Re: HSC 2016 4U Marathon - Advanced Level
Nobody ever answered this in the 2015 reg marathon and I have trouble finding someone who can do it, so I'll just put it here.
ai) We are solving the quadratic p(z)=z^2-2wz+1=0, where w is in the open upper half unit circle.
The complex numbers u+i and v+i then must be roots of q(z)=p(z-i)=z^2-2(i+w)z+2wi.
It suffices then to prove that the roots of q lie on the same open ray from the origin.
A useful substitution here is z=2(i+w)y. If you cannot see from the form of q why this should be useful, one can observe that if the roots of q both lie on the same open ray, then their sum 2(i+w) must lie on this same ray.
Our new polynomial is r(y)=q(z)=q(2(i+w)y)=4y(y-1)(w+i)^2+2wi.
Dividing by (w+i)^2, we see that the roots of r satisfy 4y^2-4y+2wi/(w+i)^2=0.
But 2wi/(w+i)^2=((w/i+i/w)/2+1)^(-1). Since w is of unit modulus and lies in the upper half plane, we can conclude that 2wi/(w+i)^2 is a real number between 0 and 1.
The quadratic formula then implies that the roots of r are both positive real numbers. Since our change of variables from z to y just corresponded to a rotation about zero, we have proven that the roots of q lie on the same open ray from the origin, and hence that arg(u+i)=arg(v+i).
ii) We proceed similarly. Our definition of q is now:
q(z)=p(z+i)=z^2+2(i-w)z-2wi.
Our substitution is z=2(i-w)y, so
q(z)=4y(y+1)(i-w)^2-2wi.
Dividing the equation be (i-w)^2, we are left with
4y^2+4y-2wi/(w-i)^2.
Since w has unit modulus and is in the upper half plane, we have 2wi/(w-i)^2=((w/i+i/w)/2-1)^(-1) which is real and strictly less than -1.
Hence our equation in y has negative discriminant. As it's coefficients are real, we must conclude that its roots are a pair of conjugate complex numbers, and hence of equal modulus. (Again, this property is invariant under the rotational change of variables.)
b) It doesn't quite make sense to talk individually about the locii of u and v, as for each the alpha there is an arbitrary choice in which to label u. We can however talk about the locus of z satisfying the original equation.
We have |z+1/z|=2 so
|(x+iy)+(x-iy)/(x^2+y^2)|^2=4
=> |x(1+1/(x^2+y^2))+iy(1-1/(x^2+y^2))|^2=4
=> x^2(x^2+y^2+1)^2+y^2(x^2+y^2-1)^2=4(x^2+y^2)^2
=> x^4+y^4+2x^2y^2-2x^2-6y^2+1=0.
=> (x^2+(y-1)^2-2)(x^2+(y+1)^2-2)=0.
That is, the locus lies on the union of the two circles centred at i and -i respectively with radius sqrt(2).
However, note that as alpha varies, we do not hit this full union of two circles, as we have the additional constraint that z+1/z has positive imaginary part, that is y(x^2+y^2-1)>0. So we have to exclude the region outside the open unit disk and below the real axis, as well as the region inside the unit disk and above the real axis.
This leaves us with the circle centred at i with radius sqrt(2), excluding the points +-1.