Chapter 7 Problem Solutions
7.1 Solving quadratic equations
\(z = 2 \pm \sqrt{3}i\)
\(z = \pm 2\sqrt{2}i\)
\(z = -12 \pm 3\sqrt{13}\)
\(z = \frac{1}{6} \pm \frac{\sqrt{59}}{6}i\)
7.5 Complex conjugate and division
:
\(\overline{z} = -7i\)
\(\overline{z} = 2 + 3i\)
\(\overline{z} = -7 - 6i\)
:
\(\frac{3 + 2i}{1 - 5i} = \frac{3 + 2i}{1 - 5i} \times \frac{1 + 5i}{1 + 5i} = -\frac{7}{26} + \frac{17}{26}i\)
\(\frac{2 + 3i}{8 - i} = \frac{2 + 3i}{8 - i}\times\frac{8 + i}{8 + i} = \frac{1}{5} +\frac{2}{5}i\)
\(\frac{(2 - i)(2 + 5i)}{1 + i} = \frac{9 - 8i}{1 + i}\times \frac{1 - i}{1 - i}= \frac{17}{2} - \frac{1}{2}i\)
\(\frac{1}{4 - 2i} + \frac{1}{3 + i} = \frac{2 + i}{10} + \frac{3 - i}{10} = \frac{1}{2}\)
\(\frac{1 - i}{2 + i} + \frac{4 - i}{1 - i} = \frac{(1-i)^2 + (4 - i)(2 + i)}{(2 + i)(1 - i)} = \frac{9}{3 - i} \times \frac{3 + i}{3 + i} = \frac{9}{10}(3 + i)\)
\(\frac{11i}{2 + 3i} = \frac{11i}{2 + 3i} \times \frac{2 - 3i}{2 - 3i} = \frac{33}{13} + \frac{22i}{13}\)
7.6 Polar co-ordinates
For \(z = a + bi\) with \(a,b \in \mathbb{R}\) the modulus is \(\left|z\right| = \sqrt{a^2 + b^2}\). To find the principal value of the argument, \(\text{Arg}(z)\), use the Argand diagram for \(z\). Remember that it is not enough to simply compute \(\tan^{-1}\left(\frac{b}{a}\right)\), since this value will always lie in the interval \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\). If we see that \(z\) lies in the first or fourth quadrant of the Argand diagram, then we can safely take \(\tan^{-1}\left(\frac{b}{a}\right)\) as the principal value of the argument. However, if \(z\) lies in the second quadrant then the principal value of the argument is \(\pi + \tan^{-1}\left(\frac{b}{a}\right)\). Furthermore, if \(z\) lies in the third quadrant then the principal value of the argument is \(-\pi + \tan^{-1}\left(\frac{b}{a}\right)\).
Modulus: \(\left|z\right| = \sqrt{1^2 + 1^2} = 1\).
Since \(z\) is in the first quadrant, \(\text{Arg}(z) = \tan^{-1}\left(\frac{1}{1}\right) =\frac{\pi}{4}\)Modulus: \(\left|z\right| = \sqrt{3 + 1} = 2\).
Since \(z\) is in the fourth quadrant, \(\text{Arg}(z) = \tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) = -\frac{\pi}{6}\)Modulus: \(\left|z\right| = \sqrt{4^2 + \left(4\sqrt{3}\right)^2} = 8\).
Since \(z\) is in the third quadrant, \(\text{Arg}(z) = -\pi + \tan^{-1}\left(\frac{\sqrt{3}}{1}\right) = -\frac{2\pi}{3}\).Modulus: \(\left|z\right| = \sqrt{16 + 16} = 4\sqrt{2}\).
Since \(z\) is in the second quadrant, \(\text{Arg}(z) = \pi + \tan^{-1}\left(-\frac{4}{4}\right) = \pi - \frac{\pi}{4} = \frac{3\pi}{4}\).Modulus: \(\left|z\right| = \sqrt{\cos^2(\frac{\pi}{3}) + \sin^2(\frac{\pi}{3})} = 1\).
