Extra (Optional) Exercises – Long Day#
These exercises are optional extra material for students who are done with the exercises of the day but have the time for and the interest in more.
Extra Exercise 1: Ordering the Complex Numbers#
For the real numbers we have the well-known less than order relation \(\,<\,\) that for all \(\,a,b\,\) and \(\,c\,\) in \(\mathbb R\) satisfy:
Only one of the claims \(\,a<b,\) \(\,b<a\) 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#
Test these four claims on some examples.
Question b#
Show that the order relation \(\,<\,\) from 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 \(\,<\,\) from the real numbers while satisfying the four above bullets for all \(\,a,b\,\) og \(\,c\,\) i \(\mathbb C\).
Hint
Attempt a proof by contradiction. That is, assume that the extension does exist and then try reaching a contradiction.
Hint
If the extension exists then either \(0<i\) or \(i<0\) holds due to bullet 1. Try reaching a contradiction in each case.
Extra Exercise 2: Yet another Equation with Modulus#
In Exercise 9 we showed that all complex numbers \(z\) that are solutions to the equation \(|z-1|=|2z-3|\) form a circle in the complex plane. We are now given complex numbers \(z_0,z_1\) and \(z_2\). Show that the solution set to the equation \(|z-z_0|=|z_2\cdot z-z_1|\) in the complex plane forms either one single point, a straight line, a circle or fills the entire complex plane.