Triangle Inequality/Complex Numbers/Proof 2

From ProofWiki
Jump to navigation Jump to search

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

\(\ds \cmod {z_1 + z_2}\) \(\le\) \(\ds \cmod {z_1} + \cmod {z_2}\)
\(\ds \leadstoandfrom \ \ \) \(\ds \paren {\paren {a_1 + b_1}^2 + \paren {a_2 + b_2}^2}^{\frac 1 2}\) \(\le\) \(\ds \paren { {a_1}^2 + {a_2}^2}^{\frac 1 2} + \paren { {b_1}^2 + {b_2}^2}^{\frac 1 2}\) Definition of Complex Modulus
\(\ds \leadstoandfrom \ \ \) \(\ds \paren {a_1 + b_1}^2 + \paren {a_2 + b_2}^2\) \(\le\) \(\ds {a_1}^2 + {a_2}^2 + {b_1}^2 + {b_2}^2 + 2 \paren { {a_1}^2 + {a_2}^2}^{\frac 1 2} \paren { {b_1}^2 + {b_2}^2}^{\frac 1 2}\) squaring both sides
\(\ds \leadstoandfrom \ \ \) \(\ds {a_1}^2 + 2 a_1 b_1 + {b_1}^2 + {a_2}^2 + 2 a_2 b_2 + {b_2}^2\) \(\le\) \(\ds {a_1}^2 + {a_2}^2 + {b_1}^2 + {b_2}^2 + 2 \paren { {a_1}^2 + {a_2}^2}^{\frac 1 2} \paren { {b_1}^2 + {b_2}^2}^{\frac 1 2}\) multiplying out
\(\ds \leadstoandfrom \ \ \) \(\ds a_1 b_1 + a_2 b_2\) \(\le\) \(\ds \paren { {a_1}^2 + {a_2}^2}^{\frac 1 2} \paren { {b_1}^2 + {b_2}^2}^{\frac 1 2}\) simplifying
\(\ds \leadstoandfrom \ \ \) \(\ds \paren {a_1 b_1 + a_2 b_2}^2\) \(\le\) \(\ds \paren { {a_1}^2 + {a_2}^2} \paren { {b_1}^2 + {b_2}^2}\) Cauchy's Inequality
\(\ds \leadstoandfrom \ \ \) \(\ds {a_1}^2 {b_1}^2 + 2 a_1 b_1 a_2 b_2 + {a_2}^2 {b_2}^2\) \(\le\) \(\ds {a_1}^2 {b_1}^2 + {a_2}^2 {b_2}^2 + {a_1}^2 {b_2}^2 + {a_2}^2 {b_1}^2\)
\(\ds \leadstoandfrom \ \ \) \(\ds 2 a_1 b_1 a_2 b_2\) \(\le\) \(\ds {a_1}^2 {b_2}^2 + {a_2}^2 {b_1}^2\)
\(\ds \leadstoandfrom \ \ \) \(\ds 0\) \(\le\) \(\ds \paren {a_1 b_2}^2 - 2 \paren {a_1 b_2} \paren {a_2 b_1} + \paren {a_2 b_1}^2\)
\(\ds \leadstoandfrom \ \ \) \(\ds 0\) \(\le\) \(\ds \paren {a_1 b_2 - a_2 b_1}^2\)

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

Hence the result.

$\blacksquare$


Sources