Using geometric considerations, prove the following…
Using geometric considerations, prove the following inequalities: $$ 1^{\circ}\,\, |z_1+z_2|\leq|z_1|+|z_2|; \quad 2^{\circ}\,\, |z_1-z_2|\geq\big||z_1|-|z_2|\big|; \quad 3^{\circ}\,\, |z-1|\leq\big||z|-1\big|+|z||\arg(z)|. $$
Explanation of Method
All three inequalities follow from the same geometric fact: in a triangle, one side is never longer than the sum of the other two, and never shorter than their difference (a straight path is the shortest connector between two points). Interpreting $z_1$, $z_2$, and $z_1\pm z_2$ as vertices of a triangle in the complex plane, and $|z-1|$ as a chord compared against a radial-then-arc path, turns each algebraic inequality into this one geometric principle.
Part $1^{\circ}$ · Triangle inequality $|z_1+z_2|\leq|z_1|+|z_2|$
Step 1 · Vector interpretation
Let $z_1$ be the vector from $O$ to $A$, and $z_2$ the vector from $A$ to $C$ (i.e. $z_2$ translated to start at $A$). Then $C$ is at position $z_1+z_2$.
Step 2 · The triangle $OAC$
Triangle $OAC$ has sides $OA=|z_1|$, $AC=|z_2|$, and $OC=|z_1+z_2|$.
Triangle $OAC$: side $OC$ (the direct path) is never longer than $OA+AC$.
Step 3 · Apply the triangle inequality
In any triangle, one side is at most the sum of the other two: $|OC|\leq|OA|+|AC|$. Hence:
$$ \boxed{|z_1+z_2|\leq|z_1|+|z_2|} $$
Step 4 · Equality condition
Equality holds exactly when $O$, $A$, $C$ are collinear with $A$ between $O$ and $C$ — that is, when $z_1$ and $z_2$ point in the same direction: $\arg(z_1)=\arg(z_2)$, equivalently $z_2=tz_1$ for some $t>0$.
Step 5 · Interpretation
The direct distance from $O$ to $z_1+z_2$ is at most the distance via the intermediate point $z_1$ — a straight line is the shortest path between two points.
Part $2^{\circ}$ · Reverse triangle inequality $|z_1-z_2|\geq\big||z_1|-|z_2|\big|$
Step 1 · Vector interpretation
Let $z_1$ be the vector $O\to A$ and $z_2$ the vector $O\to B$. Then $z_1-z_2$ is the vector $B\to A$, so triangle $OAB$ has sides $OA=|z_1|$, $OB=|z_2|$, $BA=|z_1-z_2|$.
Triangle $OAB$: side $BA$ is at least the difference of sides $OA$ and $OB$.
Step 2 · Apply the triangle inequality twice
From triangle $OAB$: $|z_1|\leq|z_2|+|z_1-z_2| \Rightarrow |z_1|-|z_2|\leq|z_1-z_2|$.
Symmetrically: $|z_2|\leq|z_1|+|z_1-z_2| \Rightarrow -(|z_1|-|z_2|)\leq|z_1-z_2|$.
Step 3 · Combine
Both $|z_1|-|z_2|$ and its negative are $\leq|z_1-z_2|$, so:
$$ \boxed{|z_1-z_2|\geq\big||z_1|-|z_2|\big|} $$
Step 4 · Geometric meaning and equality
The distance between two points is at least the difference of their distances from the origin. Equality holds when $O$, $A$, $B$ are collinear with $A$, $B$ on the same ray from $O$, i.e. $z_1=tz_2$ for some $t>0$ (equivalently $\arg(z_1)=\arg(z_2)$); otherwise the inequality is strict.
Part $3^{\circ}$ · $|z-1|\leq\big||z|-1\big|+|z||\arg(z)|$
Step 1 · Polar form
Write $z=re^{i\theta}$ with $r=|z|$, $\theta=\arg(z)$. Consider the path from the point $1$ to $z$ that goes first radially from $1$ to $r$ along the real axis (length $|r-1|$), then along the circle of radius $r$ from angle $0$ to angle $\theta$ (arc length $r|\theta|$).
Radial step (green) plus arc step (purple) form a path from $1$ to $z$; the direct chord (red) is never longer than this path.
Step 2 · Apply the shortest-path principle
The straight chord from $1$ to $z$ is at most the length of any path connecting them, in particular this radial-then-arc path:
$$ \boxed{|z-1|\leq\big||z|-1\big|+|z||\theta|=\big||z|-1\big|+|z||\arg(z)|} $$
Step 3 · Why the arc bounds the chord on the circle
The remaining piece to justify is that the chord from $|z|$ to $z$ (both at radius $r=|z|$) is no longer than the arc between them. Using $|e^{i\theta}-1| = 2\left|\sin\frac{\theta}{2}\right|$ and the standard bound $|\sin t|\leq|t|$:
$$ |z-|z|\,| = r\,|e^{i\theta}-1| = 2r\left|\sin\frac{\theta}{2}\right| \leq r|\theta| $$
so the chord-on-circle length never exceeds the arc length $r|\theta|$, which is exactly the second term of the bound.
Step 4 · Assemble via the triangle inequality
Apply the ordinary triangle inequality to the path $1\to|z|\to z$:
$$ |z-1| \leq \big|\,|z|-1\,\big| + |z-|z|\,| \leq \big|\,|z|-1\,\big| + |z||\theta| $$
which is precisely the claimed inequality.
| Part | Inequality | Equality holds when |
|---|---|---|
| $1^{\circ}$ | $|z_1+z_2|\leq|z_1|+|z_2|$ | $z_1,z_2$ have the same argument |
| $2^{\circ}$ | $|z_1-z_2|\geq\big||z_1|-|z_2|\big|$ | $z_1,z_2$ lie on the same ray from $O$ |
| $3^{\circ}$ | $|z-1|\leq\big||z|-1\big|+|z||\arg z|$ | $\arg(z)=0$ (i.e. $z$ real, $z=r\geq0$) |
Result
All three inequalities are consequences of a single geometric fact — one side of a triangle (or a chord) is never longer than a path made of the other two sides (radial step + arc):