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 $$|z+w|\le |z|+|w|$$

Proof:

$$\begin{array}{lll}|z+w{|}^2& =(z+w)(\overline{z+w})& (\because |z{|}^2=z\cdot \bar{z})\\ & =(z+w)(\bar{z}+\bar{w})& (\because \overline{z+w}=\bar{z}+\bar{w})\\ & =z\cdot \bar{z}+z\cdot \bar{w}+w\cdot \bar{z}+w\cdot \bar{w}\\ & =|z{|}^2+z\cdot \bar{w}+\bar{z}\cdot w+|w{|}^2\\ & =|z{|}^2+z\cdot \bar{w}+\overline{z\cdot \stackrel{―}{w}}+|w{|}^2& (\because \overline{\stackrel{―}{z}}=z)\\ & =|z{|}^2+2\text{Re}(z\cdot \bar{w})+|w{|}^2& (\because \frac{1}{2}(z+\bar{z})=\text{Re}(z))\\ & \le |z{|}^2+2|z\cdot \bar{w}|+|w{|}^2& (\because \text{Re}(z)\le |z|)\\ & =|z{|}^2+2|z||\bar{w}|+|w{|}^2\\ & =|z{|}^2+2|z||w|+|w{|}^2& (\because |z|=|\bar{z}|)\\ & =(|z|+|w|{)}^2\\ \therefore |z+w{|}^2& \le (|z|+|w|{)}^2\\ \therefore |z+w|& \le |z|+|w|\end{array}$$

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

Hence, $$\begin{array}{lll}& |z|+|w|& \ge |z+w|\\ \therefore & |z+w|& \le |z|+|w|\end{array}$$

Graphical Representation of Triangle Inequality

If $z$ and $w$ are two complex numbers, then from Triangle Inequality, we have $$|z+w|\le |z|+|w|$$ 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, $$|z+w|\le |z|+|w|$$

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