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Proof

(related to Proposition: The distance of complex numbers makes complex numbers a metric space.)

For any elements $z,z_1,z_2\in\mathbb C$ of the field of complex numbers $(\mathbb C, + ,\cdot)$ with $z=z_1=:(a,b)$ and $z_2=(c,d)$ ($a,b,c,d\in\mathbb R$):

Ad $(1)$

• By the definition of absolute value of complex numbers, $|z|=\sqrt{a^2+b^2}=0$
• if and only if $\sqrt{a^2+b^2}=0,$
• if and only if $a=0$ and $b=0,$
• if and only if $x=(0,0).$

Ad $(2)$

• By the definition of subtraction of complex numbers, we get $|z_1-z_2|=\sqrt{(a-c)^2+(b-d)^2}$
• $=\sqrt{(c-a)^2+(d-b)^2}$ because $x^2=(-x)^2$ for all real numbers,
• $=|z_2-z_1|$ by definition.

Ad $(3)$

• Note that for every complex number $z$ we have that its real part is smaller or equal¹ than its absolute value $\Re(z)\le |z|.$
• Given the complex conjugate $z_2^*$, we get therefore the inequality $$\Re(z_1z_2^*)\le |z_1z_2^*|=|z_1||z_2^*|=|z_1||z_2|.$$
• Therefore, $$\begin{array}{rcl}|z_1+z_2|^2&=&(z_1+z_2)(z_1^*+z_2^*)\\&=&z_1z_1^*+z_1z_2^*+z_2z_2^*+z_2z_2^*\\&=&|z_1|^2+2\Re(z_1z_2^*)+|z_2|^2\\&\le&|z_1|^2+2|z_1||z_2|+|z_2|^2\\&=&(|z_1|+|z_2|)^2\end{array}$$
• Taking the square root on both sides of the inequality yields the triangle inequality $$|z_1+z_2|\le |z_1|+|z_2|.$$

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References

Bibliography

1. Forster Otto: "Analysis 1, Differential- und Integralrechnung einer Veränderlichen", Vieweg Studium, 1983

Footnotes