Answer to Question #113665 in Trigonometry for Caylin

Question

Let w be a negative real number, z a 6th root of w.

a) Show that z(k) = p^1/6 [cos ( pi + 2 * k* pi / 6) + i sin ( pi + 12 * k * pi / 6)], k = v, 1,2,3,4,5 is a formula for the 6th roots of w.

b) Hence determine 6th roots of - 729.

c) Given z = cosA + 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^2A

Answer 1

Let us assume that $w=-p$, where $p$ is a positive real number. Next we use the trigonometric form of complex number

$p= \rho(\cos\varphi + i\sin\varphi).$

For positive real numbers $\varphi = 0$ and $\rho=p$ , so

$p = p(\cos 0 + i\sin 0).$

Therefore

$w=-p=-p(\cos 0 + i\sin 0) = p(\cos\pi + i\sin\pi),$

because $\cos\pi = -\cos 0$ and $\sin\pi = \sin 0.$

a) Let us assume that

$z = \rho(\cos\theta+i\sin\theta)$

According to de Moivre's formula,

$z^6 = \rho^6(\cos(6\theta)+i\sin(6\theta))$.

Since $z$ is a 6th root of $w$,

$z^6=w$

and

$\rho^6(\cos(6\theta)+i\sin(6\theta)) = p(\cos\pi +i\sin\pi).$

Therefore we get $\rho^6=p$ , $\cos(6\theta)=\cos\pi$ , $\sin(6\theta)=\sin\pi.$ Then we obtain

$\rho=p^{1/6}, 6\theta=\pi +2\pi k, k \in \mathbb{Z}.$

Finally, we get

$\rho=p^{1/6}, \theta(k)=\dfrac{\pi +2\pi k}{6}, k \in \mathbb{Z}.$

We can see that we will obtain the same $\sin\theta, \cos\theta$ for $k\equiv_6 m$ , so we consider only $k=0,1,\ldots 5.$

Next we obtain

$z(k)=p^{1/6}(\cos(\frac{\pi+2\pi k}{6}) + i\sin(\frac{\pi+2\pi k}{6})), k = 0,\,\ldots, 5.$

b) Here $w=-729$ . In trigonometric form

$w=729(\cos\pi + i\sin\pi),$

according to the formula in a) 6th roots are

$z(k)= 729^{1/6}(\cos(\frac{\pi+2\pi k}{6}) + i\sin(\frac{\pi+2\pi k}{6})), k = 0,\,\ldots, 5.$

Therefore,

$z(0)=3(\cos(\frac{\pi}{6}) + i\sin(\frac{\pi}{6})) = 3(\frac{\sqrt{3}}{2}+\frac12i),$

$z(1)= 3(\cos(\frac{\pi+2\pi}{6}) + i\sin(\frac{\pi+2\pi}{6})) = 3(0+i)= 3i,$

$z(2)=3(\cos(\frac{\pi+4\pi}{6}) + i\sin(\frac{\pi+4\pi}{6})) = 3(-\frac{\sqrt3}{2} +\frac12i),$

$z(3)=3(\cos(\frac{\pi+6\pi}{6}) + i\sin(\frac{\pi+6\pi}{6})) = 3(-\frac{\sqrt3}{2} -\frac12i),$

$z(4)=3(\cos(\frac{\pi+8\pi}{6}) + i\sin(\frac{\pi+8\pi}{6})) = 3(0-i) = -3i,$

$z(5)=3(\cos(\frac{\pi+10\pi}{6}) + i\sin(\frac{\pi+10\pi}{6})) = 3(\frac{\sqrt3}{2} -\frac12i).$

c) Let us consider $w_1=1+z = (1+\cos A) + i\sin A.$ The modulus of $w_1$ is

$\rho_1=\sqrt{(1+\cos A)^2+(\sin A)^2} = \sqrt{1+2\cos A + \cos^2A + \sin^2A} = \sqrt{2+2\cos A}.$

Let us consider $w_2=1+z^2 = (1+ \cos2A) + i\sin 2A.$ The modulus of $w_2$ is

$\rho_2=\sqrt{(1+\cos 2A)^2+(\sin2 A)^2} = \sqrt{1+2\cos 2A + \cos^22A + \sin^22A} = \sqrt{2+2\cos 2A}.$

If we use the trigonometric form, the modulus of a product is equal to the product of moduli.

The sum $u^2+v^2$ is the squared modulus of $w_1\cdot w_2,$ therefore

$u^2+v^2 = \rho_1^2\rho_2^2 = (2+2\cos A)\cdot(2+2\cos 2A) = 16(\dfrac{1+\cos A}{2})\cdot(\dfrac{1+\cos 2A}{2}) = 16\cos^2\frac{A}{2}\cos^2A.$

Next we'll simplify the formula

$u+iv = (1+z)(1+z^2) = 1+z+z^2+z^3.$

According to de Moivre's formula,

$1+z+z^2+z^3 = 1+ (\cos A + i\sin A) + (\cos 2A+i\sin 2A) + (\cos 3A+i\sin3A) = (1+\cos A+ \cos 2A + \cos 3A) + i\cdot(\sin A+ \sin2A+\sin3A).$

Therefore,

$u = 1+\cos A+ \cos 2A + \cos 3A, v=\sin A+ \sin2A+\sin3A.$

Using several trigonometric identities, we obtain

$u = 1 + \cos A+ \cos 2A + \cos 3A = 1 + \cos A + 2\cos^2A-1+ 4\cos^3A-3\cos A= 4\cos^3A+2\cos^2A-2\cos A = 2\cos A \cdot(\cos A+1)\cdot(2\cos A -1).$

Next we obtain

$v=\sin A+ \sin2A+\sin3A = \sin A + 2\sin A\cos A + 3\sin A(1-\sin^2A) - \sin^3A = 2 \cos A\cdot\sin A\cdot( 2\cos A +1).$

Let us consider $\tan(3A/2)$:

$\tan\dfrac{3A}{2} = \dfrac{\sin3A}{1+\cos3A} = \dfrac{(2\cos A+1)\cdot\sin A}{(\cos A+1)\cdot(2\cos A-1)}.$

We should prove that $v = u \tan(\frac{3A}{2})$ . Let us substitute the expressions obtained above:

$u\tan(\frac{3A}{2}) = 2\cos A\cdot(\cos A +1) \cdot(2\cos A - 1) \cdot \dfrac{(2\cos A+1)\cdot\sin A}{(\cos A+1)\cdot(2\cos A-1)}= 2\cos A\cdot (2\cos A+1)\cdot\sin A = v.$