Answer to Question #113650 in Linear Algebra for Morufe

Question

Given z=cos A + i sin A and u+iv =(1+z)(1+z^2).Prove that v=u tan (3A/2),u^2+ v^2=16 cos^2(A/2)cos^2(A).Hint:A=theta

Answer 1

u+iv = (1+z)(1+z²) = 1 +z² +z + z³

Since we know that if z = cosA = i sinA, then, zⁿ = cos(nA) + i sin(nA)

u+iv = 1 + cos2A + cosA + cos3A + i( sin2A + sinA + sin3A)

Comparing both sides, we get,

u= 1 + cos2A + cosA + cos3A

v= sin2A + sinA + sin3A

To prove: v = u tan(3A/2), it is enough to prove u. sin (3A/2) = v. cos (3A/2)

u. sin (3A/2)= (1 + cos2A + cosA + cos3A)sin (3A/2)

= sin (3A/2) + cos2A. sin (3A/2) + cosA. sin (3A/2) + cos3A. sin (3A/2) equation 1

Using identity 2.sin(x) . cos(y) = sin (x+y) + sin (x-y)

= (1/2) sin (3A/2) + (1/2) sin (7A/2) + (1/2) sin (5A/2) + (1/2) sin (9A/2)

v. cos (3A/2) = (sin2A + sinA + sin3A) . cos (3A/2)

= sin2A . cos (3A/2)+ sinA. cos (3A/2) + sin3A. cos (3A/2)

Using identity 2.sin(x) . cos(y) = sin (x+y) + sin (x-y)

= (1/2) sin (3A/2) + (1/2) sin (7A/2) + (1/2) sin (5A/2) + (1/2) sin (9A/2) equation 2

From equation1 and equation2,

u. sin (3A/2) = v. cos (3A/2)

v = u tan(3A/2), hence, proved.

To prove: u² + v² = 16 cos²(A/2) cos²(A/2)

u² = (1 + cos2A + cosA + cos3A)²

Using identities (a+b+c+d)² = a² +b² + c² +d² + 2(ab+bc+cd+da+ac+bd) and

2.cos(x) . cos(y) = cos (x+y) + cos (x-y)

=1 + cos²⁽2A) + cos²A + cos²(3A) + 3 cos (2A) + 3 cos(3A) + cos(5A) + 4 cos(A) equation3

v² = ( sin2A + sinA + sin3A)²

Using identities (a+b+c)² = a² +b² + c² + 2(ab+bc+ac) and

2.sin(x) . sin(y) = cos (x-y) - cos (x+y)

= sin² (2A) + sin²(A) + sin²(3A) + 2cosA - cos(3A) + cos(2A) - cos(4A) - cos(5A) equation4

Using identity sin²(x) + cos²(x) = 1,from equations 3 and 4 ,

u² + v² = 4+4cos(2A)+2cos(3A)+6cos(A)

= 4(1+cos(2A)) + 4cos(A)(1+cos(2A))

= 4 (1+cos(A)) . (1+cos(2A))

Using 2cos²(x) = 1+ cos(2x)

= 16 cos²(A) cos²(A/2)