Complex number inequality verification

Question

If $|z| \leq 1$ and $|w| \leq 1$, show that $$ |z-w|^{2} \leq(|z|-|w|)^{2}+(\operatorname{Arg} z-\operatorname{Arg} w)^{2}. $$ I would like to know the meaning of the inequality here; does it mean that for any specific $z$ and $w$ having modulus less than $1$, we always have a positive difference between $(|z|-|w|)^2 +\operatorname{Arg} z - \operatorname{Arg} w$ and $|z-w|^2$? Except the case where it's zero when the $\operatorname{Arg} z = \operatorname{Arg} w$? Shouldn't we also give the condition that $z$ is not equal to zero? As $\operatorname{Arg}(0)$ is not defined? Also is it correct to say inequality is definitely correct by just saying max of LHS can be $4$ but max in RHS can be $1 +4\pi^2$ so as $4\pi^2 +1 >4$ hence inequality is correct?

Answer 1

This is a convoluted application of:

• Taylor Series for Cosine

• Law of Cosines.

• Inequality around convergent alternating series (discussed below).

• The fact that $|z|,|w| \leq 1.$

---

First consider how the Law of Cosines applies.

Consider the triangle formed by the origin, and the vertices $z,w$. Here, since the assertion refers to Arg$(z)$ and Arg$(w)$, you can assume that these $3$ vertices are $3$ distinct points.

Edit
The case of $z = w \neq [0 + i(0)]$ may be discounted because it forces the assertion to automatically be true. This is because then, the LHS of the assertion would be equal to $0$, while the RHS of the assertion would be equal to the sum of two squares.

Let the side of the triangle opposite the origin be the side $c$ in the Law of Cosines.

Let the angle between the other two sides be designated as $\theta$.

Then, the equation

$$c^2 = a^2 + b^2 - 2ab\cos(\theta)$$

may be re-expressed as

$$|z - w|^2 = |z|^2 + |w|^2 - 2|z| ~|w| \cos(\theta). \tag1 $$

Therefore, the assertion has been reduced to demonstrating that the RHS of (1) above is

$$\leq |z|^2 + |w|^2 - 2|z| ~|w| + (\theta)^2. \tag2 $$

The first two terms in the Taylor Series for $\cos(\theta)$ are

$$1 - \frac{(\theta)^2}{2}. \tag3 $$

There is a rule that applies to convergent alternating series that says that (in effect)

$$\cos(\theta) \geq 1 - \frac{(\theta)^2}{2}. \tag4 $$

Edit
For what it's worth, I avoided asserting strict inequality in (4) above, because you could have $\theta = 0$.

So, the challenge will be to use the principle in (4) above to show that the RHS of (1) above implies the assertion in (2) above.

From (4), you know that

$$ |z - w|^2 \leq |z|^2 + |w|^2 - 2|z| ~|w| \left[1 - \frac{(\theta)^2}{2}\right]. \tag5 $$

The RHS of (5) above can be re-expressed as

$$|z|^2 + |w|^2 - 2|z| ~|w| $$

$$+ 2|z|~|w| \frac{\left[\text{Arg}(z) - \text{Arg}(w)\right]^2}{2}. \tag6 $$

Putting this all together, you can conclude that

$$|z - w|^2 \leq \left[~|z| - |w| ~\right]^2 + |z| ~|w| \left[\text{Arg}(z) - \text{Arg}(w)\right]^2. \tag7 $$

Consequently, the desired inequality is implied by (7) above, since $|z|,|w|$ are each $\leq 1.$

Answer 2

This is essentially @user2661923's answer but sidesteps the $\operatorname{arg}$ issue by using polar notation.

Let $z=r_1 e^{i \theta_1}, w=r_2 e^{i \theta_2} $, with $r_k \in [0,1]$. Note that $f(x) = {1 \over 2} x^2+\cos x$ has a (global) $\min$ of $1$ at $x=0$ and so $\cos x \ge 1 - {1 \over 2} x^2$ for all $x$.

We have \begin{eqnarray} |r_1e^{i \theta_1} - r_2 e^{i \theta_2} |^2 &=& r_1^2+r_2^2 - 2 r_1 r_2 \cos (\theta_1-\theta_2) \\ &\le & r_1^2+r_2^2 + 2 r_1 r_2 ({1 \over 2} (\theta_1-\theta_2)^2 -1 ) \\ &=& (r_1-r_2)^2 + r_1 r_1 (\theta_1-\theta_2)^2 \\ &\le& (r_1-r_2)^2 + (\theta_1-\theta_2)^2 \end{eqnarray}