Since \(z\) is in the second quadrant \(\text{Arg}(z) = \pi + \tan^{-1}\left(-\frac{\sin(\frac{\pi}{3})}{\cos(\frac{\pi}{3})}\right) = \frac{2\pi}{3}\).
Use the strategy above to find \(\left|z\right|\) and \(-\pi <\text{Arg}(z)\leq \pi\).
The polar form is then \(z = \left|z\right|\text{cis}\left(\text{Arg}(z)\right)\),
where \(\text{cis}\left(\text{Arg}(z)\right) = \cos\left(\text{Arg}(z)\right) + i\sin\left(\text{Arg}(z)\right)\).\(z = 4\text{cis}\left(-\frac{\pi}{3}\right)\)
\(z = 9\text{cis}\left(\pi\right)\)
\(z = 7\sqrt{2}\text{cis}\left(\frac{\pi}{4}\right)\)
\(z =94\sqrt{2}\text{cis}\left(\frac{3\pi}{4}\right)\)
\(z =6\text{cis}\left(-\frac{2\pi}{3}\right)\)
From the polar form \(z = r\text{cis}\left(\theta\right)\) the rectangular form is \(z = r\cos\left(\theta\right) + r\sin\left(\theta\right)i\).
\(z = \sqrt{3} + i\)
\(z = \frac{5\sqrt{3}}{2} - \frac{5}{2}i\)
\(z = -3 - 3i\)
\(z = - 2\sqrt{3} + 6i\)
7.7 de Moivre’s theorem
For complex numbers \(z_1 = r_1\text{cis}(\theta_1)\) and \(z_2 = r_2\text{cis}{\theta_2}\) the rules of multiplication and division in polar form state are:
\(z_1z_2 = r_1r_2\text{cis}(\theta_1 + \theta_2)\)
\(\frac{z_1}{z_2} = \frac{r_1}{r_2}\text{cis}\left(\theta_1 - \theta_2\right)\)
For the complex number \(z = r\text{cis}(\theta)\), and integer \(n\in\mathbb{Z}\), de Moivre’s theorem states that \(z^{n} = r^{n}\text{cis}(n\theta)\).
The respective polar forms are \(z_1 = 2\sqrt{2}\text{cis}\left(\frac{3\pi}{4}\right)\) and \(z_2 = 2\text{cis}\left(-\frac{\pi}{6}\right)\).
\(z_1z_2 = 4\sqrt{2}\text{cis}(\frac{3\pi}{4} - \frac{\pi}{6}) = 4\sqrt{2}\text{cis}\left(\frac{7\pi}{12}\right)\).
\(z_1^3 = (2\sqrt{2})^3\text{cis}\left(3\times\frac{3\pi}{4}\right) = 16\sqrt{2}\text{cis}\left(\frac{9\pi}{4} - 2\pi\right) = 16\sqrt{2}\text{cis}\left(\frac{\pi}{4}\right)\)
\(\frac{1}{z_1^3} = z_1^{-3} = (2\sqrt{2})^{-3}\text{cis}\left(-\frac{\pi}{4}\right) = \frac{1}{16\sqrt{2}}\text{cis}\left(-\frac{\pi}{4}\right)\)
\(z_1^3z_2^5 = (2\sqrt{2})^32^5\text{cis}\left(\frac{9\pi}{4} - \frac{5\pi}{6}\right) = 2^9\sqrt{2}\text{cis}\left(\frac{17\pi}{12}\right) = 2^9\sqrt{2}\text{cis}\left(\frac{-7\pi}{12}\right)\)
\(\frac{z_1^8}{z_2^6} = \frac{(2\sqrt{2})^8\text{cis}\left(6\pi\right)}{2^6\text{cis}\left(\pi\right)} = 2^6\text{cis}\left(\pi\right) = -2^6\)
Apply the rules stated at the start to this section, along with de Moivre’s theorem.
