right|} $

First let's use the technique:

  As $t\overset{∈ℂ}→0$:
    $\lfrac1{(1+i+t)^2+1} = \lfrac1{2i+2·(1+i)·t+t^2+1}$.
    $⊆ \lfrac1{2i+1+O(t)} ⊆ \lfrac1{2i+1}·\lfrac1{1+O(t)} ⊆ \lfrac1{2i+1}·(1+O(t)) → \lfrac1{2i+1}$.

That's it!

Now how to convert it into an ε-δ proof? Easy. Simply use concrete bounds in place of the asymptotic bounds:

  Let $r = \abs{t}$.
  Then $\lfrac1{(1+i+t)^2+1} = \lfrac1{2i+1}·\lfrac1{1+u}$   where $u = 2·\lfrac{1+i}{2i+1}·t+\lfrac1{2i+1}·t^2$.
  And $\abs{u} ≤ 2·r+r^2 ≤ 3r$   if $r ≤ 1$.
  And $\abs{\lfrac1{1+u}-1} = \lfrac{\abs{u}}{\abs{1+u}} ≤ \lfrac{\abs{u}}{1-\lfrac12} ≤ 6r$   if $r ≤ 1$ and $\abs{u}≤\lfrac12$.
  Thus $\abs{ \lfrac1{2i+1}·(\lfrac1{1+u}-1) } ≤ \lfrac12·6r = 3r$   if $r ≤ 1$ and $3r ≤ \lfrac12$.
  Thus $\abs{ \lfrac1{(1+i+t)^2+1} - \lfrac1{2i+1} } ≤ 3r$   if $r ≤ \lfrac16$.

It is now trivial to turn this into an ε-δ proof.