Triangle Inequality/Complex Numbers
Contents
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 $z_1, z_2, z_3 \in \C$:
- $\cmod {z_1 + z_2 + z_3} \le \cmod {z_1} + \cmod {z_2} + \cmod {z_3}$
Sources
- 1981: Murray R. Spiegel: Theory and Problems of Complex Variables (SI ed.) ... (previous) ... (next): $1$: Complex Numbers: Absolute Value: $3$