\(z = \frac{(\sqrt{8})^5\text{cis}\left(\frac{5\pi}{4}\right)}{\sqrt{2}\text{cis}\left(-\frac{\pi}{4}\right)} = 2^7\text{cis}\left(\frac{3\pi}{2}\right) = -128i\)
\(z = \frac{\left(2\text{cis}\left(\frac{\pi}{3}\right)\right)^2}{\left(2\sqrt{2}\text{cis}\left(\frac{\pi}{4}\right)\right)^3} = \frac{1}{4\sqrt{2}}\text{cis}\left(\frac{2\pi}{3} - \frac{3\pi}{4}\right) = \frac{1}{16}\text{cis}\left(- \frac{\pi}{12}\right)\)
\(z = \sqrt{2}^{10}\text{cis}\left(\frac{10\pi}{4}\right) + \sqrt{2}^{10}\text{cis}\left(-\frac{10\pi}{4}\right) = 2(\sqrt{2})^{10}\left(\text{cis}(\frac{\pi}{2})+\text{cis}(\frac{-\pi}{2})\right) = 2(\sqrt{2})^{10}(i-i) = 0\)
\(z = \left(2\text{cis}(\frac{5\pi}{6})\right)^{-5} = \frac{1}{32}\text{cis}\left(-\frac{25\pi}{6}\right) = \frac{1}{32}\text{cis}\left(-\frac{\pi}{6}\right)\)
\[\begin{aligned} z &= \frac{\left(\cos\left(\theta\right) + i\sin\left(\theta\right)\right)^4}{\left(\sin\left(\theta\right) + i\cos\left(\theta\right)\right)^2} = \frac{\text{cis}\left(4\theta\right)}{\left[i\left(\cos\theta - i\sin\theta\right)\right]^2}\\ &= -\frac{\text{cis}\left(4\theta\right)}{\left(\cos(-\theta) + i\sin(-\theta)\right)^2} = -\frac{\text{cis}\left(4\theta\right)}{\text{cis}\left(-2\theta\right)} = -\text{cis}\left(6\theta\right) \end{aligned}\]
7.8 Applications to trigonometric identities
Using de Moivre’s theorem we have that \[\begin{aligned} \cos\left(3\theta\right) + i\sin\left(3\theta\right) &= \left(\cos(\theta) + i\sin(\theta)\right)^3\\ & =\cos^3\left(\theta\right) + 3\cos^2(\theta) i\sin(\theta) + 3\cos(\theta)(i\sin(\theta)^2 + (i\sin(\theta)^3\\ & = \cos(\theta)\left(\cos^2(\theta)-3\sin^2(\theta)\right) + i\sin(\theta)\left(3\cos^2(\theta)-\sin^2(\theta)\right) \end{aligned}\] Now by Equating real and imaginary parts we get \[\begin{aligned} \cos(3\theta) = \cos(\theta)\left(\cos^2(\theta)-3\sin^2(\theta)\right),\\ \sin\left(3\theta\right) = \sin(\theta)\left(3\cos^2(\theta)-\sin^2(\theta)\right). \end{aligned}\]
Using de Moivre’s theorem we have that. \[\begin{aligned} \cos(2\theta) + i\sin(2\theta)&= \left(\cos(\theta) + i\sin(\theta)\right)^2\\ & = \cos^2(\theta) - \sin^2(\theta) + 2i\sin(\theta)\cos(\theta)\\ & = 2\cos^2(\theta) - 1 + 2i\sin(\theta)\cos(\theta). \end{aligned}\] Now by equating real and imaginary parts we get \[\begin{aligned} \cos^2(\theta) &= \frac{1}{2}\left(\cos(2\theta) + 1\right),\\ \sin(2\theta) &= 2\sin(\theta)\cos(\theta). \end{aligned}\]
