Week 4 Worksheet - Solutions

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Contours

1. We calculate $\gamma'(t) = 3 + i$ so that $|\gamma'(t)| = \sqrt{10}$. Then \[ \ell(\gamma) = \int\limits_3^5 \sqrt{10} \intd t = 2 \sqrt{10} \]

2. The following parameterization works. \begin{align*} \gamma_1(t) = 2t \textup{ on } [0,1] \\ \gamma_2(t) = 2 + 3ti \textup{ on } [0,1] \\ \gamma_3(t) = 3i + 2(1-t) \textup{ on } [0,1] \\ \gamma_4(t) = 3i(1-t) \textup{ on } [0,1] \end{align*}

3. 1. A straight line from $i$ to $2$.

   2. The reverse of $\Gamma = (\gamma_1,\gamma_2,\gamma_3)$ should be $(\tilde{\gamma_3},\tilde{\gamma_2},\tilde{\gamma_1})$.

Contour Integration

4. 1. We take $\gamma(t) = t + it$ on $[0,1]$ so that $x = y = t$ and $\gamma'(t) = 1 + i$. Thus the integral is \[ \int\limits_0^1 (t - t + it^2)(1+i) \intd t = (i-1) \int\limits_0^1 t^2 \intd t = \dfrac{i-1}{3} \]
   2. Here $\gamma(t) = it$ on $[0,1]$ so $x = 0$ and $y = t$. Also $\gamma'(t) = i$. \[ \int\limits_0^1 (-t)i \intd t = \frac{-i}{2} \]
   3. Use $\gamma(t) = i + t$ on $[0,1]$. Here $x = t$ and $y = 1$ and $\gamma'(t) = 1$. \[ \int\limits_0^1 (t - 1 + it^2) \intd t = \frac{1}{2} - 1 + \frac{i}{3} \]
5. 1. $\gamma_1'(t) = 2ie^{it}$ and $\gamma_1(t) - 2 = 2e^{it}$ \[ \int\limits_0^{2\pi} \dfrac{1}{2e^{it}} 2ie^{it} \intd t = \int\limits_0^{2\pi} i \intd t = 2 \pi i \]
   2. $\gamma_2'(t) = -ie^{-it}$ and $(\gamma_2(t) - i)^3 = e^{-3it}$ \[ \int\limits_0^{\pi/2} \dfrac{1}{e^{-3it}} (-ie^{-it}) \intd t = -i \int\limits_0^{\pi/2} e^{2it} \intd t = -i \left( \dfrac{e^{i\pi}}{2i} - \dfrac{1}{2i} \right) = 1 \]
6. Take $\gamma(t) = 1 + e^{it}$ on $[0,2\pi]$. Then $\gamma'(t) = ie^{it}$. \[ |\gamma(t)|^2 = (1 + e^{it})(1 + e^{-it}) = 2 + e^{it} + e^{-it} \] and the integral is \[ \int\limits_0^{2\pi} (2 + e^{it} + e^{-it}) i e^{it} \intd t = i \int\limits_0^{2 \pi} 2e^{it} + e^{2it} + 1 \intd t = 2 \pi i \]

