# Max Value of |bc| for Complex Numbers in Inequality

• MHB
• anemone
In summary, the max value of |bc| in complex numbers is a measure of the magnitude or size of the product of two complex numbers. It is calculated using the formula |b||c|, where |b| and |c| are the magnitudes of the two complex numbers b and c. This value cannot be negative due to the absolute value function and it is commonly used in complex number inequalities to determine the range of values that satisfy the inequality or identify the maximum/minimum values of a complex number expression. Additionally, there are real-life applications of this value in fields such as engineering, physics, and economics. It can be used to analyze electric circuits, calculate maximum force, and determine optimal production levels.
anemone
Gold Member
MHB
POTW Director
$a,\,b,\,c$ are the complex numbers in $|az^2+bz+c| \le 1$ for any complex number $z$ where $|z| \le 1$. Find the maximum value of $|bc|$.

If $|b|$ and $|c|$ are to be as large as possible, it looks likely that $|a|$ will have to be small. So a first attempt would be to see what happens if $a=0$. In that case, the function becomes $|bz+c|$. That function takes the unit disc to a disc of radius $|b|$ with its centre at a distance $|c|$ from the origin. The condition for that image to lie in the unit disc is $|b| + |c| \leqslant 1$. The maximum value of $|bc|$ is then $\frac14$, occurring when $|b|$ and $|c|$ are both $\frac12$.

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.

[TIKZ][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)});[/TIKZ]

This is a solution for the case where the parameters $a,b,c$ are all real. I suspect that the answer may be the same if they are allowed to be complex, but the proof would have to be quite different.

For the condition $|az^2 + bz + c| \leqslant 1$ to hold whenever $|z|\leqslant1$, it is sufficient (by the maximum modulus principle) to check that it holds when $|z|=1$. So we want to ensure that $|ae^{2i\theta} + be^{i\theta} + c| \leq 1$ (for all $\theta$). But if $a,b,c$ are real then \begin{aligned}|ae^{2i\theta} + be^{i\theta} + c|^2 &= (ae^{2i\theta} + be^{i\theta} + c)(ae^{-2i\theta} + be^{-i\theta} + c) \\ &= 2ac\cos(2\theta) + 2b(a+c)\cos\theta + a^2 + b^2 + c^2 \\ &= 2ac(2\cos^2\theta - 1) + 2b(a+c)\cos\theta + a^2 + b^2 + c^2 \\ &= 4acx^2 + 2b(a+c)x + (a-c)^2 + b^2, \text{ where }x = \cos\theta. \end{aligned} So we want $f(x) = 4acx^2 + 2b(a+c)x + (a-c)^2 + b^2$ to have maximum value $1$ on the interval $-1\leqslant x \leqslant 1$. Differentiate to get $f'(x) = 8acx + 2b(a+c)$. That is zero when $x = -\dfrac{b(a+c)}{4ac}$, and at that point $$f(x) = \frac{b^2(a+c)^2}{4ac} - \frac{2b^2(a+c)^2}{4ac} + (a-c)^2 + b^2.$$ Put that equal to $1$ and solve for $b^2$, to get $$b^2 = 4ac\left(1 - \frac1{(a-c)^2}\right).$$ But for fixed $c$ we want to choose $a$ so as to maximise $b$. So differentiate $b^2$ with respect to $a$, getting $$4c\left(1 - \frac1{(a-c)^2}\right) + \frac{8ac}{(a-c)^3}.$$ Put that equal to zero to get $(a-c)^3 = -(a+c).$

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}

## 1. What is the maximum value of |bc| for complex numbers in inequality?

The maximum value of |bc| for complex numbers in inequality is 1. This means that the magnitude of the product of two complex numbers b and c cannot exceed 1.

## 2. How is the maximum value of |bc| calculated for complex numbers in inequality?

The maximum value of |bc| is calculated by taking the absolute values of the complex numbers b and c and multiplying them together. This value cannot exceed 1.

## 3. What is the significance of the maximum value of |bc| in complex number inequalities?

The maximum value of |bc| is significant because it represents the upper limit of the product of two complex numbers in an inequality. It helps to determine the range of possible solutions for the inequality.

## 4. Can the maximum value of |bc| be greater than 1?

No, the maximum value of |bc| cannot be greater than 1. This is because the absolute value of a complex number can never be greater than its magnitude, which is always equal to or less than 1.

## 5. How does the maximum value of |bc| relate to the graph of a complex number inequality?

The maximum value of |bc| is represented by the boundary line on the graph of a complex number inequality. Any solutions that lie on or within this boundary line satisfy the inequality, while solutions outside the boundary line do not.

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