Using the properties of real numbers, verify that complex numbers are associative and there exists an additive inverse

Question

I am self-learning, so I need guidance, as I am unsure whether my approach is sufficient. There are two questions, both asking to verify a property of the complex numbers using the properties of real numbers.

(1) Verify that $z_{1}+(z_{2}+z_{3})=(z_{1}+z_{2})+z_{3}$ for all $z_{1}, z_{2}, z_{3} \in \mathbb{C}$

My approach: \begin{align}z_{1}+(z_{2}+z_{3}) &= (a+bi)+[(c+di)+(e+fi)]\\ &=(a+bi)+[(c+e)+(di+fi)]\\ &=[a+(c+e)]+[bi+(di+fi)]\\ &=[(a+c)+e]+[(bi+di)+fi]\\ &=[(a+c)+(bi+di)]+(e+fi)\\ &=[(a+bi)+(c+di)]+(e+fi)\\ &=(z_{1}+z_{2})+z_{3} \end{align}
I justify step 1 by the definition of complex numbers, step 2 and 3 by commutativity in R, step 4 by associativity in R, step 5 and 6 by commutativity in R.

(2) Additive inverse for every $z\in\mathbb{C}$, $\exists$w ∈ C: z + w = 0.

My approach:

Let $w=|(-z)|$. By the definition of complex numbers $\left|(-z)\right|=\left|-(a+bi)\right|$, and we have \begin{align}\left|(-z)\right| &= \left|-(a+bi)\right|\\ &=\left|-a-bi\right|\\ &=\sqrt{(-a)^2-(bi)^2}\\ &=\sqrt{a^2-b^2*i^2}\\ &=\sqrt{a^2-b^2*-1}\\ &=\sqrt{a^2+b^2}\\ &=\left|a+bi\right|\\ &=\left|z\right|\\ \end{align}

so there exists a number $w\in\mathbb{C}$ such that $z+w=0$.

Answer 1

Your first part is correct. For part two, you are presenting a rather circular argument. The goal is to prove that for any complex number $z$, the additive inverse, denoted $-z$, exists. But, on your second line, you let $w=|(-z)|$, which assumes the very statement you are trying to prove. Also, there are many complex numbers $x$ that satisfy $|z| = |x|$, but not all $x$ are additive inverses of $z$.

For example, $|2+4i| = |4+2i|$, but $(2+4i)+(4+2i) \neq 0$.

Here is my proof for (2):

Let $z$ be an arbitrary complex number. Define $a = \Re (z)$, $b= \Im (z)$, and let

$$w= -a + -b i.$$

Then, $$z+w = (a + b i) + (-a -b i) = (a - a) + (b-b)i = 0 + 0 i = 0.$$

That is, $w$ is the additive inverse of $z$. Now, we show that $w$ is unique.

Suppose that $w$ is not unique. Then, there exists $y$ so that $z + w = 0$ and $z + y = 0$ but $w \neq y$. Then,

\begin{align} z + w &= 0 \\ z + w + y &= y \\ z + y + w &= y \\ (z + y) + w &= y \\ 0 + w &= y \\ w &= y, \end{align}

a contradiction.