Antiderivatives

7. 1. We can get an antiderivative $F$ from integration by parts \begin{align*} F(z) = \int z^2 \sin(z) = & - z^2 \cos(z) + \int 2z \cos(z) \\ = & -z^2 \cos(z) + 2z \sin(z) - \int 2 \sin(z) \\ = & (2-z^2) \cos(z) + 2z \sin(z) \end{align*} and the fundamental theorem of contour integration lets us evaluate the contour integral easily. \[ \int\limits_0^i f = F(i) - F(0) = 3 \cos(i) + 2i \sin(i) - 2 \]
   2. Integration by parts again helps with the antiderivative \[ F(z) = \int z \exp(iz) = \frac{1}{i} z \exp(iz) - \int \frac{1}{i} \exp(iz) = -iz \exp(iz) + \exp(iz) \] and \[ \int\limits_0^i f = F(i) - F(0) = \exp(-1) + \exp(-1) - 1 = \frac{2}{e} - 1 \]
8. 1. $\Gamma_1 = (\gamma_1,\gamma_2)$ where $\gamma_1(t) = it$ on $[0,1]$ and $\gamma_2(t) = i + t$ on $[0,1]$. Then \[ \int\limits_{\Gamma_1} f = \int\limits_{\gamma_1} f + \int\limits_{\gamma_2} f = \int\limits_0^1 t^2 i \intd t + \int\limits_0^1 (1 + t^2) \intd t = \dfrac{i}{3} + 1 + \frac{1}{3} = \dfrac{4 + i}{3} \]
   2. $\Gamma_2 = (\gamma_3,\gamma_4)$ where $\gamma_3(t) = t$ on $[0,1]$ and $\gamma_4(t) = 1 + it$ on $[0,1]$. Then \[ \int\limits_{\Gamma_2} f = \int\limits_{\gamma_3} f + \int\limits_{\gamma_4} f = \int\limits_0^1 t^2 \intd t + \int\limits_0^1 (1 + t^2) i \intd t = \dfrac{1}{3} + i + \frac{i}{3} = \dfrac{4i + 1}{3} \]
   3. There is no possibility that $f$ has an antiderivative. If it did, then the contour integral of $f$ over the closed contour $\Gamma = (\gamma_1,\gamma_2,\tilde{\gamma_4},\tilde{\gamma_3})$ would be zero and our answers to a) and b) would have to be equal.
9. 1. $\gamma(t) = 3e^{it}$ on $[0,\pi]$
   2. $f(z) = 1/z^2$ has an antiderivative $F(z) = -1/z$ on $\C \setminus \{0\}$. Therefore \[ \int\limits_\gamma f = F(\gamma(\pi)) - F(\gamma(0)) = \dfrac{-1}{-3} - \dfrac{-1}{3} = \dfrac{2}{3} \]
   3. In this case $\tilde{\gamma}(t) = 3e^{i(\pi - t)}$. The function $f$ still has an antiderivative, so \[ \int\limits_{\tilde{\gamma}} f = F( \tilde{\gamma}(\pi) ) - F( \tilde{\gamma}(0)) = F(3) - F(-3) = \dfrac{-1}{3} - \dfrac{-1}{-3} = - \dfrac{2}{3} \]
   4. The answer to c) is the negative of the answer to b)
10. The function $h(z) = f'(z) g(z) + f(z) g'(z)$ has an antiderivative $H(z) = f(z) g(z)$ by the product rule. Therefore \[ \int\limits_\gamma f'g + \int\limits_\gamma fg' = \int\limits_\gamma f'g + fg' = \int\limits_\gamma h = H(z_1) - H(z_0) = f(z_1) g(z_1) - f(z_0) g(z_0) \] and subtracting the contour integral of $f'g$ gives the result.

The Principal Logarithm

11. 1. $\Log(z) = \ln(|z|) + i \Arg(z)$ can take any real value and any imaginary value between $-\pi$ and $\pi$. Therefore the range is $\{ x + iy : x \in \R, -\pi < y \le \pi \}$.
    2. In order to calculate $\Log(\exp(z))$ we need the absolute value and the principal argument of $\exp(z)$. Write $z = x+iy$ so that $\exp(z) = e^x( \cos(y) + i \sin(y) )$.
12. 1. Take $a = b = e^{3 \pi i / 4}$. Then $ab = e^{3 \pi i /2}$ and $\Log(e^{3\pi i /2}) = \ln(1) - i \frac{\pi}{2}$ because we must use the principal argument, but this is not twice $\Log(e^{3\pi i /4}) = \ln(1) + i \frac{3\pi}{4}$.
    2. Let $\theta$ and $\psi$ be the principal arguments of $a$ and $b$ respectively. We know from polar multiplication that $\theta + \psi$ is an argument of $ab$. It may not be the principal argument, but there is $n \in \Z$ such that $\Arg(ab) = \theta + \psi + 2 \pi n$. Thus \begin{align*} \Log(a) + \Log(b) + 2 \pi i n & = \ln(|a|) + i \Arg(a) + \ln(|b|) + i\Arg(b) + 2 \pi i n \\ & = \ln(|ab|) + i(\theta + \psi + 2 \pi n) \\ & = \ln(|ab|) + i \Arg(ab) = \Log(ab) \end{align*}

13. $|1+i| = \sqrt{2}$ and $\Arg(1+i) = \pi/4$, so $\Log(1+i) = \ln(\sqrt{2}) + i \frac{\pi}{4}$.

    $|-1-i| = \sqrt{2}$ and $\Arg(-1-i) = -3 \pi / 4$ so $\Log(-1-i) = \ln(\sqrt{2}) - i \frac{3 \pi}{4}$.

14. 1. $i^i = \exp(i \Log(i)) = \exp(i (\ln(1) + i \pi/2)) = \exp(- \pi/2)$
    2. $a^{b+c} = \exp((b+c) \Log(a)) = \exp(b \Log(a)) + \exp(c \Log(a)) = a^b a^c$
    3. $(a^{1/2})^2 = (\exp(\frac{1}{2} \Log(a))^2 = \exp(\frac{1}{2} \Log(a) + \frac{1}{2} \Log(a)) = \exp(\Log(a)) = a$
    4. Take $a = b = e^{i 3 \pi / 4}$. Then $a^{1/2} = e^{i 3 \pi / 8}$ but $ab = e^{3 \pi i/2}$ and $(ab)^{1/2} = e^{-i \pi/4}$.