# Triangle Inequality Equality Conditions

I am looking for the conditions on two complex numbers $z_1$ and $z_2$ such that $$|z_1+z_2|=|z_1|+|z_2|$$ Letting $z_n=a_n+ib_n$ and using $|z_n|^2=a_n^2+b_n^2$ yields $$(a_1+a_2)^2+(b_1+b_2)^2=a_1^2+b_1^2+2\sqrt{(a_1^2+b_1^2)(a_2^2+b_2^2)}+a_2^2+b_2^2$$ Expanding and simplifying gives $$a_1a_2+b_1b_2=\sqrt{(a_1^2+b_1^2)(a_2^2+b_2^2)}$$ Squaring, expanding, and simplifying gives $$2a_1a_2b_1b_2=a_1^2b_2^2+a_2^2b_1^2$$ Rearranging gives $$(a_1b_2-a_2b_1)^2=0$$ Then $a_1b_2=a_2b_1$ I think this is the condition found algebraically. Intuitively, thinking geometrically, those complex numbers when adding should have the same direction to yield equality. Are there any other conditions that need to be met, or are there other methods other than algebraically or geometrically that ealisy show these conditions?

## 1 Answer

Consider $z_1,z_2$ as vectors in the complex plane. Then $z_1+z_2$ is their vector sum, and so $|z_1|,|z_2|,|z_1+z_2|$ are three sides of a triangle (which is possible degenerate - having two angles of $0$), and by the Cosine Rule:

$$|z_1+z_2|^2=|z_1|^2+|z_2|^2-2|z_1||z_2|\cos\theta \tag{1}$$ where $\theta$ is the supplement of the angle between $z_1$ and $z_2$ (or is the angle between $-z_1$ and $z_2$).

Now by the condition stated in the question

$$|z_1+z_2|=|z_1|+|z_2| \implies |z_1+z_2|^2=|z_1|^2+|z_2|^2+2|z_1||z_2|$$

and by comparison with (1) we have

$$\cos\theta=-1 \implies \theta=\pi$$

So the angle between the two vectors must be $0$, and so:

• either at least one of $z_1,z_2$ is zero, or
• $z_2=k \cdot z_1$ with $k$ a positive real number