Let \(z = \text{cis}\left(\theta\right)\) and so \(\frac{1}{z} = (\text{cis}(\theta))^{-1} = \cos\theta - i\sin\theta\). It follows that \(\sin\left(\theta\right) = -\frac{i}{2}\left(z - \frac{1}{z}\right)\). Therefore, starting with the binomial expansion: \[\begin{aligned} \sin^5(\theta) &= \left(-\frac{i}{2}\right)^5\left(z -\frac{1}{z}\right)^5\\ &= -\frac{i}{32}\left[z^5\left(-\frac{1}{z}\right)^0 + 5z^4\left(-\frac{1}{z}\right) + 10z^3\left(-\frac{1}{z}\right)^2 + 10z^2\left(-\frac{1}{z}\right)^3 + 5z\left(-\frac{1}{z}\right)^4\right.\\ &+ \left. z^0\left(-\frac{1}{z}\right)^5\right]\\ &=-\frac{i}{32}\left[z^5 - 5z^3 + 10z - \frac{10}{z} + \frac{5}{z^3} - \frac{1}{z^5}\right]\\ &=-\frac{i}{32}\left[\left(z^5 - \frac{1}{z^5}\right) - 5\left(z^3 - \frac{1}{z^3}\right) + 10\left(z - \frac{1}{z}\right)\right]\\ &= \frac{1}{16}\left[\sin(5\theta) - 5\sin(3\theta) + 10\sin(\theta)\right]. \end{aligned}\]
This means that \[\begin{aligned} \int\sin^5\left(\theta\right)\text{d}\theta & = \frac{1}{16}\int\left(\sin(5\theta) - 5\sin\left(3\theta\right) + 10\sin\left(\theta\right)\right)\text{d}\theta\\ & = -\frac{1}{80}\cos\left(5\theta\right) + \frac{5}{48}\cos(3\theta) - \frac{5}{8}\cos(\theta) \end{aligned}\] Alternatively this can be solved by using \(z^5 = (\text{cis}(\theta))^5\) with the binomial expansion followed by equating real and imaginary parts, as in Q1 of this subsection.
Let \(z = \text{cis}\left(\theta\right)\). It follows that \[\begin{aligned} \cos^2(\theta)\sin^4(\theta) &= \cos^2(\theta)\left(i\sin(\theta)\right)^4\\ &=\left(\frac{1}{2}\left(z + \frac{1}{z}\right)\right)^2\left(\frac{1}{2}\left(z - \frac{1}{z}\right)\right)^4\\ &=\frac{1}{64}\left[\left(z^6 + \frac{1}{z^6}\right) - 2\left(z^4 + \frac{1}{z^4}\right) - \left(z^2 + \frac{1}{z^2}\right) + 4\right]\\ & = \frac{1}{32}\left(\cos(6\theta) - 2\cos(4\theta) - \cos(2\theta) + 2\right) \end{aligned}\]
From the solution to the previous question we have that \[\begin{aligned} \int_{0}^{\frac{\pi}{6}}\cos^2(\theta)\sin^4(\theta)\text{d}\theta &= \frac{1}{32}\int_{0}^{\frac{\pi}{6}}\left(\cos(6\theta) - 2\cos(4\theta) - \cos(2\theta) + 2\right)\text{d}\theta. \end{aligned}\] It therefore follows that \[\begin{aligned} \int_{0}^{\frac{\pi}{6}}\cos^2(\theta)\sin^4(\theta)\text{d}\theta &= \frac{1}{32}\left[\frac{1}{6}\sin(6\theta) - \frac{1}{2}\sin(4\theta) - \frac{1}{2}\sin(2\theta)+ 2\theta\right]_0^{\frac{\pi}{6}}\\ & = \frac{1}{32}\left[-\frac{\sqrt{3}}{2} + \frac{\pi}{3}\right]. \end{aligned}\]
7.9 Roots
In polar form \[\begin{aligned} 4i = 4\text{cis}\left(\frac{\pi}{2} + 2k\pi\right),\quad k \in\mathbb{Z}\implies w_k = \left(4i\right)^{\frac{1}{2}} = 2\text{cis}\left(\frac{\pi}{4} + k\pi\right), \quad k = 0, 1. \end{aligned}\] Therefore \[\begin{aligned} w_0 &= 2\text{cis}\left(\frac{\pi}{4}\right) = \sqrt{2} + \sqrt{2}i,\\ w_1 & = 2\text{cis}\left(\frac{5\pi}{4}\right) = -\sqrt{2} - \sqrt{2}i. \end{aligned}\]
