Problems

If \(z=x+\mathrm{i}y\) where \(x\) and \(y\) are real, define \(\left|z\right|\) in terms of \(x\) and \(y\). Show, using your definition, that if \(z_{1},z_{2}\in\mathbb{C}\) then \(\left|z_{1}z_{2}\right|=\left|z_{1}\right|\left|z_{2}\right|.\) Explain, by means of a diagram, or otherwise, why \(\left|z_{1}+z_{2}\right|\leqslant\left|z_{1}\right|+\left|z_{2}\right|.\) Suppose that \(a_{j}\in\mathbb{C}\) and \(\left|a_{j}\right|\leqslant1\) for \(j=1,2,\ldots,n.\) Show that, if \(\left|z\right|\leqslant\frac{1}{2},\) then \[ \left|a_{n}z^{n}+a_{n-1}z^{n-1}+\cdots+a_{1}z\right|<1, \] and deduce that any root \(w\) of the equation \[ a_{n}z^{n}+a_{n-1}z^{n-1}+\cdots+a_{1}z+1=0 \] must satisfy \(\left|x\right|>\frac{1}{2}.\)

Sketch the following subsets of the complex plane using Argand diagrams. Give reasons for your answers.

1. \(\{z:\mathrm{Re}((1+\mathrm{i})z)\geqslant0\}.\)
2. \(\{z: |z^{2}| \leqslant2,\mathrm{Re}(z^{2})\geqslant0\}.\)
3. \(\{z=z_{1}+z_{2}:\left|z_{1}\right|=2,\left|z_{2}\right|=1\}.\)

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. \(|\dfrac{z-1}{z}|=1.\)

Show Solution

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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