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- #1

- Feb 14, 2012

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- Thread starter anemone
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- #1

- Feb 14, 2012

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- Feb 7, 2012

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So my first guess was that the answer to the problem might be $\frac14$. But I then did some numerical experiments with graphs, and I found that if $a = 0.16$, $b = 0.56$ and $c = -0.56$, then the function $0.16z^2 + 0.56z - 0.56$ takes the unit disc to the green region in the diagram below, which clearly lies inside the (red) unit circle. But it has the property that $|bc| = 0.56^2 = 0.3136$, which is quite a bit larger than $\frac14$.

That is as far as I can go with this problem. I don't even have a guess as to what the maximum value of $|bc|$ might be.

\begin{tikzpicture}

[scale=6]\draw [help lines, ->] (-1.1,0) -- (1.1,0) ;

\draw [help lines, ->] (0,-1.1) -- (0,1.1) ;

\draw [red] circle (1) ;

\fill [green, domain=0:6.285, samples=100] plot ({0.16*cos(2*\x r) + 0.56*cos(\x r) - 0.56}, {0.16*sin(2*\x r) + 0.56*sin(\x r)});

\end{tikzpicture}

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- Feb 7, 2012

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Now let $t = a-c$. Then $t^3 = -a-c$, so that $c = \frac12(t^3 + t)$. Also, $t^6 - t^2 = (a+c)^2 - (a-c)^2 = 4ac$. Therefore $$b^2c^2 = (t^6 - t^2)\left(1 - \frac1{t^2}\right)\cdot\frac14(t^3+t)^2 = \frac14t^2(t^2-1)^2(t^2+1)^3.$$ On the interval $-1\leqslant t\leqslant 1$, that function has a maximum value $\dfrac{27}{256}$ (attained when $t =\pm \dfrac1{\sqrt2}$). So the maximum value of $|bc|$ is $\dfrac{3\sqrt3}{16} \approx 0.3248.$

From that, it is easy to see that this maximum value occurs when $a = \dfrac{\sqrt2}8 \approx 0.177$, $b = \dfrac{\sqrt6}4 \approx 0.612$ and $c =

-\dfrac{3\sqrt2}8 \approx -0.53$. Using those numbers, the diagram from the previous comment looks like this, with the green image of the unit disc under the function $|az^2 + bz + c|$ fitting snugly inside the unit circle.

\begin{tikzpicture} [scale=6]\draw [help lines, ->] (-1.1,0) -- (1.1,0) ; \draw [help lines, ->] (0,-1.1) -- (0,1.1) ; \draw [red] circle (1) ; \fill [green, domain=0:6.285, samples=100] plot ({0.177*cos(2*\x r) + 0.612*cos(\x r) - 0.53}, {0.177*sin(2*\x r) + 0.612*sin(\x r)}); \end{tikzpicture}