Complex number inequality verification 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?
 A: 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 $\leq 1.$
A: 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}
