Complex numbers proof problem If $|z|,|w| \leq 1$, show that 
$|z-w|^2 \leq (|z|-|w|)^2 + (\arg(z)-\arg(w))^2$, where $z,w$ are complex numbers.
How can I solve such a problem?
 A: Let $z=r_1e^{\varphi_1},w=r_2e^{\varphi_2}$. We need to prove $$|r_1\cos\varphi_1-r_2\cos\varphi_2+i(r_1\sin\varphi_1-r_2\sin\varphi_2)|\leq (r_1-r_2)^2+(\varphi_1-\varphi_2)^2\iff$$$$ r_1^2+r_2^2-2r_1r_2\cos(\varphi_1-\varphi_2)\leq r_1^2+r_2^2-2r_1r_2+(\varphi_1-\varphi_2)^2\iff$$$$r_1r_2(1-\cos(\varphi_1-\varphi_2))\leq\frac{(\varphi_1-\varphi_2)^2}{2}$$ Since $r_1,r_2\leq 1$ we obtain $r_1r_2(1-\cos(\varphi_1-\varphi_2))\leq1-\cos(\varphi_1-\varphi_2)$. $1-\cos x\leq\frac{x^2}{2}$ is a well-known inequality wich can be proven using derivatives, substituting $x=\varphi_1-\varphi_2$ we obtain the desired inequality. 
A: Let $z=r_1(cos\theta_1+isin\theta_1)$ and $w=r_2(cos\theta_2+isin\theta_2)$. Then $arg(z)=\theta_1$, $arg(w)=\theta_2$. It suffices to prove $-2r_1r_2cos(\theta_1-\theta_2)\leq-2r_1r_2+(\theta_1-\theta_2)^2$. To prove this, just need to prove $2r_1r_2(1-cos(\theta_1-\theta_2))\leq(\theta_1-\theta_2)^2$.Since $r_1\leq 1, r_2\leq 1$, we just need to prove $2(1-cos(\theta_1-\theta_2))\leq(\theta_1-\theta_2)^2$. This is obvious since it's easy to prove $2(1-cosx)\leq x^2.$ We finish the proof.
A: Hint:
Let $z = (a+bi)$ and $w= (c+di)$ and work with the LHS !!
