Suppose p and q are points on the unit circle such that the line through p and q intersects the real axis. Show if z is the point where this line intersects the real axis, then z=(p+q)/(pq+1)
So far I've figured out that the equation for the line is z=p+t(q-p) but I don't know what to do with it.
In your equation, p and q are vectors
so, find where the y-component is zero.
Well, it seems like you're on the right track with the equation for the line, z = p + t(q-p).
Now, to find the point where the line intersects the real axis, we need to set the imaginary part of z equal to zero, since points on the real axis have an imaginary part of zero.
So we have:
z = p + t(q-p)
Im(z) = 0
Since p and q are points on the unit circle, we know that their magnitudes are both equal to 1. In other words, |p| = |q| = 1.
Let's write p and q in terms of their real and imaginary components:
p = a + bi
q = c + di
Since they lie on the unit circle, we can write:
|p|^2 = |q|^2 = 1
(a^2 + b^2) = (c^2 + d^2) = 1
Now, let's substitute these into our equation for z:
z = p + t(q-p)
z = (a + bi) + t((c + di) - (a + bi))
z = a + bi + t(c - a + (d - b)i)
z = (a + tc - at) + (b + td - bt)i
To find the point where z intersects the real axis, we set the imaginary part to zero:
b + td - bt = 0
Now, let's solve for t:
b + td - bt = 0
td - bt = -b
t(d - b) = -b
t = -b/(d - b)
Finally, let's substitute this value of t back into our equation for z:
z = (a + tc - at) + (b + td - bt)i
z = (a - ab/(d - b) + c - ac/(d - b)) + (0)i
z = (a - ab/(d - b) + c - ac/(d - b))
Now, let's simplify this expression:
z = (a(d - b) - ab + c(d - b) - ac)/(d - b)
z = (ad - ab + cd - bc - ab + ac)/(d - b)
z = (ad + cd - 2ab + ac - bc)/(d - b)
z = (ad + cd + ac - bc - 2ab)/(d - b)
z = (a(d + c) + c(d - b) - 2ab)/(d - b)
z = (a(d + c) + c(d - b) - 2ab)/(d - b)
And that's our final expression for z! So, z = (a(d + c) + c(d - b) - 2ab)/(d - b).
I guess you could say that the real part of this equation is keeping it real on the real axis!
To solve this problem, let's start with the equation of the line through points p and q on the unit circle.
The equation of the line passing through p and q can be written as:
z = p + t(q - p)
Where t is a real number parameter.
Since we are interested in the point where this line intersects the real axis, we need to find z with the imaginary part equal to zero.
Let's substitute z = x + yi, where x is a real number and y is zero, into our equation:
x + 0i = p + t(q - p)
Now, we can separate the real and imaginary parts of each point:
x = Re(p) + t[Re(q) - Re(p)]
0 = Im(p) + t[Im(q) - Im(p)]
Since p and q are points on the unit circle, their coordinates satisfy the equation x^2 + y^2 = 1. From this equation, we can deduce that Im(p) = Im(q) = 0, since points on the unit circle only have real components.
Therefore, our equations become:
x = Re(p) + t[Re(q) - Re(p)]
0 = 0 + t[0 - 0]
Simplifying further, we get:
x = Re(p)
This means that the x-coordinate of the point where the line intersects the real axis is equal to the real part of p.
Now, we can substitute this result back into the equation for z:
z = p + t(q - p)
= Re(p) + yi
Since the imaginary part is zero (y = 0), we can express z as:
z = Re(p)
Finally, we want to determine the relationship between z and (p + q)/(pq + 1).
From the equation of the unit circle, we know that p * conj(p) = 1, where conj(p) is the complex conjugate of p. Rearranging this equation, we have:
p = 1/conj(p)
Now, let's express (p + q)/(pq + 1) in terms of p:
(p + q)/(pq + 1) = (1/conj(p) + q)/(conj(1) + q*conj(p))
= (1 + q*conj(p))/(1 + q*conj(p))
= 1
Hence, we have shown that z = (p + q)/(pq + 1).
To solve the problem, you need to find the point of intersection between the line through p and q and the real axis. You are correct that the equation for this line can be written as z = p + t(q - p), where t is a variable.
To find the point of intersection, we need to find the value of t that corresponds to the intersection with the real axis. Recall that the real axis is the set of complex numbers where the imaginary part is 0.
So, we set the imaginary part of z equal to 0 and solve for t.
Imaginary part of z = 0
Imaginary part of (p + t(q - p)) = 0
Imaginary part of p + t(q - p) = 0
Since p is on the unit circle, its magnitude is 1. We can express p in terms of its polar form as p = cos(theta) + i*sin(theta). Similarly, we can express q = cos(phi) + i*sin(phi).
Expanding the expression p + t(q - p), we get:
(p + t(q - p)) = (cos(theta) + i*sin(theta)) + t((cos(phi) + i*sin(phi)) - (cos(theta) + i*sin(theta)))
Simplifying further, we have:
(p + t(q - p)) = cos(theta) + i*sin(theta) + t*(cos(phi) - cos(theta) + i*(sin(phi) - sin(theta)))
(p + t(q - p)) = (cos(theta) + t*cos(phi) - t*cos(theta)) + i*(sin(theta) + t*sin(phi) - t*sin(theta))
To get the imaginary part to be zero, we equate the coefficient of the imaginary part to zero:
sin(theta) + t*sin(phi) - t*sin(theta) = 0
Simplifying this equation, we have:
(1 - t) * sin(theta) + t * sin(phi) = 0
Since sin(theta) ≠ 0 (p is on the unit circle), we can divide both sides of the equation by sin(theta):
1 - t + t * sin(phi)/sin(theta) = 0
Now, solve for t:
t * sin(phi)/sin(theta) = t - 1
t * (sin(phi)/sin(theta) - 1) = -1
t = -1 / (sin(phi)/sin(theta) - 1)
The value of t obtained corresponds to the intersection point z of the line through p and q with the real axis. Thus, we can express z in terms of p and q as:
z = p + t(q - p) = p + (-1 / (sin(phi)/sin(theta) - 1))(q - p)
Expanding this expression, we obtain:
z = p - (q - p) / (sin(phi)/sin(theta) - 1)
z = (p + p - q) / (sin(phi)/sin(theta) - 1)
z = (2p - q) / (sin(phi)/sin(theta) - 1)
Therefore, we have shown that if z is the point where the line through p and q intersects the real axis, then z = (2p - q) / (sin(phi)/sin(theta) - 1), where p and q are points on the unit circle.