MATH3401
Complex Analysis
Lecture 13
Limits at \(\infty\)
Recall that in \(\R\), \(\ds \lim_{x\ra x_0}f(x)=\infty\) means:
Given \(M>0\), there exists \(\delta>0\) such that \[ 0 < |x - x_0| < \delta \Rightarrow f(x) > M. \]
Limits at \(\infty\)
In \(\C\), a neighbourhood of \(z_0\in \C\) is
Limits at \(\infty\)
A neighbourhood of \(\infty\) in \(\C\) has the form \(\{z: |z|>M\}.\)
Limits at \(\infty\)
"Close to \(\infty\)" \(\iff\) \(|z|\) is large \(\iff\) \(\dfrac{1}{|z|}\) is small.
Keeping that in mind:
\((*)\) \(\ds \lim_{z\ra z_0}f(z) = \infty\) means \(\ds \lim_{z\ra z_0}\frac{1}{f(z)} = 0.\)
Limits at \(\infty\)
\((*)\) \(\ds \lim_{z\ra \infty}f(z) = w_0 \) means \(\ds \lim_{z\ra 0}f\left(\frac{1}{z}\right) = w_0.\)
\((*)\) \(\ds \lim_{z\ra \infty}f(z) = \infty \) means \(\ds \lim_{z\ra 0}\frac{1}{f\left(\frac{1}{z}\right)} = 0.\)
Limits at \(\infty\)
Example 1: Show \(\ds \lim_{z\ra -1}\frac{iz+3}{z+1}= \infty.\)
Let \(f(z)=\ds \frac{iz+3}{z+1}\). We want to show \(\ds \lim_{z\ra -1}\frac{1}{f\left(z\right)}= 0.\)
LHS \( =\ds \lim_{z\ra -1} \left(\frac{z+1}{iz+3}\right) \) \(= \frac{\ds \lim_{z\ra -1} (z+1) }{\ds \lim_{z\ra -1} (iz+3) } \) \(= \ds \frac{0}{3-i} = 0 \) \(=\) RHS.
(via Lect 12, Theorem 2, part 4)
Limits at \(\infty\)
Example 2: Calculate \(\ds \lim_{z\ra \infty}\frac{2z^3-1}{z^2-1}.\) Let \(g(z)=\ds \frac{2z^3-1}{z^2-1}\).
Claim \(\ds \lim_{z\ra \infty}g(z)= \infty.\) So we want to show \(\ds \lim_{z\ra 0}\frac{1}{g\left(\frac{1}{z}\right)} = 0.\)
\(\ds \frac{1}{g\left(\frac{1}{z}\right)} \) \(=\ds \frac{1}{\frac{2\left(\frac{1}{z}\right)^3-1}{\left(\frac{1}{z}\right)^2-1}} \) \(= \ds \frac{\frac{1}{z^2}-1}{\frac{2}{z^3}-1} \) \(= \ds \frac{z-z^3}{2-z^3} \) \(\ra \ds\frac{0}{2}\) as \(z\ra 0\).
Continuity
Let \(f\) be a \(\C\)-valued function defined in a neighbourhood of \(z_0\).
"\(f\) is continuous at \(z_0\)" means: \(\ds\lim_{z\ra z_0 } f(z) = f(z_0).\)
i. e. given \(\varepsilon >0\) there exists \(\delta >0\) such that \[|z-z_0|\lt\delta \implies |f(z)-f(z_0)|\lt\varepsilon.\]
Continuity
Basic results
1. If \(f: \Omega \ra U\) and \(g:U\ra W\) are continuous, so is \(g
\circ f : \Omega \ra W\)
\(g\) composed with \(f\):
\(\left(g\circ f\right)(z) = g\left(\,f(z)\right)\).
2. If \(f\) is continuous and nonzero at \(z_0\) then there exists \(\varepsilon>0\) such that \(f(z)\neq 0\) on \(B_{\varepsilon}(z_0)\). Proof in tutes.
3. \(f: x+iy \mapsto u(x,y) + iv(x,y)\) is continuous if and only if \(u\) and \(v\) are continuous.
4. Obvious analogues of Theorems 1 & 2 from Lecture 12.
Differentiability in \(\R\)
\(f: \Omega \subseteq \R \ra \R \), \[ \lim_{h\ra 0 } \frac{f(x+h)-f(z)}{h}. \]
If this exists, defines \(f'(x)\). Limit in \(\R\).
Differentiability in \(\C\)
\(f: \Omega \subseteq \C \ra \C \), \[ \lim_{\xi\ra 0 } \frac{f(z_0+\xi)-f(z_0)}{\xi}. \qquad (1) \]
If this exists, defines \(f'(z_0)\).
Differentiability in \(\C\)
Write \(\Delta z\) for \(\xi\).
\((1)\) implies \(f'(z_0) = \ds \lim_{\Delta z \ra 0} \frac{f(z_0+\Delta z)- f(z_0)}{\Delta z}.\quad (2)\)
Write \(w = f(z)\): \(\Delta w = f(z_0 + \Delta) - f(z) \).
So \((2)\) implies \(f'(z_0) = \ds \lim_{\Delta z \ra 0} \frac{\Delta w}{\Delta z} = \frac{dw}{dz}(z_0).\quad (3)\)
Note (1)-(3) are equivalent.