We have \((1,1)*(a,b) = (a - b,a + b) = (0,-2)\). This gives us the simultaneous linear equations \(a - b = 0\) and \(a + b = -2\). The first equation gives \(a = b\). Substituting into the second equation we get \(2a = -2\) and so \(a = b = -1\). Therefore \((a,b) = (-1,-1)\).
Since \((a,b)*(c,d) = (c,d)\) for all points \((c,d)\), we must have \((a,b)*(1,0) = (1,0)\). The definition of the product gives \((a,b)*(1,0) = (a,b)\). Therefore \((a,b) = (1,0)\) is a good candidate for an answer! Let’s check that it works. We have \[(1,0)*(c,d) = (1c - 0d,1d + 0c) = (c,d),\] so the answer is \((a,b) = (1,0)\).
Before we get stuck into the solution, there is a subtle point about the product \(*\) that we need to sort out. The expression \((a,b)^4\) is defined to mean the product of \((a,b)\) with itself four times. But what does this mean exactly? Does it mean \(\left[(a,b)*(a,b)\right]*\left[(a,b)*(a,b)\right]\) or \((a,b)*\left[(a,b)*\left[(a,b)*(a,b)\right]\right]\), or anything else? Does it even matter?
We will see in a moment, just like with multiplication of real numbers, that how we group such a product doesn’t affect the outcome. Let’s prove that now. Let \((a,b), (c,d), (e,f)\) be three points in the Cartesian plane. Then \[\begin{align*} *(e,f) &= (ac - bd, ad + bc)*(e,f) \\ &= ((ac-bd)e - (ad + bc)f,(ac - bd)f + (ad + bc)e) \\ &= (ace - bde - adf - bcf, acf - bdf + ade + bce).\end{align*}\] Also, \[\begin{align*} (a,b)*[(c,d)*(e,f)] &= (a,b)*(ce - df,cf + de) \\ &= (a(ce - df) - b(cf + de), a(cf + de) + b(ce - df)) \\ &= (ace - adf - bcf - bde, acf + ade + bce - bdf) \\ &= [(a,b)*(c,d)]*(e,f).\end{align*}\] We have just proved that for a product of three points, the grouping of the three points does not matter. That is, it doesn’t matter if you first take the product of the first two, and then the third, or if you first take the product of the last two, and then the first. In fancy math terminology, we have proved that the product \(*\) is associative.
Since the product \(*\) is associative, we can take powers without worrying about how we’re grouping the product. We can now safely move on to the question.
Let’s compute the first few powers of \(\left(\frac{1}{2},\frac{\sqrt 3}{2}\right)\). \[\begin{align*} \left(\frac{1}{2},\frac{\sqrt 3}{2}\right)^2 &= \left(-\frac{1}{2},\frac{\sqrt 3}{2}\right) \\ \left(\frac{1}{2},\frac{\sqrt 3}{2}\right)^3 = \left(-\frac{1}{2},\frac{\sqrt 3}{2}\right)*\left(\frac{1}{2},\frac{\sqrt 3}{2}\right) &= (-1,0) \\ \left(\frac{1}{2},\frac{\sqrt 3}{2}\right)^4 = (-1,0)*\left(\frac{1}{2},\frac{\sqrt 3}{2}\right) &= \left(-\frac{1}{2},-\frac{\sqrt 3}{2}\right) \\ \left(\frac{1}{2},\frac{\sqrt 3}{2}\right)^5 = \left(-\frac{1}{2},-\frac{\sqrt 3}{2}\right)*\left(\frac{1}{2},\frac{\sqrt 3}{2}\right) &= \left(\frac{1}{2},-\frac{\sqrt 3}{2}\right) \\ \left(\frac{1}{2},\frac{\sqrt 3}{2}\right)^6 = \left(\frac{1}{2},-\frac{\sqrt 3}{2}\right)*\left(\frac{1}{2},\frac{\sqrt 3}{2}\right) &= (1,0).\end{align*}\] We know from 1(b) above that \((1,0)*(c,d) = (c,d)\) for any point \((c,d)\). Therefore, for any integer \(k > 6\), we have \[\left(\frac{1}{2},\frac{\sqrt 3}{2}\right)^k = \left(\frac{1}{2},\frac{\sqrt 3}{2}\right)^6*\left(\frac{1}{2},\frac{\sqrt 3}{2}\right)^{k-6} = (1,0)*\left(\frac{1}{2},\frac{\sqrt 3}{2}\right)^{k-6} = \left(\frac{1}{2},\frac{\sqrt 3}{2}\right)^{k-6}.\] So, to compute a power of \(2025\), we can keep subtracting \(6\) from \(2025\) until we get to a power that we have already computed. Since \(2025 = 337\cdot 6 + 3\) we have \[\left(\frac{1}{2},\frac{\sqrt 3}{2}\right)^{2025} = \left(\frac{1}{2},\frac{\sqrt 3}{2}\right)^{2019} = \left(\frac{1}{2},\frac{\sqrt 3}{2}\right)^{2013} = \cdots = \left(\frac{1}{2},\frac{\sqrt 3}{2}\right)^3 = (-1,0).\]
We first compute \((0,2)*(1,1) = (-2,2)\). Then \[|(0,2)*(1,1)| = |(-2,2)| = \sqrt{(-2)^2 + 2^2} = \sqrt 8 = 2\sqrt 2.\] Also, \(\theta((0,2)*(1,1)) = \theta(-2,2)\). The point \((-2,2)\) lies in the second quadrant of the Cartesian plane on the line \(y = -x\), which is at \(45\degree\) to the \(y\)-axis. Therefore \(\theta(-2,2) = 90\degree + 45\degree = 135\degree\).