In polar form \[\begin{aligned} 2 + 2\sqrt{3}i = 4\text{cis}\left(\frac{\pi}{3} + 2k\pi\right),\quad k \in\mathbb{Z}\implies w_k = \left(2 + 2\sqrt{3}i\right)^{\frac{1}{2}} = 2\text{cis}\left(\frac{\pi}{6} + k\pi\right), \quad k = 0, 1. \end{aligned}\] Therefore \[\begin{aligned} w_0 &= 2\text{cis}\left(\frac{\pi}{6}\right) = \sqrt{3} + i,\\ w_1 & = 2\text{cis}\left(\frac{7\pi}{6}\right) = -\sqrt{3} -i. \end{aligned}\]
In polar form \[\begin{aligned} -27 = 27\text{cis}\left(\pi + 2k\pi\right),\quad k \in\mathbb{Z}\implies w_k = \left(-27\right)^{\frac{1}{3}} = 3\text{cis}\left(\frac{\pi}{3} + \frac{2k\pi}{3}\right), \quad k = 0, 1, 2. \end{aligned}\] Therefore \[\begin{aligned} w_0 &= 3\text{cis}\left(\frac{\pi}{3}\right) = \frac{3}{2} + \frac{3\sqrt{3}}{2}i,\\ w_1 & = 3\text{cis}\left(\pi\right) = -3,\\ w_2 & = 3\text{cis}\left(\frac{5\pi}{3}\right) = \frac{3}{2} - \frac{3\sqrt{3}}{2}i, \end{aligned}\]
In polar form \[\begin{aligned} 8i = 8\text{cis}\left(\frac{\pi}{2} + 2k\pi\right),\quad k \in\mathbb{Z}\implies w_k = \left(8i\right)^{\frac{1}{3}} = 2\text{cis}\left(\frac{\pi}{6} + \frac{2k\pi}{3}\right), \quad k = 0, 1, 2. \end{aligned}\] Therefore \[\begin{aligned} w_0 &= 2\text{cis}\left(\frac{\pi}{6}\right) = \sqrt{3} + i,\\ w_1 & = 2\text{cis}\left(\frac{5\pi}{6}\right) = -\sqrt{3} + i,\\ w_2 & = 2\text{cis}\left(\frac{3\pi}{2}\right) = - 2i, \end{aligned}\]
In polar form \[\begin{aligned} &-8 - 8\sqrt{3}i = 16\text{cis}\left(\frac{4\pi}{3} + 2k\pi\right)\quad k \in\mathbb{Z}\\ \implies& w_k = \left(-8 - 8\sqrt{3}i\right)^{\frac{1}{4}} = 2\text{cis}\left(\frac{\pi}{3} + \frac{k\pi}{2}\right), \quad k = 0, 1, 2, 3. \end{aligned}\] Therefore \[\begin{aligned} w_0 &= 2\text{cis}\left(\frac{\pi}{3}\right) = 1 + \sqrt{3}i,\\ w_1 & = 2\text{cis}\left(\frac{5\pi}{6}\right) = -\sqrt{3} + i,\\ w_2 & = 2\text{cis}\left(\frac{4\pi}{3}\right) = - 1 - \sqrt{3}i,\\ w_3 & = 2\text{cis}\left(\frac{11\pi}{6}\right) = \sqrt{3} - i. \end{aligned}\]
In polar form \[\begin{aligned} & 729 = 729\text{cis}\left(2k\pi\right)\quad k \in\mathbb{Z} \implies w_k = 729^{\frac{1}{6}} = 3\text{cis}\left(\frac{k\pi}{3}\right), \quad k = 0, 1, 2, 3, 4, 5. \end{aligned}\] Therefore \[\begin{aligned} w_0 &= 3\\ w_1 & = 3\text{cis}\left(\frac{\pi}{3}\right) = \frac{3}{2}\left(1 + \sqrt{3}i\right),\\ w_2 & = 3\text{cis}\left(\frac{2\pi}{3}\right) = \frac{3}{2}\left(-1 + \sqrt{3}i\right),\\ w_3 & = 3\text{cis}\left(\pi\right) = -3,\\ w_4 & = 3\text{cis}\left(\frac{4\pi}{3}\right) = \frac{3}{2}\left(-1 - \sqrt{3}i\right),\\ w_5 & = 3\text{cis}\left(\frac{5\pi}{3}\right) = \frac{3}{2}\left(1 - \sqrt{3}i\right). \end{aligned}\]