# Inequality related to modulus of complex numbers

How to show: for any $\alpha >0$, there is constant depends on $\alpha$, say $C=C(\alpha)$, such that,

$$\mid \mid w \mid ^{\alpha} w - \mid z \mid ^{\alpha} z \mid \leq C (\mid w \mid ^{\alpha} + \mid {z}\mid ^{\alpha}) \mid w -z \mid$$

for all $z, w \in \mathbb C$

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It seems the following.

If we wish to show the existence of the required number $C(\alpha)$, it suffices to find a number $D=D(\alpha)\ge 1$ such that

(*) $$\mid \mid w \mid ^{\alpha} w - \mid z \mid ^{\alpha} z \mid^2 \leq D (\mid w \mid ^{\alpha} + \mid {z}\mid ^{\alpha})^2 \mid w -z \mid^2$$ for all $z, w \in \mathbb C$.

We shall use the geometric interpretation of complex numbers. Let $w$ be a vector of length $r$, $z$ be a vector of length $s$, and $\varphi$ be the angle between these vectors. Then by Law of cosines, $$|w-z|^2=r^2+s^2-2rs\cos\varphi,$$ $$||w|^\alpha w-|z|^\alpha z|^2=r^{2+2\alpha}+s^{2+2\alpha}-2r^{1+\alpha}s^{1+\alpha}\cos\varphi.$$

By routine equivalent transformations we can transform the inequality (*) to the form:

$$(D-1)r^{2+2\alpha}+2Dr^{2+\alpha}s^\alpha+Dr^2s^{2\alpha}+ Dr^{2\alpha}s^2+2Dr^\alpha s^{2+\alpha}+(D-1)s^{2+2\alpha}\ge$$ $$(2Dr^{1+2\alpha}s+ 2(D-1)r^{1+\alpha}s^{1+\alpha}+2Drs^{1+2\alpha}s)\cos\varphi.$$

It suffices to prove this inequality only for the case $\varphi=1$. But in this case the vectors $w$ and $z$ are collinear, and therefore it suffice to prove the initial inequality only for the case $w,z\in\mathbb R_+$ and some $C=C(\alpha)\ge 1$.

Without loss of generality, we can assume that $w\ge z$. Then we have to prove that $$f(w,z)=(C-1)w^{1+\alpha}+Cwz^\alpha-Cw^\alpha z-(C-1)z^{1+\alpha}\ge 0.$$ If $w=z$ then $f(w,z)=0$. Moreover, $$\frac{\partial f}{\partial w}= (C-1)(1+\alpha)w^\alpha+Cz^\alpha-C\alpha w^{\alpha-1}z\ge Cz^\alpha\ge 0$$ provided $C\ge 1+\alpha$. So it suffices to take $C(\alpha)=1+\alpha.$

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