Show that the modulus of a complex number is a norm

Question

For a function $\|\cdot\|$ to be a norm, it must satisfy the following: Let $X$ be a vector space over the field $F$. For any $x, y\in X$ and $r\in F$

N1) $\|x\|\geqslant 0$

N2) $\|x\|=0 \rightarrow x=0$

N3) $\|rx\|= |r|\cdot ||x||$

N4) $\|x+y\|\leqslant \|x\| +\|y\|$

We let $X$ and $F$ in this case to be the set complex numbers $\mathbb C$

I managed N1) to N3) but N4) the triangular inequality just doesn't seem to work for modulus of a complex number, how can I show this?

$N1$ as stated is incorrect. It should be $|x| \ge 0$. $N4$ is also incorrect, it should be $|x+y| \le |x| + |y|$. — stoic-santiago, May 29 '21 at 12:49
Hint: Use $|x_1+x_2|^2 = (x_1+x_2)\overline{(x_1+x_2)}$ and $Re(x_1\overline{x_2}) \leq |x_1x_2|$. — Yathi, May 29 '21 at 12:50
https://math.stackexchange.com/questions/1836349/triangle-inequality-for-complex-numbers — Shubham Johri, May 29 '21 at 12:53

score 2 · Answer 1 · answered May 29 '21 at 13:00

Let $z,w\in\mathbb{C}$, $z=a+ib$, $w=c+id$. \begin{align} &|z|+|w|=\sqrt{a^2+b^2}+\sqrt{c^2+d^2},\\ &|z+w|=\sqrt{(a+c)^2+(d+b)^2} \end{align} then \begin{align} |z+w|\le|z|+|w|&\iff|z+w|^2\le(|z|+|w|)^2\\ &\iff(a+c)^2+(b+d)^2\le a^2+b^2+c^2+d^2+2\sqrt{(a^2+b^2)(c^2+d^2)}\\ &\iff ac+bd\le\sqrt{(a^2+b^2)(c^2+d^2)}\\ &\iff (ac+bd)^2\le(a^2+b^2)(c^2+d^2)\\ &\iff 2acbd\le(ad)^2+(bc)^2\\ &\iff 0\le(ad-bc)^2 \end{align} which is always true.

Show that the modulus of a complex number is a norm

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