Beweis. (1). Sei $z = a + i b .$ Dann gilt

$| z | = \sqrt{a^{2} + b^{2}} \geq 0; \sqrt{a^{2} + b^{2}} = 0 \Leftrightarrow a = b = 0 \Leftrightarrow z = 0 .$

(2). Trivial !

(3). Sei $z_{n} = a_{n} + i b_{n}, n = 1, 2 .$ Dann gilt

$| z_{1} \cdot z_{2} | = | a_{1} a_{2} + i^{2} \cdot b_{1} b_{2} + i a_{1} b_{2} + i b_{1} a_{2} |$

$= | a_{1} a_{2} - b_{1} b_{2} + i (a_{1} b_{2} + b_{1} a_{2}) |$

$= \sqrt{{(a_{1} a_{2} - b_{1} b_{2})}^{2} + {(a_{1} b_{2} + b_{1} a_{2})}^{2}}$

$= \sqrt{a_{1}^{2} a_{2}^{2} - 2 a_{1} a_{2} b_{1} b_{2} + b_{1}^{2} b_{2}^{2} + a_{1}^{2} b_{2}^{2} + 2 a_{1} b_{2} b_{1} a_{2} + b_{1}^{2} a_{2}^{2}}$

$= \sqrt{a_{1}^{2} (a_{2}^{2} + b_{2}^{2}) + b_{1}^{2} (b_{2}^{2} + a_{2}^{2})}$

$= \sqrt{(a_{1}^{2} + b_{1}^{2}) \cdot (a_{2}^{2} + b_{2}^{2})}$

$= \sqrt{a_{1}^{2} + b_{1}^{2}} \cdot \sqrt{a_{2}^{2} + b_{2}^{2}} = | z_{1} | \cdot | z_{2} | .$

(4). Es genügt zu zeigen: $|\frac{1}{z}| = \frac{1}{| z |},$ denn

$|\frac{u}{z}| = |u \cdot \frac{1}{z}| = | u | \cdot |\frac{1}{z}| = | u | \cdot \frac{1}{| z |} = \frac{| u |}{| z |} .$

Wir wissen schon, daß für $z = a + i b$ und $c : = a^{2} + b^{2}$ gilt:

$\frac{1}{z} = \frac{a}{c} + i \frac{- b}{c} \Rightarrow$

$| \frac{1}{z} | = \sqrt{\frac{a^{2}}{c^{2}} + \frac{b^{2}}{c^{2}}} = \sqrt{\frac{a^{2} + b^{2}}{c^{2}}} = \sqrt{\frac{1}{c}} = \frac{1}{\sqrt{a^{2} + b^{2}}} = \frac{1}{| z |} .$

(5). Es sei $z_{n} = a_{n} + i b_{n}, n = 1, 2 .$ Dann ist die linke Seite $l s$ von (5):

$l s : = | a_{1} + i b_{1} + a_{2} + i b_{2} | = | a_{1} + a_{2} + i (b_{1} + b_{2}) | = \sqrt{{(a_{1} + a_{2})}^{2} + {(b_{1} + b_{2})}^{2}}$

und die rechte Seite $r s$ ist:

$r s : = | a_{1} + i b_{1} | + | a_{2} + i b_{2} | = \sqrt{a_{1}^{2} + b_{1}^{2}} + \sqrt{a_{2}^{2} + b_{2}^{2}} .$

Offenbar sind $l s, r s \geq 0 .$ Folglich ist

$l s \leq r s \Leftrightarrow l s^{2} \leq r s^{2} \Leftrightarrow$

${(a_{1} + a_{2})}^{2} + {(b_{1} + b_{2})}^{2} \leq (a_{1}^{2} + b_{1}^{2}) + 2 \sqrt{a_{1}^{2} + b_{1}^{2}} \cdot \sqrt{a_{2}^{2} + b_{2}^{2}} + (a_{2}^{2} + b_{2}^{2}) \Leftrightarrow$

$2 (\underset{: = (⋆)}{\underset{︸}{a_{1} a_{2} + b_{1} b_{2}}}) \leq 2 \cdot \underset{: = (⋆ ⋆)}{\underset{︸}{\sqrt{a_{1}^{2} + b_{1}^{2}} \cdot \sqrt{a_{2}^{2} + b_{2}^{2}}}} .$

Ist $(⋆) < 0,$ dann gilt offenbar die letzte Ungleichung und damit $l s \leq r s .$

Es sei jetzt $(⋆) \geq 0 .$ dann ist

$(⋆) \leq (⋆ ⋆) \Leftrightarrow {(⋆)}^{2} \leq {(⋆ ⋆)}^{2} \Leftrightarrow$

${(a_{1} a_{2} + b_{1} b_{2})}^{2} \leq (a_{1}^{2} + b_{1}^{2}) (a_{2}^{2} + b_{2}^{2}) \Leftrightarrow$

$a_{1}^{2} a_{2}^{2} + 2 a_{1} a_{2} b_{1} b_{2} + b_{1}^{2} b_{2}^{2} \leq a_{1}^{2} a_{2}^{2} + a_{1}^{2} b_{2}^{2} + b_{1}^{2} a_{2}^{2} + b_{1}^{2} b_{2}^{2} \Leftrightarrow$

$0 \leq {(a_{1} b_{2})}^{2} - 2 a_{1} b_{2} b_{1} a_{2} + {(b_{1} a_{2})}^{2} = {(a_{1} b_{2} - b_{1} a_{2})}^{2} .$

Damit gilt insgesamt $l s \leq r s .$

(6). Der Beweis hierzu erfolgt durch ähnliche Überlegungen. <mi
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