Chapter 14.4, Problem 32WE

To determine

To show: $\left|w+z\right|\le \left|w\right|+\left|z\right|$

Explanation of Solution

Given information:

For two complex numbers, $\left|w+z\right|\le \left|w\right|+\left|z\right|$ .

Calculation:

Let us consider two complex numbers $w\text{ and }z$ represented by the points $A\text{ and }B$ in a complex plane as below.

  

Using geometrical complex addition, a parallelogram $OACB$ is constructed where $OA\text{ and }OB$ represents $w\text{ and }z$ and $OC$ represents $w+z$ .

Since $OACB$ is a parallelogram, $OB=AC=z$ .

So, $\left|OA\right|=\left|w\right|,\left|AC\right|=\left|z\right|,\left|OC\right|=|w+z|.$

Using properties of triangle, sum of two sides of triangle is always greater than the third side.

Hence, $\left|w+z\right|<\left|w\right|+\left|z\right|$ .

Further when two complex numbers are on complex axes. Sum of magnitude of two complex number will be less than or equal to magnitude of addition of two complex number.

  

Hence, in every case $\left|w+z\right|\le \left|w\right|+\left|z\right|$