Problems

Four complex numbers \(u_1\), \(u_2\), \(u_3\) and \(u_4\) have unit modulus, and arguments \(\theta_1\), \(\theta_2\), \(\theta_3\) and \(\theta_4\), respectively, with \(-\pi < \theta_1 < \theta_2 < \theta_3 < \theta_4 < \pi\). Show that \[ \arg \l u_1 - u_2 \r = \tfrac{1}{2} \l \theta_1 + \theta_2 -\pi \r + 2n\pi \] where \(n = 0 \hspace{4 pt} \mbox{or} \hspace{4 pt} 1\,\). Deduce that \[ \arg \l \l u_1 - u_2 \r \l u_4 - u_3 \r \r = \arg \l \l u_1 - u_4 \r \l u_3 - u_2 \r \r + 2n\pi \] for some integer \(n\). Prove that \[ | \l u_1 - u_2 \r \l u_4 - u_3 \r | + | \l u_1 - u_4 \r \l u_3 - u_2 \r | = | \l u_1 - u_3 \r \l u_4 - u_2 \r |\;. \]

1. Prove that the equations $$ \left|z - (1 + \mathrm{i}) \right|^2 = 2 \eqno(*) $$ and $$ \qquad \quad \ \left|z - (1 - \mathrm{i}) \right|^2 = 2 \left|z - 1 \right|^2 $$ describe the same locus in the complex \(z\)--plane. Sketch this locus.
2. Prove that the equation $$ \arg \l {z - 2 \over z} \r = {\pi \over 4} \eqno(**) $$ describes part of this same locus, and show on your sketch which part.
3. The complex number \(w\) is related to \(z\) by \[ w = {2 \over z}\;. \] Determine the locus produced in the complex \(w\)--plane if \(z\) satisfies \((*)\). Sketch this locus and indicate the part of this locus that corresponds to \((**)\).

The variable non-zero complex number \(z\) is such that \[ \left|z-\mathrm{i}\right|=1. \] Find the modulus of \(z\) when its argument is \(\theta.\) Find also the modulus and argument of \(1/z\) in terms of \(\theta\) and show in an Argand diagram the loci of points which represent \(z\) and \(1/z\). Find the locus \(C\) in the Argand diagram such that \(w\in C\) if, and only if, the real part of \((1/w)\) is \(-1\).

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Solution:

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\(z\) is a point on the circle shown: Therefore using the cosine rule \(|z|^2 = 1^2 + 1^2 - 2\cdot 1 \cdot 1 \cdot \cos (2 \theta) = 2 -2\cos 2\theta = 2\sin^2 \theta \Rightarrow |z| = \sqrt{2}|\sin \theta|\) \(\frac{1}{z}\) has modulus \(\frac{1}{\sqrt{2}|\sin \theta|}\) and argument \(-\theta\). \(|\frac{1}{z} - i| = 1 \Rightarrow |1-iz| = |z| \Rightarrow |-i-z| = |z|\) ie we're looking for the points on the perpendicular bisector of \(0\) and \(-i\). \(\textrm{Re}\left (\frac{1}{w}\right) = -1 \Rightarrow -1 = \textrm{Re} \left (\frac{1}{a+ib} \right) = \frac{a-ib}{a^2+b^2} = \frac{a}{a^2+b^2} \Rightarrow a^2+b^2 = -a \Rightarrow (a+\tfrac12)^2+b^2 = \tfrac14\) so we are looking at a circle radius \(\tfrac12\) centre \(-\frac12\)

\textit{In this question, the argument of a complex number is chosen to satisfy \(0\leqslant\arg z<2\pi.\)} Let \(z\) be a complex number whose imaginary part is positive. What can you say about \(\arg z\)? The complex numbers \(z_{1},z_{2}\) and \(z_{3}\) all have positive imaginary part and \(\arg z_{1}<\arg z_{2}<\arg z_{3}.\) Draw a diagram that shows why \[ \arg z_{1}<\arg(z_{1}+z_{2}+z_{3})<\arg z_{3}. \] Prove that \(\arg(z_{1}z_{2}z_{3})\) is never equal to \(\arg(z_{1}+z_{2}+z_{3}).\)

Let \(\alpha\) be a fixed angle, \(0 < \alpha \leqslant\frac{1}{2}\pi.\) In each of the following cases, sketch the locus of \(z\) in the Argand diagram (the complex plane):

1. \({\displaystyle \arg\left(\frac{z-1}{z}\right)=\alpha,}\)
2. \({\displaystyle \arg\left(\frac{z-1}{z}\right)=\alpha-\pi,}\)
3. \(|\dfrac{z-1}{z}|=1.\)

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Solution:

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\begin{align*} \arg w &= \arg \frac{(z_{1}-z_{2})(z_{3}-z_{4})}{(z_{4}-z_{1})(z_{2}-z_{3})} \\ &= \arg \frac{(z_{1}-z_{2})(z_{3}-z_{4})}{(z_{2}-z_{3})(z_{4}-z_{1})} \\ &= \arg \frac{(z_{1}-z_{2})}{(z_{3}-z_{2})}\frac{(z_{3}-z_{4})}{(z_{1}-z_{4})} \\ &= \arg \frac{(z_{1}-z_{2})}{(z_{3}-z_{2})} + \arg \frac{(z_{3}-z_{4})}{(z_{1}-z_{4})}\\ &= \beta + \pi - \beta = \pi \end{align*} Therefore \(w\) is real

1. If \(z=x+\mathrm{i}y,\) with \(x,y\) real, show that \[ \left|x\right|\cos\alpha+\left|y\right|\sin\alpha\leqslant\left|z\right|\leqslant\left|x\right|+\left|y\right| \] for all real \(\alpha.\)
2. By considering \((5-\mathrm{i})^{4}(1+\mathrm{i}),\) show that \[ \frac{\pi}{4}=4\tan^{-1}\left(\frac{1}{5}\right)-\tan^{-1}\left(\frac{1}{239}\right). \] Prove similarly that \[ \frac{\pi}{4}=3\tan^{-1}\left(\frac{1}{4}\right)+\tan^{-1}\left(\frac{1}{20}\right)+\tan^{-1}\left(\frac{1}{1985}\right). \]