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Let $f(w)=\frac13+\frac12 w+\frac16 w^3$. If $\vert f(w)\vert\leq1$ or simply $\vert f(w)\vert=1$, show that $\vert \frac{w}2 f(\frac{w}2)\vert\leq1$. Here, $w$ is a complex number.

What happens if we give the problem in its full generality? That is, let $f(w)=\frac13+\frac12 w+\frac16 w^3$. If $\vert f(w)\vert\leq1$ or simply $\vert f(w)\vert=1$, show that $\left\vert \frac{tw+1-t}{1+t^2} f(\frac{w}{1+t})\right\vert\leq1$ for any real number $t\geq0$. Here, $w$ is a complex number.

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  • $\begingroup$ Homework in complex analysis? $\endgroup$ – Alex Degtyarev Jan 19 '15 at 16:10
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    $\begingroup$ Not at all. It arose in some ODE research. $\endgroup$ – T. Amdeberhan Jan 19 '15 at 16:17
  • $\begingroup$ @robert: Really cool. The actual problem has a free parameter and $2$ in $\frac{w}2$ is a special case. Perhaps I can work out the rest similarly, with a possible snag on having no solutions at the end with the resultant. Thanks. Of course, any other method is still appreciated. $\endgroup$ – T. Amdeberhan Jan 20 '15 at 1:16
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If $w = x + i y$, the curve $|f(w)|^2=1$ can be written as $Q(x,y) = 1$ where $$ Q(x,y) = \dfrac{{x}^{6}}{36}+\dfrac{{x}^{4}{y}^{2}}{12}+\dfrac{{x}^{2}{y}^{4}}{12}+\dfrac{{y}^{6}}{36}+ \dfrac{{x}^{4}}6-\dfrac{{y}^{4}}{6}+\dfrac{{x}^{3}}{9}-\dfrac{x{y}^{2}}{3}+\dfrac{{x}^{2}}{4}+\dfrac{{y}^{2}}4+\dfrac{x}{3}+\dfrac{1}{9} $$ The curve $\left|\frac{w}{2} f\left(\frac{w}{2}\right)\right|^2 = 1$ can be written as $R(x,y) = 1$ where $$ R(x,y) = {\frac {{x}^{8}}{9216}}+{\frac {{x}^{6}{y}^{2}}{2304}}+{\frac {{x}^{4} {y}^{4}}{1536}}+{\frac {{x}^{2}{y}^{6}}{2304}}+{\frac {{y}^{8}}{9216}} +{\frac {{x}^{6}}{384}}+{\frac {{x}^{4}{y}^{2}}{384}}-{\frac {{x}^{2}{ y}^{4}}{384}}-{\frac {{y}^{6}}{384}}+{\frac {{x}^{5}}{288}}-{\frac {{x }^{3}{y}^{2}}{144}}-{\frac {x{y}^{4}}{96}}+{\frac {{x}^{4}}{64}}+\frac{{x}^{2}{y}^{2}}{32}+{\frac {{y}^{4}}{64}}+\frac{{x}^{3}}{24}+\dfrac{x{y}^{2}}{24}+\frac{{x}^{2}}{36}+\frac{{y}^{2}}{36} $$ The resultant of $Q(x,y)-1$ and $R(x,y)-1$ with respect to $y$ is $$ \left( -{\frac {293\,{x}^{12}}{10030613004288}}-{\frac {293\,{x}^{10} }{835884417024}}-{\frac {3661\,{x}^{9}}{20061226008576}}-{\frac {4919 \,{x}^{8}}{990677827584}}+{\frac {44923\,{x}^{7}}{8916100448256}}-{ \frac {378475\,{x}^{6}}{11888133931008}}+{\frac {90211\,{x}^{5}}{ 3962711310336}}-{\frac {3355225\,{x}^{4}}{31701690482688}}+{\frac { 15202897\,{x}^{3}}{47552535724032}}-{\frac {452345\,{x}^{2}}{ 7044820107264}}+{\frac {57373\,x}{1761205026816}}-{\frac {857353}{ 1761205026816}} \right) ^{2} $$ and this has no real roots (as confirmed using Maple's sturm function). Therefore the sign of $R(x,y) - 1$ is constant on the curve $Q(x,y)=1$ (which happens to be connected). Evaluating at one point, say $(x,y) = (1,0)$, you find that sign is negative.

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