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Prove that $ |z_{1} - z_{2}| \le |z_{1}| + |z_{2}| $ for all $ z_{1},z_{2} \in \mathbb{C} $.

I don’t quite get how to deal with these triangle-inequality proofs, so any extra insight would be greatly appreciated.

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are you asking for a proof? –  Git Gud Feb 3 '13 at 21:24
    
Yes and insight on how to approach similar problems. –  DJ_ Feb 3 '13 at 21:26
    
Is it something like what Dimitri did that you were looking for? –  Git Gud Feb 3 '13 at 21:34
    
Yes i think so. –  DJ_ Feb 3 '13 at 21:37
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2 Answers 2

up vote 2 down vote accepted

Or similarly : $$z_1 = a_1 + i b_1, z_2 = a_2 + i b_2, \Rightarrow $$ $$ |z_1 - z_2 |^2 = |(a_1 - a_2) + i(b_1 - b_2)|^2 = (a_1 - a_2)^2 + (b_1 - b_2)^2 = a_1^2 + a_2^2 + b_1^2 + b_2^2 - 2a_1a_2 - 2b_1b_2 \leq ( |z_1|+|z_2| )^2 $$

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$$|z_1 - z_2| = |z_1 + (-z_2)| \leq |z_1| + |-z_2| = |z_1| + |z_2|$$

where the inequality is exactly the triangle inequality.

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