This is exercise 4.(b) from Theory of Complex Functions, GTM 122, page 17. I was able to solve exercise 4.(a), but I don't see how this could help solving 4.(b):
a) Show that from $(1+\lvert v\rvert^2) u=(1+\lvert u\rvert^2)v,\,u,v\in\mathbb{C}$, it follows that either $u=v$ or $\bar{u}v=1$.
b) Show that for $u,v\in\mathbb{C}$ with $\lvert u\rvert<1, \lvert v\rvert<1$ and $\bar{u}v\ne u\bar{v}$, we always have $$\lvert(1+\lvert u\rvert^2) v-(1+\lvert v\rvert^2)u\rvert > \lvert u\bar{v}-\bar{u}v\rvert$$
For a), I took modulus of both sides and get either $\lvert u\rvert=\lvert v\rvert$ or $\lvert u\rvert \lvert v\rvert=1$, where the conclusion follows easily. For b), I tried substituting $u=x+iy$ and $v=p+iq$ but the resulting inequality is too complicated and it's difficult to see where the condition $\lvert u\bar{v}-\bar{u}v\rvert=2\lvert xq-yp\rvert\ne 0$ is used. Any advice?
For Part (a), we note that $\big(1+|v|^2\big)\,u=\big(1+|u|^2\big)\,v$ implies $\big(1+|v|^2\big)\,|u|=\big(1+|u|^2\big)\,|v|$. That is, $$\big(|u|-|v|\big)\,\big(1-|u|\,|v|\big)=0\,.$$ If $|u|=|v|$, then we obtain $u=v$. If $u\neq v$, then $|u|\,|v|=1$, whence $u\neq 0$ and $v=\dfrac{1}{|u|^2}\,u$. Ergo, $$\bar{u}\,v=\bar{u}\,\left(\frac{1}{|u|^2}\,u\right)=\frac{|u|^2}{|u|^2}=1\,.$$
For Part (b), we shall prove that $$\Big|\big(1+|v|^2\big)\,u-\big(1+|u|^2\big)\,v\Big|\geq \big|u\,\bar{v}-\bar{u}\,v\big|\tag{*}$$ for all $u,v\in\mathbb{C}$, and the equality occurs if and only if $u=v$. If $u=0$ or $v=0$, then (*) trivially holds. From now on, we assume that $u\neq 0$ and $v\neq 0$.
Observe that $$\begin{align}\Big|\big(1+|v|^2\big)\,u-\big(1+|u|^2\big)\,v\Big|^2&=\Big(\big(1+|v|^2\big)\,u-\big(1+|u|^2\big)\,v\Big)\,\Big(\big(1+|v|^2\big)\,\bar{u}-\big(1+|u|^2\big)\,\bar{v}\Big) \\ &=\big(1+|v|^2\big)^2\,|u|^2+\big(1+|u|^2\big)^2\,|v|^2-2\,\big(1+|u|^2\big)\,\big(1+|v|^2\big)\,\text{Re}\big(u\,\bar{v}\big)\,. \end{align}$$ Using the AM-GM Inequality on $\big(1+|v|^2\big)^2\,|u|^2+\big(1+|u|^2\big)^2\,|v|^2$, we obtain \begin{align}\Big|\big(1+|v|^2\big)\,u-\big(1+|u|^2\big)\,v\Big|^2&\geq2\,\big(1+|u|^2\big)\,\big(1+|v|^2\big)\,|u|\,|v|-2\,\big(1+|u|^2\big)\,\big(1+|v|^2\big)\,\text{Re}\big(u\,\bar{v}\big)\\ &= 2\,\big(1+|u|^2\big)\,\big(1+|v|^2\big)\,\Big(|u|\,|v|-\text{Re}\big(u\,\bar{v}\big)\Big)\,. \end{align} Because $\big(\text{Im}(z)\big)^2=z^2-\big(\text{Re}(z)\big)^2$ for every $z\in\mathbb{C}$, we get \begin{align}\Big|\big(1+|v|^2\big)\,u-\big(1+|u|^2\big)\,v\Big|^2&\geq2\,\left(\frac{\big(1+|u|^2\big)\,\big(1+|v|^2\big)}{|u|\,|v|+\text{Re}\big(u\,\bar{v}\big)}\right)\,\Big(\text{Im}\big(u\,\bar{v}\big)\Big)^2 \\ &\geq 2\,\left(\frac{\big(1+|u|^2\big)\,\big(1+|v|^2\big)}{2\,|u|\,|v|}\right)\,\Big(\text{Im}\big(u\,\bar{v}\big)\Big)^2\,, \end{align} where we have applied the trivial inequality $\text{Re}(z)\leq |z|$ for each $z\in\mathbb{C}$. Finally, using the AM-GM Inequality $t+\dfrac1t\geq 2$ for all $t>0$, we get \begin{align} \Big|\big(1+|v|^2\big)\,u-\big(1+|u|^2\big)\,v\Big|^2&\geq\left(|u|+\frac{1}{|u|}\right)\,\left(|v|+\frac{1}{|v|}\right)\,\Big(\text{Im}\big(u\,\bar{v}\big)\Big)^2 \\ &\geq 4\,\Big(\text{Im}\big(u\,\bar{v}\big)\Big)^2=\big|u\,\bar{v}-\bar{u}\,v\big|^2\,,\end{align} and the claim is proven. (The equality case is easy to verify.)