Extra (optional) exercises – Long Day#
These exercises are optional extra material for students who have finished the exercises of the day and have the time and interest for more.
Extra exercise 1: Ordering the complex numbers#
For the real numbers we have the well-known order relation \(<\) (less than), which for all \(a\), \(b\) and \(c\) in \(\mathbb R\) satisfies:
Exactly one of the statements \(a<b\), \(b<a\) and \(a=b\) is true.
If \(a<b\) and \(b<c\), then \(a<c\).
If \(a<b\), then \(a+c<b+c\).
If \(a<b\) and \(0<c\), then \(ac<bc\).
Question a#
Check these four properties with some examples.
Question b#
Show that the order relation \(<\) on the real numbers cannot be extended to apply to the complex numbers. More precisely, show that no order relation \(<\) exists on \(\mathbb C\) that extends the order relation \(<\) on the real numbers while satisfying the four properties above for all \(a\), \(b\) and \(c\) in \(\mathbb C\).
Hint
Attempt a proof by contradiction. That is, assume that the extension does exist and then try to reach a contradiction.
Hint
If the extension exists, then by property 1 either \(0<i\) or \(i<0\) holds, since \(i\neq 0\). Try to reach a contradiction in each case.
Extra exercise 2: Another equation involving the modulus#
In Exercise 9 from the day’s main exercises, we showed that all complex numbers \(z\) that are solutions to the equation \(|z-1|=|2z-3|\) form a circle in the complex plane. Now let \(z_0\), \(z_1\) and \(z_2\) be complex numbers. Show that the solution set to the equation \(|z-z_0|=|z_2\cdot z-z_1|\) is either a single point, a straight line, a circle or the entire complex plane.