Question

Given z = cos θ + isin θ and u + iv = (1 + z)(1 + z
2
). Prove that v = u tan( 3θ
2
) and
u
2 + v
2 = 16 cos2
(
θ
2
) cos2
(θ)

Answer 1

$z=\cos\theta+i\sin\theta\\ u+iv=(1+z)(1+z^2)=\\ =((1+\cos\theta)+i\sin\theta)\cdot\\(1+\cos^2\theta+2i\cos\theta\sin\theta-\sin^2\theta)=\\ =((1+\cos\theta)+i\sin\theta)\cdot\\((1+\cos^2\theta-\sin^2\theta)+2i\cos\theta\sin\theta)=\\ =((1+\cos\theta)+i\sin\theta)\cdot\\(2\cos^2\theta+2i\cos\theta\sin\theta)=\\ =2\cos^2\theta(1+\cos\theta)-2\cos\theta\sin^2\theta+\\ +i(2\cos\theta\sin\theta(1+\cos\theta)+2\sin\theta\cos^2\theta)\\ u=2\cos^2\theta(1+\cos\theta)-2\cos\theta\sin^2\theta=\\ =2\cos\theta(\cos\theta+\cos2\theta)\\ v=2\cos\theta\sin\theta(1+\cos\theta)+2\sin\theta\cos^2\theta=\\ =2\sin\theta\cos\theta(1+2\cos\theta)=2\cos\theta(\sin\theta+\sin2\theta)\\ u\tan\frac{3\theta}{2}=2\cos\theta(\cos\theta+\cos2\theta)\frac{\sin\frac{3\theta}{2}}{\cos\frac{3\theta}{2}}=\\ =\frac{2\cos\theta}{\cos\frac{3\theta}{2}}\frac{\sin\frac{\theta}{2}+\sin\frac{5\theta}{2}-\sin\frac{\theta}{2}+\sin\frac{7\theta}{2}}{2}=\\ =\frac{2\cos\theta}{\cos\frac{3\theta}{2}}\frac{2\sin3\theta\cos\frac{\theta}{2}}{2}=\\ =2\cos\theta(2\sin\frac{3\theta}{2}\cos\frac{\theta}{2})=\\ =2\cos\theta(\sin\theta+\sin2\theta)\\ u^2+v^2=u=4\cos^2\theta(\cos\theta+\cos2\theta)^2+\\ +4\cos^2\theta(\sin\theta+\sin2\theta)^2=\\ =4\cos^2\theta(\cos^2\theta+2\cos\theta\cos2\theta+\cos^22\theta+\\ +\sin^2\theta+2\sin\theta\sin2\theta+\sin^22\theta)=\\ =4\cos^2\theta(2+2\cos\theta)=8\cos^2\theta(1+\cos\theta)=\\ =16\cos^2\theta\cos^2\frac{\theta}{2}$