Vtyjkyuk

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 $(1+lvert vrvert^2) u=(1+lvert urvert^2)v,,u,vinmathbbC$, it follows that either $u=v$ or $baruv=1$.

b) Show that for $u,vinmathbbC$ with $lvert urvert<1, lvert vrvert<1$ and $baruvne ubarv$, we always have $$lvert(1+lvert urvert^2) v-(1+lvert vrvert^2)urvert > lvert ubarv-baruvrvert$$

For a), I took modulus of both sides and get either $lvert urvert=lvert vrvert$ or $lvert urvert lvert vrvert=1$, 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 $lvert ubarv-baruvrvert=2lvert xq-yprvertne 0$ is used. Any advice?

For Part (a), we note that $big(1+|v|^2big),u=big(1+|u|^2big),v$ implies $big(1+|v|^2big),|u|=big(1+|u|^2big),|v|$. That is,
$$big(|u|-|v|big),big(1-|u|,|v|big)=0,.$$
If $|u|=|v|$, then we obtain $u=v$. If $uneq v$, then $|u|,|v|=1$, whence $uneq 0$ and $v=dfrac1^2,u$. Ergo,
$$baru,v=baru,left(frac1^2,uright)=frac^2^2=1,.$$

For Part (b), we shall prove that
$$Big|big(1+|v|^2big),u-big(1+|u|^2big),vBig|geq big|u,barv-baru,vbig|tag*$$
for all $u,vinmathbbC$, 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 $uneq 0$ and $vneq 0$.

Observe that
$$beginalignBig|big(1+|v|^2big),u-big(1+|u|^2big),vBig|^2&=Big(big(1+|v|^2big),u-big(1+|u|^2big),vBig),Big(big(1+|v|^2big),baru-big(1+|u|^2big),barvBig)
\
&=big(1+|v|^2big)^2,|u|^2+big(1+|u|^2big)^2,|v|^2-2,big(1+|u|^2big),big(1+|v|^2big),textRebig(u,barvbig),.
endalign$$
Using the AM-GM Inequality on $big(1+|v|^2big)^2,|u|^2+big(1+|u|^2big)^2,|v|^2$, we obtain
beginalignBig|big(1+|v|^2big),u-big(1+|u|^2big),vBig|^2&geq2,big(1+|u|^2big),big(1+|v|^2big),|u|,|v|-2,big(1+|u|^2big),big(1+|v|^2big),textRebig(u,barvbig)\
&= 2,big(1+|u|^2big),big(1+|v|^2big),Big(|u|,|v|-textRebig(u,barvbig)Big),.
endalign
Because $big(textIm(z)big)^2=z^2-big(textRe(z)big)^2$ for every $zinmathbbC$, we get
beginalignBig|big(1+|v|^2big),u-big(1+|u|^2big),vBig|^2&geq2,left(frac^2big),big(1+,right),Big(textImbig(u,barvbig)Big)^2
\
&geq 2,left(frac^2big),big(1+right),Big(textImbig(u,barvbig)Big)^2,,
endalign
where we have applied the trivial inequality $textRe(z)leq |z|$ for each $zinmathbbC$. Finally, using the AM-GM Inequality $t+dfrac1tgeq 2$ for all $t>0$, we get
beginalign
Big|big(1+|v|^2big),u-big(1+|u|^2big),vBig|^2&geqleft(|u|+frac1uright),left(|v|+frac1right),Big(textImbig(u,barvbig)Big)^2
\
&geq 4,Big(textImbig(u,barvbig)Big)^2=big(u,barv-baru,vbig)^2,,endalign
and the claim is proven. (The equality case is easy to verify.)

Notice that both sides remain the same if we substitute $u to cu, v to cv$ for $|c| = 1$. Thus without loss of generality, we can assume $u$ is real. Then

$$beginalign*Left&= |(1+|u|^2)v - (1+|v|^2)u| \
&= |left((1+|u|^2) mathrmRe (v) - (1+|v|^2)uright) + left((1+|u|^2) mathrmIm(v)right) i|\
&textSince norm of a complex number is at least absolute value of its imaginary part, \
&ge |(1+|u|^2) mathrmIm (v)| \
&= (1+|u|^2) |mathrmIm(v)| \
&ge 2|u| |mathrmIm(v)| \
&= |u| |overlinev - v| \
&= |uoverlinev - overlineuv| text (since $u$ is real) \
&= Right
endalign*
$$