\[ \newcommand{\Arg}{\mathsf{Arg}} \newcommand{\C}{\mathbb{C}} \newcommand{\R}{\mathbb{R}} \newcommand{\N}{\mathbb{N}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\Im}{\mathsf{Im}} \newcommand{\intd}{\,\mathsf{d}} \newcommand{\Re}{\mathsf{Re}} \newcommand{\ball}{\mathsf{B}} \newcommand{\wind}{\mathsf{wind}} \]

Week 1 Worksheet - Solutions

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Complex Numbers

    1. $|2i| = 2$. The arguments are $\frac{\pi}{2} + 2n \pi$ for any $n \in \Z$. $\Arg(2i) = \frac{\pi}{2}$.
    2. $|-1 - i \sqrt{3}| = \sqrt{(-1)^2 + (\sqrt{3})^2} = 2$. The arguments are $\frac{4 \pi}{3} + 2 n \pi$ for any $n \in \Z$. The principal argument is $\frac{-2 \pi}{3}$.
    3. $|-4| = 4$. The arguments are $-\pi + 2 n \pi$ for any $n \in \Z$. $\Arg(-4) = \pi$.

Arithmetic

  1. Calculate using the rules of arithmetic. \[ \begin{align*} & (2+3i) * (-5 + i) \\ = & \Big( 2 \times (-5) - 3 \times 1 \Big) + i \Big( 2 \times 1 + 3 \times (-5) \Big) \\ = & -13 - 13i \end{align*} \]
    1. $-5 + 24i$
    2. $\dfrac{2+3i}{3-4i} * \dfrac{3+4i}{3+4i} = \dfrac{-6 + 17i}{25} = \dfrac{-6}{25} + \dfrac{17}{25}i$
    3. $\dfrac{1-5i}{3i-1} * \dfrac{-3i-1}{-3i-1} = \dfrac{-16 + 2i}{10} = -\dfrac{8}{5} + \dfrac{1}{5} i$
    4. $\dfrac{1-i}{1+i} * \dfrac{1-i}{1-i} - i + 2 = \dfrac{-2i}{2} - i + 2 = 2 - 2i$
    5. $-i$ (becuase $i * i = -1$)

    1. Write $z = x+iy$ so that the equation becomes $x^2 - y^2 + 2xyi = -5 - 12i$. Real and imaginary parts must be equal, so $x^2 - y^2 = -5$ and $2xyi = 12i$. By inspection we can take $x = 2, y = 3$ or $x = -2, y = -3$. The solutions are $z = 2 + 3i$ and $z = -2 - 3i$.

    2. Completing the square gives $(z + 2)^2 - 4 = -12 + 6i$ so $(z+2)^2 = -8 + 6i$. Writing $z+2 = x+iy$ gives the equations $x^2 - y^2 = -8$ and $2xyi = 6i$. By inspection, we can take $x = 1, y = 3$ or $x = -1, y = -3$. The solutions are therefore $-1+3i$ and $-3 - 3i$.

    1. Calclate \[ \begin{align*} \Re(z+w) & = \Re((x+iy) + (u + iv)) \\ & = \Re((x+u) + i(y+v)) \\ & = x+u \\ & = \Re(z) + \Re(w) \end{align*} \]
    2. Calculate \[ \begin{align*} \Im(z-w) & = \Re((x+iy) - (u + iv)) \\ & = \Re((x-u) + i(y-v)) \\ & = y-v \\ & = \Im(z) - \Im(w) \end{align*} \]

  5. Take $z = w = i$. Then \[ \Re(z) = \Re(w) = 0 \] but $\Re(zw) = \Re(-1) = -1$. Also $\Im(z) = \Im(w) = 1$ but $\Im(zw) = \Im(-1) = 0$.

    1. Calcuate \[ \begin{align*} \overline{z+w} & = \overline{(x+iy) + (u + iv)} \\ & = \overline{(x+u) + i(y+v)} \\ & = (x+u) - i(y+v) \\ & = \overline{z} + \overline{w} \end{align*} \]

    2. Calcuate \[ \begin{align*} \overline{zw} & = \overline{(xu - yv) + i(xv + yu)} \\ & = (xu - yv) - i(xv + yu) \\ & = (x-iy)(u-iv) \\ & = \overline{z} \overline{w} \end{align*} \]

    3. Calcuate \[ \begin{align*} \overline{1/z} & = \overline{\dfrac{x}{x^2+y^2} - i \dfrac{y}{x^2+y^2}} \\ & = \dfrac{x}{x^2+y^2} + i \dfrac{y}{x^2+y^2} \\ & = \dfrac{x}{x^2+(-y)^2} - i \dfrac{(-y)}{x^2+(-y)^2} \\ & = 1 / \overline{z} \end{align*} \]

