Complex Region

Calculus Level 3

Consider two complex numbers:

$N_1 = 1 + j 0 \\ N_2 = A + j B$

Form a Cartesian coordinate system with $A$ on one axis and $B$ on the other. Within this $AB$ plane, what is the area of the region described by the following inequality?

$\frac{|N_1 + N_2|}{|N_1| + |N_2|} \leq \frac{1}{4}$

Notations:

$j= \sqrt{-1}$ denotes the imaginary unit .
$|\cdot|$ denotes the absolute value function .

The answer is 0.9215.

2 solutions

Chris Lewis
Sep 17, 2020

Putting the conditions together, we're interested in the region $4|1+A+Bi| \le 1+|A+Bi|$

Let's try to find an equation for the boundary of this region - ie where the two sides are equal. Expanding the absolute values, this is $4\sqrt{(1+A)^2+B^2}=1+\sqrt{A^2+B^2}$

Note that $B$ only appears in the form $B^2$ . This suggests looking for a biquadratic equation in $B$ . With some unpleasant manipulations... $\begin{aligned} 4\sqrt{(1+A)^2+B^2} &=1+\sqrt{A^2+B^2} \\ 16\left((1+A)^2+B^2 \right) &= 1+2\sqrt{A^2+B^2}+A^2+B^2 \\ 16+32A+16A^2+16B^2 &=1+2\sqrt{A^2+B^2}+A^2+B^2 \\ 15+32A+15A^2+15B^2 &=2\sqrt{A^2+B^2} \\ \left(15+32A+15A^2+15B^2\right)^2 &=4\left(A^2+B^2\right) \\ 225B^4+\left(450A^2+960A+446\right)B^2+\left(225 A^4 + 960 A^3+ 1470 A^2 + 960 A + 225\right)&=0 \end{aligned}$

This is our hoped for-biquadratic (ie a quadratic equation in $B^2$ ). Plugging it in to the quadratic formula gives $B^2=\frac{1}{225} \left(-223 \pm 8 \sqrt2 \sqrt{-7 - 15 A} - 480 A - 225 A^2\right)$

$B=\pm\frac{1}{15} \sqrt {-223 \pm 8 \sqrt2 \sqrt{-7 - 15 A} - 480 A - 225 A^2}$

To get real solutions, we need the second $\pm$ to be a $+$ , so finally $B=\pm\frac{1}{15} \sqrt {-223 + 8 \sqrt2 \sqrt{-7 - 15 A} - 480 A - 225 A^2}$

Some quick points about this: the curve is roughly elliptical (in fact, this is a Cartesian oval). The region we're after extends in the $A$ direction from $-\frac53$ to $-\frac35$ , and in the $B$ direction from $-\frac{1}{15}\sqrt{9+24\sqrt6}$ to $\frac{1}{15}\sqrt{9+24\sqrt6}$ . If we assume the shape is an ellipse (it's not), and use these values to find its axes, the area we get is $R' \approx 0.91967$ , which is very close to the true answer.

However, Wolfram|Alpha can actually handle the integration we need it to do. Plugging in $R=2\int_{-\frac53}^{-\frac35} \frac{1}{15} \sqrt {-223 + 8 \sqrt2 \sqrt{-7 - 15 A} - 480 A - 225 A^2} \; dA$

we find $R=\boxed{0.921534}$

Wolfram|Alpha can actually do the indefinite integral as well, but weirdly doesn't seem to be able to cope with the output. If I can get it into a manageable form, I'll post the exact value here.

Thank you for sharing this. Very interesting

Karan Chatrath - 8 months, 4 weeks ago

Engineer vs mathematician: Fight!

Krishna Karthik - 8 months, 3 weeks ago

Haha - I wondered if anyone would notice that subtle change...

Chris Lewis - 8 months, 3 weeks ago

Karan Chatrath
Sep 15, 2020

clear all
clc

% Numerical resoloution:
dA    = 1e-3;
dB    = 1e-3;

% Area initialisation:
S     = 0;

% N1 definition:
N1    = 1;

% Nested looping for area calculation:
for A = -2:dA:2

    for B = -2:dB:2


        % N2 definition:
        N2 = A + j*B;

        % Given quantity definition:
        N  = abs(N1 + N2)/(abs(N1) + abs(N2));

        % Numerical integration of area:
        if N <= 0.25            
            S = S + dA*dB;
        end

    end
end

% ANSWER = 0.9215

This is a brute force approach to the problem. Looking forward to seeing more insightful solutions

Karan Chatrath - 9 months ago

Very nice problem and a very nice solution.

Talulah Riley - 9 months ago

You can get an explicit formula for $B$ in terms of $A$ , and you can integrate it, but it's horrible. I'm trying to see if there's a neat way to do it but haven't found one yet.

It might help to know that this shape is a "Cartesian oval" - though I haven't found any area formulas online.

Chris Lewis - 8 months, 4 weeks ago

That's interesting. I hadn't heard of the Cartesian oval before.

Steven Chase - 8 months, 4 weeks ago

I spent some time trying to deduce a neat expression in terms of $A$ and $B$ . I did plot the region (plot not included), and just by eyeballing it, it did look like a circular shaped curve. Please do share your findings if you have any.

Karan Chatrath - 8 months, 4 weeks ago

I've posted them as a solution, such as they are. As I say, if I can get the final form any neater I'll post an update.

The fact it's so close to an ellipse is enticing. It would be nice to show that more formally - kind of approximating a quartic curve with a quadratic one - but I'm not sure if there's a good way to do that.

Chris Lewis - 8 months, 4 weeks ago

Complex Region

The answer is 0.9215.

2 solutions

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