Similar Triangles in the Complex Plane
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- Опубліковано 12 чер 2024
- We explore conditions for triangles in the complex plane to be similar to each other. The formulation is slightly different if the triangles have different orientations (i.e. mirror images).
I came across the result for triangles with the same orientation in Hahn's book:
Hahn, L., 1994. Complex numbers and geometry. Cambridge University Press.
00:00 Intro
00:29 Similar triangles
01:24 Using complex numbers
04:33 Bonus equivalent condition
06:20 Different orientations
The determinant condition is interesting! One way of rewriting the conjugate version as a determinant is to say that w_1*,w_2*,w_3* form a triangle similar to the one formed by z_1,z_2,z_3, and with the same orientation. But then w_2* corresponds to z_3 and w_3* to z_2, so we get det(A)=0, where A has rows (1,1,1),(z_1,z_2,z_3),(w_1*,w_3*,w_2*). This can also be checked by comparing the fractions in the two cases: w_1 gets conjugated, and w_2,w_3 are conjugated and swapped.
Very neat! This makes sense, since conjugation changes the orientation, so one triangle would be similar with the same orientation as the conjugate version of the other.
That determinant matrix reminds me of the matrix form of the cross product 🤔, lovely video as always!
triangles are positive similar (no reflection) if and only if there exists a rotation-dilation (rotation+stretching) that maps one to the other. Since rotation-dilations are multiplicatiins by a complex number c one could state:
A triangle z1z2z3 is positive similar to a triangle w1w2w3 if and only if there exists a constant complex number c with z1=c*w1, z2=c*w2, z3=c*w3. This is equivalent to your determinant description.
For negative similarities (i.e. plus reflection) you take w1*, w2*, w3* instead w1,w2,w3. Or w1*, w3*, w2* if you orient the points anticlockwise such that z1z2z3 is similar to w1w3w2.
This is a really nice, concise explanation for why the determinant condition holds - essentially just that the complex numbers (corresponding to vertices of the triangles) are multiples of each other. And we know that the determinant is zero when one row is a multiple of another.
I am thinking about writing a geometry/trig book that emphasizes the relationship between trig and the complex numbers by defining the product of points as a geometric construction in the plane. The basic operation is to create the triangle 0,1,z1 and construct a similar triangle with the 0 and 1 side being 0 and z2. The image of z1 is now z1z2. This is essentially the radial multiplication formula for complex numbers.
Note that the image of a 45 45 90 triangle formed by (0,1) multiplied by itself in this manner is the point (-1,0).
Now if we suspect z1,z2,z3 to be similar to w1,w2,w3 we can translate them by subtracting the first point of each, to align to 0, and normalizing the 0,1 edge. Now the image of z3 and w3 are uniquely determined. If this is equal they are similar, if these are conjugate they are mirrors.
Indeed, this gives us (z3-z1)/(z2-z1)=(w3-w1)/(w2-w1) or, for the opposite orientation: (z3-z1)/(z2-z1)=(w3-w1)/(w2-w1)^*=(w3^*-w1^*)/(w2^*-w1^*).
That this results in the determinant formula is very nice!
If w1 w2 and w3 is similar to z1 z2 and z3 but flipped, then surely w1* w2* and w3* is also similar to z1 z2 and z3 but oriented the same; in which case surely the determinant criterion still works but with row 3 conjugated?
though you'd have to label the vertices consistently (so z1 and w1 are same angle, z2 and w2 same angle, etc, rather than labelling in the same direction on both triangles)
Yes, this works! I think we need to have z_2 with w_3, and z_3 with w_2, if they were the labels of the original triangle.
An interesting consequence of the determinant condition is that if you multiply the matrix
(1 z1 w1
1 z2 w2
1 z3 w3)
which is singular [i.e. has det=0] (it's the transpose of the matrix in the video) by any other 3×3 matrix A we still get a singular matrix (since det is multiplicative). Now if A satisfies that each of it's rows sum to 1 then the product of the matrices (where we multiply A from the left [that's why I took the transpose of the matrix in the video]) then A transforms our two similar triangles into another pair of similar triangles.
For example the linear transformation defined by the matrix
(1 -1 1
2 -3 2
-1 4 -2)
Somewhat miraculously transforms every pair of similar triangles to another pair of similar triangles!
3:00 why do the angles can be expressed in that way? (it's probably obvious, but I'm too sleepy for complex analysis ahah)
You can rewrite a complex number z as |z| * exp(i * arg(z)) with arg(z) representing the angle.
@@Bl00drav3nz that I know (I mean, it's the basis of complex numbers, I'm not that sleepy ahah), I just don't get why arg(z)-arg(w)=arg(z/w) (leaving aside constants)
arg(z/w)
= arg(|z| exp(i arg(z)) / |w| exp(i arg(w)))
= arg(exp(i arg(z)) / exp(i arg(w)))
= arg(exp(i (arg(z) - arg(w))))
= arg(z) - arg(w)
In the first equality, I just entered the definition for z and w. In the second equality, I crossed out |z| and |w| since they are positive real numbers and don't affect the arg. In the third equality, I used the exponential property that exp(a)exp(b)=exp(a+b). The fourth equation is just the definition of the arg-function
Also, if you look at what the log-function does for z = |z| exp(i arg(z)), then log(z/w) = log(z) - log(w) is basically all you need to show :)
@@robinros2595 oh, I knew that was that simple lol
Thank you for the answer :)
UR VIDEOS are complex to be understood ☺
no u