Triangle Inequality/Complex Numbers

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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 $z_1 + z_2$.


Triangle-Inequality-Complex.png


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

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


Sources