\[
\newcommand{\Arg}{\mathsf{Arg}}
\newcommand{\C}{\mathbb{C}}
\newcommand{\R}{\mathbb{R}}
\newcommand{\N}{\mathbb{N}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\Im}{\mathsf{Im}}
\newcommand{\intd}{\,\mathsf{d}}
\newcommand{\Re}{\mathsf{Re}}
\newcommand{\ball}{\mathsf{B}}
\newcommand{\wind}{\mathsf{wind}}
\]
Week 2 Worksheet - Solutions
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Paths
Fix $1 \le c \le 2$. We calculate
\begin{align*}
\lim_{t \to c} \gamma(t) & = \lim_{t \to c} 4t^2 + 2it \\
& = 4 \lim_{t \to c} t^2 + 2i \lim_{t \to c} t \\
& = 4c^2 + 2ic = \gamma(c)
\end{align*}
using the limit laws, which proves $\gamma$ is continuous at $c$.
Since $1 \le c \le 2$ was arbitrary $\gamma$ is continuous on $[1,2]$.
- Could be the image of a path.
- Could be the image of a path.
- Could not be the image of a path as it is not contiguous.
- Could be the image of a path.
Take
\[
\gamma(t) = (1-t)2 + t(i-1) = 2 + t(i-3)
\]
which is continuous and satisfies has $\gamma(0) = 2$ and $\gamma(1) = i-1$.
Domains
-
This is a domain as it is open and path connected.
-
This is a domain as it is open and path connected.
-
This is not a domain as it is not path connected.
This is a domain as it is open and path connected.
- Fix $z \in E \cap F$. Since $E$ is open we can find $r > 0$ such that $\mathsf{B}(z,r) \subset E$. Since $F$ is open we can find $s > 0$ such that $\mathsf{B}(z,s) \subset F$. Put $t = \min \{t,s\}$. Then $\mathsf{B}(z,t) \subset E \cap F$. Since $z \in E \cap F$ was arbitrary, the intersection is open.
-
The red and blue sets above are both domains. However, their intersection is not path connected. The intersection of two domains need not be a domain.
Functions
Write $z = x+iy$. For $f(z) = \Re(z)$ we have $u(x,y) = x$ and $v(x,y) = 0$. For
\[
f(z) = |z| = \sqrt{x^2 + y^2}
\]
we have $u(x,y) = \sqrt{x^2 + y^2}$ and $v(x,y) = 0$.
- $(x+iy)^2 = x^2 - y^2 + 2xyi$ so $u(x,y) = x^2 - y^2$ and $v(x,y) = 2xy$.
- $(\nabla u)(x,y) = \langle 2x, -2y \rangle$ and $(\nabla v)(x,y) = \langle 2y, 2x \rangle$.
Calculate
\[
\begin{align*}
& (\nabla u)(x,y) \bullet (\nabla v)(x,y) \\
={} & \langle 2x, -2y \rangle \bullet \langle 2y, 2x \rangle \\
={} & (2x)(2y) + (-2y)(2x) \\
={} & 0
\end{align*}
\]
- $u(x,y) = \dfrac{x}{x^2 + y^2}$ and $v(x,y) = \dfrac{-y}{x^2 + y^2}$
- Fix $k \in \R$ non-zero. The equation $u(x,y) = k$ has the form
\[
\begin{align*}
& \dfrac{x}{x^2 + y^2} = k \\
\Rightarrow{} & kx^2 - x + y^2 = 0 \\
\Rightarrow{} & k \left(x - \frac{1}{2k} \right)^2 - \frac{1}{4k} + ky^2 = 0
\end{align*}
\]
so we have
\[
\left( x - \frac{1}{2k} \right)^2 + y^2 = \left( \frac{1}{2k} \right)^2
\]
which is a circle centered at $(\frac{1}{2k},0)$ of radius $\frac{1}{2k}$.
Continuous Functions
- Fix $a \in \C$ non-zero. We need to prove that
\[
\lim\limits_{z \to a} \dfrac{1}{z} = \dfrac{1}{a}
\]
holds. Fix $\epsilon > 0$. If $|z-a| < |a|/2$ then $| |z| - |a| | < |a|/2$ so $|z| > |a|/2$. We then have
\[
\left| \frac{1}{z} - \frac{1}{a} \right| = \left| \dfrac{z-a}{za} \right| \le \dfrac{2}{|a|^2} |z-a|
\]
which will be at most $\epsilon$ if $|z-a| = \epsilon |a|^2 / 4$. Take $\delta = \min \{ |a|/2, \epsilon |a|^2 / 4 \}$.
With $z = t$ we have
\[
\lim\limits_{t \to 0} \dfrac{|t|}{t} = 1
\]
and with $z = it$ we have
\[
\lim\limits_{t \to 0} \dfrac{|it|}{it} = -i
\]
so the complex limit does not exist.
Here $\left| \dfrac{|z|^2}{z} \right| = |z|$ which does go to zero as $z \to 0$.
With $z = t > 0$ we have
\[
\lim\limits_{t \to 0} \Arg(t) = 0
\]
while with $z = it$ we have
\[
\lim\limits_{t \to 0} \Arg(t) = \dfrac{\pi}{2}
\]
so the complex limit does not exist.
One can use the fact that $|\Re(z) - \Re(a)| \le |z-a|$ to prove
\[
\lim\limits_{z \to a} \Re(z) = \Re(a)
\]
for all $a \in \C$. Therefore $f$ is continuous on $\C$.
Use the reverse triangle inequality $||z| - |a|| \le |z-a|$ to prove that
\[
\lim\limits_{z \to a} |z| = |a|
\]
holds.
Fix $a \in \C \setminus (-\infty,0]$ and put $\theta = \Arg(a)$.