A very helpful thing for this question is to first write the coordinates of each point in terms of \(r_i\) and \(\phi_i\). Let’s start with points in the first quadrant.
Suppose \(x\) and \(y\) are positive real numbers, \(|(x,y)| = r\), and \(\theta(x,y) = \phi\). Then we have the following diagram.
Therefore, \(x = r\cos(\phi)\) and \(y = r\sin(\phi)\). Miraculously, these equations hold in every quadrant, and on the \(x\)- and \(y\)-axes. It is left up to you to check the details of this.
So, the coordinates of \(D_1\) and \(D_2\) are \((r_1\cos(\phi_1),r_1\sin(\phi_1))\) and \((r_2\cos(\phi_2),r_2\sin(\phi_2))\) respectively. Let’s see what happens when we take the product of \(D_1\) and \(D_2\). We have \[\begin{align*} & (r_1\cos(\phi_1),r_1\sin(\phi_1))*(r_2\cos(\phi_2),r_2\sin(\phi_2)) \\ &= (r_1r_2\cos(\phi_1)\cos(\phi_2) - r_1r_2\sin(\phi_1)\sin(\phi_2),r_1r_2\cos(\phi_1)\sin(\phi_2) + r_1r_2\sin(\phi_1)\cos(\phi_2)) \\ &= (r_1r_2\cos(\phi_1 + \phi_2),r_1r_2\sin(\phi_1+\phi_2))\end{align*}\] where the last equality uses the angle sum formulas for sine and cosine. Therefore, \(|D_1*D_2| = r_1r_2\) and \(\theta(D_1*D_2) = \phi_1 + \phi_2\). The moral of the story is that when you take the product of two points, their distances multiply and their angles (when measured counterclockwise from the positive \(x\)-axis) add.
The three other points are \((-3,2), (-2,-3)\), and \((3,-2)\) (you can check that these are indeed solutions by computing \((-3,2)^4\), for example). Here is one way we could have found these points.
Suppose \(C\) is a solution to \(x^4 = (-119,-120)\). Then we know \(|C|^4 = |(-119,-120)|\) by Question \(3\)(b). However, \(|(2,3)|^4 = |(-119,-120)|\) and so \(|C|^4 = |(2,3)|^4\). Since both \(|C|\) and \(|(2,3)|\) are non-negative real numbers, it must be the case that \(|C| = |(2,3)|\).
So, we know how far away from the origin any solution to \(x^4 = (-119,-120)\) must be, we just need to figure out which directions to go from the origin to find solutions.
Using the result from \(3\)(b) again, we know that \(4\cdot \theta(C) = \theta(-119,-120) = 4\cdot \theta(2,3)\). It’s tempting to think that we can therefore conclude that \(\theta(C) = \theta(2,3)\), but this is not the case!
However, since \(4\cdot\theta(C)\) gives the same direction from the origin as \(4\cdot\theta(2,3)\), it must be that \(4\cdot \theta(C)\) and \(4\cdot \theta(2,3)\) differ by a multiple of \(360\degree\). This occurs precisely when \(\theta(C)\) and \(\theta(2,3)\) differ by a multiple of \(90\degree\).
Putting all of this together, to find other solutions to \(x^4 = (-119,-120)\), we can take the given solution and rotate it counterclockwise about the origin by \(90\degree\), \(180\degree\), and \(270\degree\).
The points \((-3,2)\), \((-2,-3)\), and \((3,-2)\) are the result of rotating the given solution \((2,3)\) by \(90\degree\), \(180\degree\), and \(270\degree\) counterclockwise about the origin respectively.
In fact, the argument above proves that these are all the possible solutions the given equation. Can you see why?
Although this appears to be a problem in Euclidean geometry, we can approach it using properties of the product of points on the Cartesian plane. Note that \(OF = |F|\), \(OY = |Y|\), \(\angle EOF = \theta(F)\) and \(\angle EOY = \theta(Y)\).
Therefore, the conditions given in the problem are asking us to find a point \(F\) satisfying \(|F|^2 = |Y|\) and \(2\theta(F) = \theta(Y)\). By \(3\)(b) above, a point \(F\) satisfying \(F*F = Y\) will do the trick!
We compute \(|Y| = \sqrt{7^2 + 24^2} = 25\). Therefore, we are looking for a point \(F\) satisfying \(|F| = 5\). Suppose \(F\) has coordinates \((x,y)\). The question specifies that \(F\) has integer coordinates, so we are looking for integers \(x\) and \(y\) such that \(\sqrt{x^2 + y^2} = 5\). There are not very many pairs of integers to check here! We must have that \(x\) and \(y\) are, in some order \(\pm 3\) and \(\pm 4\), or \(0\) and \(\pm 5\).
We have \((\pm 5,0)^2 = (25,0)\) and \((0,\pm 5)^2 = (-25,0)\). Neither of these are the point \((7,24)\). Trying \((x,y) = (3,4)\) we get \((3,4)^2 = (-7,24)\). Close, but not quite. Finally, we have \((4,3)^2 = (7,24)\), which is what we are after! Therefore the point \(F\) with coordinates \((4,3)\) is a solution to the problem.