A Pythagoras-like conjecture using complex numbers

Question

A physical experiment with weights, pulleys and a string led me to the following conjecture.

Let $a$ and $b$ be complex numbers having $\mathrm{Re}(a)<0$, $\mathrm{Re}(b)>0$ and $\mathrm{Im}(a)=\mathrm{Im}(b)>0.$

Let $F_a$, $F_b$ and $G$ positive real numbers.

Conjecture. If

$$F_a\frac{a}{|a|} + F_b\frac{b}{|b|} -Gi = 0\tag{1}$$

and

$$F_a^2+F_b^2=G^2\tag{2}$$

then

$$|a|^2+|b|^2=(a-b)^2.\tag{3}$$

(or in a stronger form, $\frac{F_a}{|b|}=\frac{F_b}{|a|}=\frac{G}{|a-b|}.$)

Is this conjecture true?

Answer 1

I think the most simple and elementary solution is this.

From (1) and (2), by Pythagoras, $\gamma = \frac{\pi}{2}$ on the picture, so $\pi-\gamma= \frac{\pi}{2}$ too.

Answer 2

No, this isn't true, even without considering $F_a, F_b, G$. As a counterexample, let $a = -e^{-i\pi/3} = -\frac{1}{2}+\frac{\sqrt{3}}{2}i$ and $b = e^{i\pi/3} = \frac{1}{2} + \frac{\sqrt{3}}{2}i$, so that $|a|=|b|=1$ and thus $|a|^2 + |b|^2 = 2$. On the other hand, $(a-b)^2 = (-1)^2 = 1 \ne 2$.

Answer 3

Consider $a=-x+iy$ and $b=u+iy$. Then we have that $|a|^2=x^2+y^2$ and the same for b. So from your third equation we have: $$|a|^2+|b|^2=(a-b)^2$$ $$x^2+u^2+2y^2=(-x-u)^2=x^2+u^2+2xu$$ $$xu=y^2$$ From the second equation we have that $$-F_a\frac{x}{|a|}+F_b\frac{u}{|b|}+(\frac{F_a}{|a|}+\frac{F_b}{|b|})yi=Gi$$ Since if we have two equal complex numbers their real part and imaginary part are equal ($a=x+yi, b=u+iv, a=b \Rightarrow x=u, y=v$), then we have that: $$\frac{F_a x}{|a|}=\frac{F_b u}{|b|}$$ $$G=(\frac{F_a}{|a|}+\frac{F_b}{|b|})y$$ So $$F_a^2+F_b^2=y^2(\frac{F_a^2}{|a|^2}+\frac{2F_a F_b}{|ab|}+\frac{F_b^2}{|b|^2})$$ $$F_a^2(1-\frac{1}{|a|^2})+F_b^2(1-\frac{1}{|b|^2})=\frac{2F_aF_by^2}{|ab|}$$

I could reach just this expressions.

EDIT:

$$\frac{F_a \Re(a)}{|a|}=\frac{F_b \Re(b)}{|b|}$$ $$\Re(a)\Re(b)=\Im(y)^2$$ $$G=(\frac{F_a}{|a|}+\frac{F_b}{|b|})\Im(a)$$

Answer 4

Yes, this conjecture is true. Interesting question! :)

We can make things easier for ourselves by observing that we may make a number of simplifications.

We might as well start off by dividing $a$ and $b$ by $\operatorname{Im}(a)$ in order to simplify the algebra a bit (this loses no generality due to all equations being positive-real-homogeneous in $a$ and $b$).

Your assumptions are also homogeneous in $F_a$, $F_b$, and $G$, so you might as well divide by $G$ and $G^2$ and just assume $G = 1$.

Also, $a - b$ is purely real, so $(a - b)^2 = \lvert a - b \rvert^2$. So the statement you're asking about is equivalent to $\lvert a \rvert^2 + \lvert b \rvert^2 = \lvert a - b \rvert^2$ - ie you want to know if the triangle with vertices $0$, $a$, $b$ has a right angle at the origin.

Indeed, say $a = x_a + i$ and $b = x_b + i$. Taking real parts in your first equation shows $\frac{F_a}{\lvert a \rvert} x_a + \frac{F_b}{\lvert b \rvert}x_b = 0$, and taking imaginary parts shows $\frac{F_a}{\lvert a \rvert} + \frac{F_b}{\lvert b \rvert} = 1$. This is pair of simultaneous equations in the quantities $\frac{F_a}{\lvert a \rvert}$ and $\frac{F_b}{\lvert b \rvert}$, which are not so hard to solve - we see $\frac{F_a}{\lvert a \rvert} = x_b(x_b - x_a)^{-1}$ and $\frac{F_b}{\lvert b \rvert} = x_a(x_a - x_b)^{-1}$.

The second equation now shows that $\lvert a \rvert^2 x_b^2 + \lvert b \rvert^2 x_a^2 = (x_a - x_b)^2$. Expanding, we obtain $(x_a^2 + 1) x_b^2 + (x_b^2 + 1) x_a^2 = (x_a - x_b)^2$. After cancellation, we're left with $2x_a^2 x_b^2 = -2x_a x_b$, and therefore $x_a x_b = -1$. This exactly says that the dot product of $(x_a, 1)$ and $(x_b, 1)$ is zero, ie that $a$ and $b$ are orthogonal. So we're done. (You can also just check that given this fact, we can prove $\lvert a \rvert^2 + \lvert b \rvert^2 = \lvert a - b \rvert^2$ by expanding).

PS: your stronger form is also true. The same reductions still apply, so we need to show $F_a(x_b - x_a) = \lvert b \rvert$ and $F_b (x_b - x_a) = \lvert a \rvert$. Let's just do the first one: it is equivalent to $x_b \lvert a \rvert = \lvert b \rvert$, or, after, squaring, $x_b^2 (x_a^2 + 1) = (x_b^2 + 1)$. This follows from $x_a^2 x_b^2 = 1$. The other equality is similar.

Answer 5

From (1) and (2)

$$\left|F_a\frac{a}{|a|} + F_b\frac{b}{|b|}\right|^2=F_a^2+F_b^2\tag{i}$$

So,

$$F_a^2+F_b^2+2F_aF_b\frac{a\cdot b}{|a||b|}=F_a^2+F_b^2$$

(where $\cdot$ stands for the dot product of of complex numbers as vectors, i.e. $a\cdot b=\Re(a\overline b)$), so, $a\cdot b=0$, i.e. $a$ is orthogonal to $b$, so the conjecture is true.