# Triangle Inequality

An important property of absolute values of complex numbers is Triangle Inequality. Triangle Inequality is a relation relating complex numbers with addition of absolute values. If $z$ and $w$ are any two complex numbers, then $|z+w|≤|z|+|w|$

Proof:

$\begin{array}{l} |z+w|^2 &=(z+w)(\overline{z+w}) &(\because |z|^2=z\cdot\overline{z}) \\ &=(z+w)(\overline{z}+\overline{w}) &(\because \overline{z+w}=\overline{z}+\overline{w}) \\ &= z\cdot\overline{z} + z\cdot\overline{w} + w\cdot\overline{z}+w\cdot\overline{w} \\ &= |z|^2+z\cdot\overline{w} +\overline{z}\cdot w+|w|^2 \\ &= |z|^2+z\cdot\overline{w}+\overline{z\cdot \overline{w}} + |w|^2 &(\because \overline{\overline{z}}=z) \\ &= |z|^2+2\text{Re}(z\cdot\overline{w})+|w|^2 &\left(\because\frac{1}{2}(z+\overline{z})=\text{Re}(z)\right) \\ &≤ |z|^2+2|z\cdot\overline{w}|+|w|^2 &(\because\text{Re}(z)≤|z|) \\ &\;\;\;\;= |z|^2+2|z||\overline{w}|+|w|^2 \\ &\;\;\;\; =|z|^2+2|z||w|+|w|^2 &(\because |z|=|\overline{z}|) \\ &\;\;\;\; =(|z|+|w|)^2 \\ \therefore |z+w|^2 &≤(|z|+|w|)^2 \\ \therefore |z+w| &≤ |z|+|w| \end{array}$

Alternative Method:

Let $z=a+ib$ and $w=c+id$ then $z=\sqrt{a^2+b^2}$ and $w=\sqrt{c^2+d^2}$. Also, $z+w=(a+c)+i(b+d)$ $\therefore|z+w|=\sqrt{(a+c)^2+(b+d)^2}$

Now, $|z|+|w|≥|z+w|$ will be true if $\begin{array}{l} & \sqrt{a^2+b^2}+\sqrt{c^2+d^2} &≥ \sqrt{(a+c)^2+(b+d)^2} \\ \text{i.e.} & a^2+b^2+c^2+d^2+2\sqrt{(a^2+b^2)(c^2+d^2)} &≥ (a+c)^2+(b+d)^2 \\ \text{i.e.} & \sqrt{(a^2+b^2)(c^2+d^2)} &≥ ac+bd \\ \text{i.e.} & (a^2+b^2)(c^2+d^2) &≥ a^2c^2+b^2d^2+2abcd \\ \text{i.e.} & a^2d^2+b^2c^2 &≥ 2abcd \\ \text{or,} & a^2d^2+b^2c^2-2abcd &≥ 0 \\ \text{i.e.} & (ad-bc)^2 &≥ 0 \end{array}$ which is true for all real numbers $a$, $b$, $c$ and $d$.

Hence, $\begin{array}{l} & |z|+|w| &≥|z+w| \\ \therefore & |z+w| &≤ |z|+|w| \end{array}$

## Graphical Representation of Triangle Inequality

If $z$ and $w$ are two complex numbers, then from Triangle Inequality, we have $|z+w|≤|z|+|w|$ One can see this from the parallelogram law for addition. Consider a triangle whose vertices are $0$, $z$ and $w$.

One side of the triangle from $0$ to $z+w$ has length $|z+w|$. Second side of the triangle from $0$ to $z$ has length $|z|$. And, the third side of the triangle, from $z$ to $z+w$, is parallel and equal to the line from $0$ to $w$, and therefore has length $|w|$. Now, in any triangle, any one side is less than or equal to the sum of the other two sides. Hence, $|z+w|≤|z|+|w|$