Answer to Question #214416 in Complex Analysis for prince

Question #214416

Use De Moivre’s Theorem to

(7.1) derive the 4th roots of w = −8i

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

(7.3) expand cos6θ in terms of multiple powers of z based on θ

(7.4) express cos3θ sin4θ in terms of multiple angles.


1
Expert's answer
2021-07-13T09:18:43-0400

Solution.

7.1

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)=-\\frac\u03c02=-90^\u00b0"

W=8cos(-90°)+i8sin(-90°)

By De Moivres theorem,

W1/n=r1/n[cos"\\frac{(2\u03c0k+\\theta)}{n}+isin\\frac{(2\u03c0k+\\theta)}{n}]"

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

1st root ,take k=0

"8^{\\frac 1 4}[cos(-22.5\u00b0)+isin(-22.5\u00b0)]=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

7.2

Solution

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)"

Answer;

"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

Answer

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

7.3

Solution:

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;

Answer

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

7.4

Solution;

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=(z4-z-4)4=z4-4z4+6+4z-2+z-4

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

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

Therefore;

"cos^2\\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.

Answer;

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


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