    4. $z + \overline{z} = (x + iy) + (x - iy) = 2x = 2 \Re(z)$
    5. $z - \overline{z} = (x + iy) - (x - iy) = 2iy = 2i \Im(z)$

    1. The case $n = 1$ is immediate. We assume \[ (\cos \theta + i \sin \theta)^n = \cos(n \theta) + i \sin(n \theta) \] and calculate \[ \begin{align*} & (\cos \theta + i \sin \theta)^{n+1} \\ = & (\cos \theta + i \sin \theta)^n (\cos \theta + i \sin \theta) \\ = & (\cos(n \theta) + i \sin(n \theta))(\cos \theta + i \sin \theta) \\ = & (\cos(n \theta + \theta) + i \sin(n\theta + \theta)) \\ = & \cos((n+1)\theta) + i \sin((n+1)\theta) \end{align*} \] because when we multiply complex numbers we add arguments and multiply moduli.

    2. We have \[ \begin{align*} & \cos(3\theta) + i \sin(3\theta) \\ = & (\cos \theta + i \sin \theta)^3 \\ = & (\cos \theta)^3 + 3i (\cos \theta)^2 (\sin \theta) - 3 (\cos \theta)(\sin \theta)^2 - i(\sin \theta)^3 \end{align*} \] so equating real and imaginary parts tell us \[ \cos(3\theta) = (\cos \theta)^3 - 3 (\cos \theta)(\sin \theta)^2 \] and \[ \sin(3\theta) = 3 (\cos \theta)^2(\sin \theta) - (\sin \theta)^3 \] hold.

      From \[ \begin{align*} & (\cos \theta + i \sin \theta)^4 \\ = & (\cos \theta)^4 + 4 i (\cos \theta)^3(\sin \theta) -6 (\cos \theta)^2 (\sin \theta)^2 - 4i (\cos \theta) (\sin \theta)^3 + (\sin \theta)^4 \end{align*} \] we get also that \[ \cos (4\theta) = (\cos \theta)^4 - 6(\cos \theta)^2(\sin \theta)^2 - (\sin \theta)^4 \] and \[ \sin(4\theta) = 4 (\cos \theta)^3(\sin \theta) - 4 (\cos \theta)(\sin \theta)^3 \] hold.

  8. Write $z = r(\cos \theta + i \sin \theta)$. If $z^n = w_0$ then $|z|^n = |w_0|$ and $n \theta$ is an argument of $w_0$. Thus $n \theta = \Arg(w) + 2k \pi$ for some $k \in \Z$. This means \[ \theta = \frac{\Arg(w) + 2 k \pi}{n} \] for some $k \in \Z$. Taking $k \in \{0,1,\dots,n-1\}$ gives all possible distinct $z$ as after that we repeat.

  9. Take $z_1 = z_2 = -1 + i \sqrt{3}$. We have $\Arg(z_1) = \Arg(z_2) = \frac{2\pi}{3}$. But \[ (-1 + i\sqrt{3})^2 = -2 - 2 \sqrt{3} i \] has principal argument $-\frac{2\pi}{3}$.

Distance and Limits

  1. Fix $\epsilon > 0$. There is $N \in \N$ such that $n \ge N$ guarantees $|z_n - a| < \epsilon$. If $n \ge N$ then $| |z_n| - |a| | \le |z_n - a| < \epsilon$ by the reverse triangle inequality.
    1. $|z_n| = |1+i|^n = (\sqrt{2})^n$ does not converge, so $z_n$ does not converge.

    2. Calculate \[ |z_n| = \dfrac{|(1+i)|^n}{n} = \dfrac{(\sqrt{2})^n}{n} \] and note that if $n$ is large enough then $(\sqrt{2})^{n/2} > n$ so $|z_n| \ge (\sqrt{2})^{n/2}$ which does not converge, so $z_n$ does not converge.

    3. $|z_n| = \dfrac{1}{|(1+i)^n|} = \dfrac{1}{(\sqrt{2})^n}$ converges to 0 so $z_n$ converges to 0.
  3. There is $K \in \N$ such that $|z| / \sqrt{n} < 1$ for all $n \ge K$. Now for $n \ge K$ we have \[ \begin{align*} & \left| \frac{z^n}{n!} \right| \\ = & \frac{|z|}{1} \frac{|z|}{2} \cdots \frac{|z|}{K-1} \frac{|z|}{K} \cdot \frac{|z|}{K+1} \cdots \frac{|z|}{n} \\ \le & \frac{|z|}{K!} \frac{1}{\sqrt{K+1}} \cdots \frac{1}{\sqrt{n}} \end{align*} \] which converges to zero as $n \to \infty$.

Paths and Domains

  1. 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 $[1,2]$.

    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 $\gamma(1) = i-1$.

    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 $t = \min \{r,s\}$. 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.