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(section:uge3S-extra)=
# Extra (optional) exercises -- Long Day

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These exercises are optional extra material for students who have finished the exercises of the day and have the time and interest for more.

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## Extra exercise 1: Ordering the complex numbers

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For the real numbers we have the well-known order relation $<$ (less than), which for all $a$, $b$ and $c$ in $\mathbb R$ satisfies:

1. Exactly one of the statements $a<b$, $b<a$ and $a=b$ is true.

2. If $a<b$ and $b<c$, then $a<c$.

3. If $a<b$, then $a+c<b+c$. 

4. If $a<b$ and $0<c$, then $ac<bc$.

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### Question a

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Check these four properties with some examples.

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### Question b

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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}
:class: dropdown
Attempt a proof by contradiction. That is, assume that the extension does exist and then try to reach a contradiction.
```

```{hint}
:class: dropdown
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.
```

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## Extra exercise 2: Another equation involving the modulus

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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.