Answer to Question #218692 in Linear Algebra for Tshego

Question

1.)Use De Moivre’s Theorem to

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

b.) express cos(4θ) and sin(5θ) in terms of powers of cos θ and sin θ

c.) expand cos^6 θ in terms of multiple powers of z based on θ

d.) express cos3θ sin4 θ in terms of multiple angles.

(2.)
a.) Let z =z1/z2 where z1 = tan θ + i and z2 = z1. Find an expression for z^n with n ∈ N.
b.) Let z = cos θ - i(1 + sin θ). Determine

2z + i/-1 - iz

Answer 1

Solution.

1a

Derive the 4th roots of W=-8i

Write the complex number in polar form

If W=a+bi=rcos$\theta$ +irsin$\theta$

r=$\sqrt{a^2+b^2}=\sqrt{0+(-8)^2}=8$

$\theta=tan^-1(\frac{-8 }0)=-\fracπ2=-90^°$

W=8cos(-90°)+i8sin(-90°)

By De Moivres theorem,

W^1/n=r^1/n[cos$\frac{(2πk+\theta)}{n}+isin\frac{(2πk+\theta)}{n}]$

Where k=0,1,2...,n-1

1st root ,take k=0

$8^{\frac 1 4}[cos(-22.5°)+isin(-22.5°)]=1.554-0.6436i$

2nd root ,take k=1

$8^\frac1 4$ [Cos(-22.5°+90°)+iSin(-22.5°+90°)]=0.6436+1.554i

3rd root ,take k=2

$8^\frac14$[Cos(-22.5+180°)+iSin (-22.5°+180°)]=-1.554+0.6436i

4th root,take k=3

8$^\frac14$ [Cos(-22.5+270°)+iSin (-22.5°+270°)=-0.6436-1.554i

1b

De Moivres theorem;

$Cosn\theta+iSin n\theta=(cos\theta+isin\theta)^n$

Cos $4\theta=Re(cos\theta+iSin\theta)^4)$

$Cos4\theta=Re(Cos^4\theta+4Cos^3\theta(isin\theta)+6cos^2\theta(isin\theta)^2+4cos\theta(isin\theta)^3+(isin\theta)^4)$

$Cos4\theta=cos^4\theta-6cos^2\theta sin^2\theta+sin^4\theta$

sin$5\theta$

By De Moivre's Theorem;

$Cos5\theta+isin5\theta=(cos\theta+isin\theta)^5$

$Cos5\theta$ +$isin5\theta$ can now be expressed as;

$Cos5\theta+isin5\theta=cos^5\theta+5cos^4\theta isin\theta+10cos^3\theta i^2sin^2\theta+10cos^2\theta i^3sin^3\theta+5cos\theta i^4sin^4\theta+i^5sin^5\theta$

Group real parts and imaginary parts together;

$Cos5\theta+isin5\theta=(cos^5\theta-10cos^3\theta sin^2\theta+5cos\theta sin^4\theta)+i (5cos^4\theta sin\theta-10cos^2\theta sin^3\theta+sin^5\theta)$

Equate the imaginary parts of both sides together;

$isin5\theta=i(5cos^4\theta sin\theta-10cos^2\theta sin^3\theta+sin^5\theta)$

Divide out i

$Sin5\theta=5cos^4\theta sin\theta-10cos^2\theta sin^3\theta+sin^5\theta)$

1c

By De Moivre's Theorem;

zⁿ+z^-n=2cosn$\theta$

Take n=1

2cos$\theta$ =z¹+z^-1

(2cos$\theta$ )⁶=(z¹+z^-1)⁶

(2cos$\theta$)⁶ =z⁶+6z⁴+15z²+20+15z^-2+6z^-4+z^-6

64cos⁶$\theta$ =(z⁶+z^-6)+6(z⁴+z^-4)+15(z²+z^-2)+20

$64cos^6\theta=2cos6\theta+12cos4\theta+30cos2\theta+20$

Divide by 64;

$Cos^6\theta=\frac 2{32}cos6\theta+\frac3{16}cos4\theta+\frac{15}{32}cos2\theta+\frac5{16}$

1d

Find $Cos^3\theta$

(2cos$\theta$)³=(z¹+z^-1)³=z³+3z¹+3z^-1+z^-3

8cos³$\theta$ =(z³+z^-3)+3(z¹+z^-1)

=2$cos3\theta+6cos\theta$

cos$^3\theta=\frac14cos3\theta+\frac34cos\theta$

Find sin⁴$\theta$

(2isin$\theta$ )⁴=(z-z^-1)⁴=z⁴-4z²+6-4z^-2+z^-4

16sin$^4\theta$ =$2cos4\theta-6cos2\theta+6$

$Sin^4\theta=\frac18cos4\theta-\frac12cos2\theta+\frac38$

Therefore;

$cos^3\theta sin^4\theta=$($\frac14cos3\theta+\frac38cos\theta)(\frac18cos4\theta-\frac12cos2\theta+\frac38)$

Using the identity ;

CosAcosB=$\frac12$ [Cos(A+B)+Cos(A-B)

Simplify.

$\frac1{64}(cos7\theta-cos5\theta-3cos3\theta+3cos\theta)$

2a

$z =\frac{z_1}{z_2} \\ z_1 = tan θ + i \\ z_2 = \bar{z_1} = \bar{(\tan\theta+i)}=tan \theta -i\\ z =\frac{z_1}{z_2} =\frac{tan \theta +i}{tan \theta -i}\\ z =\frac{tan \theta +i}{tan \theta -i}*\frac{tan \theta +i}{tan \theta +i}\\ Simplifying \\ z=- cos 2\theta + sin2 \theta i\\ Hence\\ e^{i \theta}= cos \theta + isin \theta\\ e^{-i \theta}= cos \theta - isin \theta\\ \implies e^{i2\theta}= cos 2\theta + isin 2\theta\\ \space \space \space \space \space \space \space \space \space \space e^{-i \theta}= cos 2\theta -isin2 \theta\\ Adding \space them\\ cos 2 \theta = \frac{e^{i2 \theta }-e^{-i2 \theta }}{2}\\ Also, sin 2 \theta = \frac{e^{i2 \theta }-e^{-i2 \theta }}{2i}\\ z= \frac{e^{i2 \theta }-e^{-i2 \theta }}{2}+ \frac{e^{i2 \theta }-e^{-i2 \theta }}{2i}\\ z= -e^{i2 \theta }\\ \therefore z^n= (-e^{i2 \theta })^n = (-)^ne^{i2n \theta }$

2b

$z = cos θ - i(1 + sin θ)\\ |\frac{\bar{2z+i}}{-1-iz}|=|\frac{\bar{2(cos θ - i(1 + sin θ))+i}}{-1-i(cos θ - i(1 + sin θ))}|\\ =\frac{\bar{2cos θ - i(1 + 2sin θ)}}{(-2- sin θ)-icos \theta}*\frac{-2-sin \theta+icos \theta}{(-2- sin θ)-icos \theta}\\ = \frac{-3 cos \theta + i (5 sin \theta +3)}{4 sin \theta +5}\\ Conjugate \\ \frac{-3 cos \theta + i (5 sin \theta +3)}{4 sin \theta +5}\\ |\frac{\bar{2z+i}}{-1-iz}|= \sqrt{(\frac{-3 cos \theta }{4 sin \theta +5})^2+(\frac{5 sin \theta +3 }{4 sin \theta +5})^2}\\ = \frac{ \sqrt{16 sin^2+30sin \theta +18}}{4 sin \theta +5}$