Triangle Inequality/Complex Numbers

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Theorem

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

Let $\left\vert{z}\right\vert$ be the modulus of $z$.


Then:

$\left\vert{z_1 + z_2}\right\vert \le \left\vert{z_1}\right\vert + \left\vert{z_2}\right\vert$


General Result

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

Let $\left\vert{z}\right\vert$ be the modulus of $z$.


Then:

$\left\vert{z_1 + z_2 + \cdots + z_n}\right\vert \le \left\vert{z_1}\right\vert + \left\vert{z_2}\right\vert + \cdots + \left\vert{z_n}\right\vert$


Proof

Let $z_1 = a_1 + i a_2, z_2 = b_1 + i b_2$.

Then from the definition of the modulus, the above equation translates into:

$\left({\left({a_1 + b_1}\right)^2 + \left({a_2 + b_2}\right)^2}\right)^{\frac 1 2} \le \left({a_1^2 + a_2^2}\right)^{\frac 1 2} + \left({b_1^2 + b_2^2}\right)^{\frac 1 2}$

This is a special case of Minkowski's Inequality, with $n = 2$.

$\blacksquare$


Sources