# Triangle Inequality/Complex Numbers

## Theorem

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

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

Then:

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

### General Result

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 1

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, with $n = 2$.

$\blacksquare$

## Proof 2

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

 $\displaystyle \cmod {z_1 + z_2}$ $\le$ $\displaystyle \cmod {z_1} + \cmod {z_2}$ $\displaystyle \leadstoandfrom \ \$ $\displaystyle \paren {\paren {a_1 + b_1}^2 + \paren {a_2 + b_2}^2}^{\frac 1 2}$ $\le$ $\displaystyle \paren { {a_1}^2 + {a_2}^2}^{\frac 1 2} + \paren { {b_1}^2 + {b_2}^2}^{\frac 1 2}$ Definition of Complex Modulus $\displaystyle \leadstoandfrom \ \$ $\displaystyle \paren {a_1 + b_1}^2 + \paren {a_2 + b_2}^2$ $\le$ $\displaystyle {a_1}^2 + {a_2}^2 + {b_1}^2 + {b_2}^2 + 2 \paren { {a_1}^2 + {a_2}^2}^{\frac 1 2} \paren { {b_1}^2 + {b_2}^2}^{\frac 1 2}$ squaring both sides $\displaystyle \leadstoandfrom \ \$ $\displaystyle {a_1}^2 + 2 a_1 b_1 + {b_1}^2 + {a_2}^2 + 2 a_2 b_2 + {b_2}^2$ $\le$ $\displaystyle {a_1}^2 + {a_2}^2 + {b_1}^2 + {b_2}^2 + 2 \paren { {a_1}^2 + {a_2}^2}^{\frac 1 2} \paren { {b_1}^2 + {b_2}^2}^{\frac 1 2}$ multiplying out $\displaystyle \leadstoandfrom \ \$ $\displaystyle a_1 b_1 + a_2 b_2$ $\le$ $\displaystyle \paren { {a_1}^2 + {a_2}^2}^{\frac 1 2} \paren { {b_1}^2 + {b_2}^2}^{\frac 1 2}$ simplifying $\displaystyle \leadstoandfrom \ \$ $\displaystyle \paren {a_1 b_1 + a_2 b_2}^2$ $\le$ $\displaystyle \paren { {a_1}^2 + {a_2}^2} \paren { {b_1}^2 + {b_2}^2}$ squaring both sides $\displaystyle \leadstoandfrom \ \$ $\displaystyle {a_1}^2 {b_1}^2 + 2 a_1 b_1 a_2 b_2 + {a_2}^2 {b_2}^2$ $\le$ $\displaystyle {a_1}^2 {b_1}^2 + {a_2}^2 {b_2}^2 + {a_1}^2 {b_2}^2 + {a_2}^2 {b_1}^2$ $\displaystyle \leadstoandfrom \ \$ $\displaystyle 2 a_1 b_1 a_2 b_2$ $\le$ $\displaystyle {a_1}^2 {b_2}^2 + {a_2}^2 {b_1}^2$ $\displaystyle \leadstoandfrom \ \$ $\displaystyle 0$ $\le$ $\displaystyle \paren {a_1 b_2}^2 - 2 \paren {a_1 b_2} \paren {a_2 b_1} + \paren {a_2 b_1}^2$ $\displaystyle \leadstoandfrom \ \$ $\displaystyle 0$ $\le$ $\displaystyle \paren {a_1 b_2 - a_2 b_1}^2$

The final statement is a tautology, and all implications are reversible.

Hence the result.

$\blacksquare$

## Proof 3

Let $z_1$ and $z_2$ be represented by the points $A$ and $B$ respectively in the complex plane.

From Geometrical Interpretation of Complex Addition, we can construct the parallelogram $OACB$ where:

$OA$ and $OB$ represent $z_1$ and $z_2$ respectively
$OC$ represents $z_1 + z_2$. As $OACB$ is a parallelogram, we have that $OB = AC$.

The lengths of $OA$, $AC$ and $OC$ are:

 $\displaystyle OA$ $=$ $\displaystyle \cmod {z_1}$ $\displaystyle AC$ $=$ $\displaystyle \cmod {z_2}$ $\displaystyle OC$ $=$ $\displaystyle \cmod {z_1 + z_2}$

But $OA$, $OB$ and $OC$ form the sides of a triangle.

The result then follows directly from Sum of Two Sides of Triangle Greater than Third Side.

$\blacksquare$

## Examples

### 3 Arguments

For all $z_1, z_2, z_3 \in \C$:

$\cmod {z_1 + z_2 + z_3} \le \cmod {z_1} + \cmod {z_2} + \cmod {z_3}$