Triangle Inequality/Complex Numbers/Proof 1

Theorem

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

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

Then:

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

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:

$\paren {\paren {a_1 + b_1}^2 + \paren {a_2 + b_2}^2}^{\frac 1 2} \le \paren { {a_1}^2 + {a_2}^2}^{\frac 1 2} + \paren { {b_1}^2 + {b_2}^2}^{\frac 1 2}$

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

$\blacksquare$