Triangle Inequality/Complex Numbers/General Result

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Theorem

Let $z_1, z_2, \dotsc, z_n \in \C$ be complex numbers.

Let $\cmod z$ be the modulus of $z$.


Then:

$\cmod {z_1 + z_2 + \dotsb + z_n} \le \cmod {z_1} + \cmod {z_2} + \dotsb + \cmod {z_n}$


Proof

Proof by induction:

For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:

$\cmod {z_1 + z_2 + \dotsb + z_n} \le \cmod {z_1} + \cmod {z_2} + \dotsb + \cmod {z_n}$


$\map P 1$ is true by definition of the usual ordering on real numbers:

$\cmod {z_1} \le \cmod {z_1}$


Basis for the Induction

$\map P 2$ is the case:

$\cmod {z_1 + z_2} \le \cmod {z_1} + \cmod {z_2}$

which has been proved in Triangle Inequality for Complex Numbers.

This is our basis for the induction.


Induction Hypothesis

Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.


So this is our induction hypothesis:

$\cmod {z_1 + z_2 + \dotsb + z_k} \le \cmod {z_1} + \cmod {z_2} + \dotsb + \cmod {z_k}$


Then we need to show:

$\cmod {z_1 + z_2 + \dotsb + z_{k + 1} } \le \cmod {z_1} + \cmod {z_2} + \dotsb + \cmod {z_{k + 1} }$


Induction Step

This is our induction step:


\(\ds \cmod {z_1 + z_2 + \dotsb + z_{k + 1} }\) \(=\) \(\ds \cmod {\paren {z_1 + z_2 + \dotsb + z_k} + z_{k + 1} }\) Definition of Indexed Summation
\(\ds \) \(\le\) \(\ds \cmod {z_1 + z_2 + \dotsb + z_k} + \cmod {z_{k + 1} }\) Basis for the Induction
\(\ds \) \(\le\) \(\ds \paren {\cmod {z_1} + \cmod {z_2} + \dotsb + \cmod {z_k} } + \cmod {z_{k + 1} }\) Induction Hypothesis
\(\ds \) \(\le\) \(\ds \cmod {z_1} + \cmod {z_2} + \dotsb + \cmod {z_k} + \cmod {z_{k + 1} }\) Definition of Indexed Summation


So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.

$\blacksquare$


Also see


Sources

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