Week 2 Worksheet - Solutions

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Paths

1. Fix $1\le c\le 2$. We calculate $$\begin{array}{rl}\underset{t\to c}{lim}\gamma (t)& =\underset{t\to c}{lim}4t^2+2it\\ & =4\underset{t\to c}{lim}{t}^2+2i\underset{t\to c}{lim}t\\ & =4c^2+2ic=\gamma (c)\end{array}$$ using the limit laws, which proves $\gamma$ is continuous at $c$. Since $1\le c\le 2$ was arbitrary $\gamma$ is continuous on $[1,2]$.

2. 1. Could be the image of a path.
   2. Could be the image of a path.
   3. Could not be the image of a path as it is not contiguous.
   4. Could be the image of a path.

3. 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

4. 1. 

      This is a domain as it is open and path connected.

   2. 

      This is a domain as it is open and path connected.

   3. 

      This is not a domain as it is not path connected.

   4. 

      This is a domain as it is open and path connected.

5. 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.
6. 

   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

7. Write $z=x+iy$. For $f(z)=\mathsf{Re}(z)$ 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 $v(x,y)=0$.

8. 1. $(x+iy{)}^2=x^2-y^2+2xyi$ so $u(x,y)=x^2-y^2$ and $v(x,y)=2xy$.
   2. $(\mathrm{\nabla }u)(x,y)=⟨2x,-2y⟩$ and $(\mathrm{\nabla }v)(x,y)=⟨2y,2x⟩$.

   3. Calculate $$\begin{array}{rl}& (\mathrm{\nabla }u)(x,y)\bullet (\mathrm{\nabla }v)(x,y)\\ =& ⟨2x,-2y⟩\bullet ⟨2y,2x⟩\\ =& (2x)(2y)+(-2y)(2x)\\ =& 0\end{array}$$

9. 1. $u(x,y)={\displaystyle \frac{x}{x^2+y^2}}$ and $v(x,y)={\displaystyle \frac{-y}{x^2+y^2}}$
   2. Fix $k\in \mathbb{R}$ non-zero. The equation $u(x,y)=k$ has the form $$\begin{array}{rl}& {\displaystyle \frac{x}{x^2+y^2}}=k\\ \Rightarrow & k{x}^2-x+y^2=0\\ \Rightarrow & k{(x-\frac{1}{2k})}^2-\frac{1}{4k}+k{y}^2=0\end{array}$$ so we have $${(x-\frac{1}{2k})}^2+y^2={\left(\frac{1}{2k}\right)}^2$$ which is a circle centered at $(\frac{1}{2k},0)$ of radius $\frac{1}{2k}$.

Continuous Functions

10. Fix $a\in \mathbb{C}$ non-zero. We need to prove that $$\underset{z\to a}{lim}{\displaystyle \frac{1}{z}}={\displaystyle \frac{1}{a}}$$ holds. Fix $ϵ>0$. If $|z-a|<|a|/2$ then $||z|-|a||<|a|/2$ so $|z|>|a|/2$. We then have $$|\frac{1}{z}-\frac{1}{a}|=\left|{\displaystyle \frac{z-a}{za}}\right|\le {\displaystyle \frac{2}{|a{|}^2}}|z-a|$$ which will be at most $ϵ$ if $|z-a|=ϵ|a{|}^2/4$. Take $\delta =min\{|a|/2,ϵ|a{|}^2/4\}$.

11. 1. With $z=t$ we have $$\underset{t\to 0}{lim}{\displaystyle \frac{|t|}{t}}=1$$ and with $z=it$ we have $$\underset{t\to 0}{lim}{\displaystyle \frac{|it|}{it}}=-i$$ so the complex limit does not exist.

    2. Here $\left|{\displaystyle \frac{|z{|}^2}{z}}\right|=|z|$ which does go to zero as $z\to 0$.

    3. With $z=t>0$ we have $$\underset{t\to 0}{lim}\mathsf{Arg}(t)=0$$ while with $z=it$ we have $$\underset{t\to 0}{lim}\mathsf{Arg}(t)={\displaystyle \frac{\pi }{2}}$$ so the complex limit does not exist.

12. 1. One can use the fact that $|\mathsf{Re}(z)-\mathsf{Re}(a)|\le |z-a|$ to prove $$\underset{z\to a}{lim}\mathsf{Re}(z)=\mathsf{Re}(a)$$ for all $a\in \mathbb{C}$. Therefore $f$ is continuous on $\mathbb{C}$.

    2. Use the reverse triangle inequality $||z|-|a||\le |z-a|$ to prove that $$\underset{z\to a}{lim}|z|=|a|$$ holds.

    3. Fix $a\in \mathbb{C}\setminus (-\mathrm{\infty },0]$ and put $\theta =\mathsf{Arg}(a)$.