Answer to Question #212671 in Complex Analysis for Faith

Question

Use de moivres theorem to

1.derive the 4th roots of w=-8i.

2.express cos(4@) and sin(5@) in terms of powers of cos@ and sin@.

3.expand cos^6@ in terms of multiple powers of z based on @.

4.express cos^3@sin^4@ in terms of multiple angles.

NOTE:@ represents theta.

Answer 1

1.

$-8i=8(\cos(-\dfrac{\pi}{2})+i\sin(-\dfrac{\pi}{2}))$

$k=0:$

$\sqrt[4]{8}(\cos(\dfrac{-\dfrac{\pi}{2}+2\pi(0)}{4})+i\sin(\dfrac{-\dfrac{\pi}{2}+2\pi(0)}{4}))$

$=2^{3/4}(\cos(-\dfrac{\pi}{8})+i\sin(-\dfrac{\pi}{8}))$

$=2^{-1/4}\sqrt{2+\sqrt{2}}-i2^{-1/4}\sqrt{2-\sqrt{2}}$

$k=1:$

$\sqrt[4]{8}(\cos(\dfrac{-\dfrac{\pi}{2}+2\pi(1)}{4})+i\sin(\dfrac{-\dfrac{\pi}{2}+2\pi(1)}{4}))$

$=2^{3/4}(\cos(\dfrac{3\pi}{8})+i\sin(\dfrac{3\pi}{8}))$

$=2^{-1/4}\sqrt{2-\sqrt{2}}+i2^{-1/4}\sqrt{2+\sqrt{2}}$

$k=2:$

$\sqrt[4]{8}(\cos(\dfrac{-\dfrac{\pi}{2}+2\pi(2)}{4})+i\sin(\dfrac{-\dfrac{\pi}{2}+2\pi(2)}{4}))$

$=2^{3/4}(\cos(\dfrac{7\pi}{8})+i\sin(\dfrac{7\pi}{8}))$

$=-2^{-1/4}\sqrt{2+\sqrt{2}}+i2^{-1/4}\sqrt{2-\sqrt{2}}$

$k=3:$

$\sqrt[4]{8}(\cos(\dfrac{-\dfrac{\pi}{2}+2\pi(3)}{4})+i\sin(\dfrac{-\dfrac{\pi}{2}+2\pi(3)}{4}))$

$=2^{3/4}(\cos(\dfrac{11\pi}{8})+i\sin(\dfrac{11\pi}{8}))$

$=-2^{-1/4}\sqrt{2-\sqrt{2}}-i2^{-1/4}\sqrt{2+\sqrt{2}}$

2.

$\cos(4\alpha)+i\sin(4\alpha)=(\cos \alpha+i\sin \alpha)^4$

$=\cos^4\alpha+4i\cos^3\alpha\sin\alpha-6\cos^2\alpha\sin^2\alpha$

$-4i\cos\alpha\sin^3\alpha+\sin^4\alpha$

$\cos(4\alpha)=\cos^4\alpha-6\cos^2\alpha\sin^2\alpha+\sin^4\alpha$

$\cos(5\alpha)+i\sin(5\alpha)=(\cos \alpha+i\sin \alpha)^5$

$=\cos^5\alpha+5i\cos^4\alpha\sin\alpha-10\cos^3\alpha\sin^2\alpha$

$-10i\cos^2\alpha\sin^3\alpha+5\cos \alpha\sin^4\alpha+i\sin^5\alpha$

$\sin(5\alpha)=5\cos^4\alpha\sin\alpha-10\cos^2\alpha\sin^3\alpha+\sin^5\alpha$

3,

$(2\cos\alpha)^6=(z+z^{-1})^6=z^6+6z^5z^{-1}+15z^4z^{-2}$

$+20z^3z^{-3}+15z^2z^{-4}+6zz^{-5}+z^{-6}$

$=(z^6+z^{-6})+6(z^4+z^{-4})+15(z^2+z^{-2})+20$

$=2\cos(6\alpha)+12\cos(4\alpha)+30\cos(2\alpha)+20$

$\cos^6\alpha=\dfrac{1}{32}\cos(6\alpha)+\dfrac{3}{16}\cos(4\alpha)+\dfrac{15}{32}\cos(2\alpha)+\dfrac{5}{16}$

4.

$(2\cos\alpha)^3=(z+z^{-1})^3$

$(2i\sin\alpha)^4=(z-z^{-1})^4$

$128\cos^3\alpha\sin^4\alpha=(z+z^{-1})^3(z-z^{-1})^4$

$=(z^2-z^{-2})^3(z-z^{-1})$

$=(z^6-3z^2+3z^{-2}-z^{-6})(z-z^{-1})$

$=z^7-z^5-3z^3+3z+3z^{-1}-3z^{-3}-z^{-5}+z^{-7}$

$=(z^7+z^{-7})-(z^5+z^{-5})-3(z^3+z^{-3})+3(z+z^{-1})$

$=2\cos(7\alpha)-2\cos(5\alpha)-6\cos(3\alpha)+6\cos\alpha$

$\cos^3\alpha\sin^4\alpha=\dfrac{1}{64}\cos(7\alpha)-\dfrac{1}{64}\cos(5\alpha)$

$-\dfrac{3}{64}\cos(3\alpha)+\dfrac{3}{64}\cos\alpha$