Solution.
7.1
Derive the 4th roots of W=-8i
Write the complex number in polar form
If W=a+bi=rcosθ \theta θ θ \theta θ
r=a 2 + b 2 = 0 + ( − 8 ) 2 = 8 \sqrt{a^2+b^2}=\sqrt{0+(-8)^2}=8 a 2 + b 2 = 0 + ( − 8 ) 2 = 8
θ = t a n − 1 ( − 8 0 ) = − π 2 = − 9 0 ° \theta=tan^-1(\frac{-8 }0)=-\fracπ2=-90^° θ = t a n − 1 ( 0 − 8 ) = − 2 π = − 9 0 °
W=8cos(-90°)+i8sin(-90°)
By De Moivres theorem,
W1/n =r1/n [cos( 2 π k + θ ) n + i s i n ( 2 π k + θ ) n ] \frac{(2πk+\theta)}{n}+isin\frac{(2πk+\theta)}{n}] n ( 2 πk + θ ) + i s in n ( 2 πk + θ ) ]
Where k=0,1,2...,n-1
1st root ,take k=0
8 1 4 [ c o s ( − 22.5 ° ) + i s i n ( − 22.5 ° ) ] = 1.554 − 0.6436 i 8^{\frac 1 4}[cos(-22.5°)+isin(-22.5°)]=1.554-0.6436i 8 4 1 [ cos ( − 22.5° ) + i s in ( − 22.5° )] = 1.554 − 0.6436 i
2nd root ,take k=1
8 1 4 8^\frac1 4 8 4 1
3rd root ,take k=2
8 1 4 8^\frac14 8 4 1
4th root,take k=3
81 4 ^\frac14 4 1
7.2
Solution
De Moivres theorem;
C o s n θ + i S i n n θ = ( c o s θ + i s i n θ ) n Cosn\theta+iSin n\theta=(cos\theta+isin\theta)^n C os n θ + i S inn θ = ( cos θ + i s in θ ) n
Cos 4 θ = R e ( c o s θ + i S i n θ ) 4 ) 4\theta=Re(cos\theta+iSin\theta)^4) 4 θ = R e ( cos θ + i S in θ ) 4 )
C o s 4 θ = R e ( C o s 4 θ + 4 C o s 3 θ ( i s i n θ ) + 6 c o s 2 θ ( i s i n θ ) 2 + 4 c o s θ ( i s i n θ ) 3 + ( i s i n θ ) 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) C os 4 θ = R e ( C o s 4 θ + 4 C o s 3 θ ( i s in θ ) + 6 co s 2 θ ( i s in θ ) 2 + 4 cos θ ( i s in θ ) 3 + ( i s in θ ) 4 )
Answer;
C o s 4 θ = c o s 4 θ − 6 c o s 2 θ s i n 2 θ + s i n 4 θ Cos4\theta=cos^4\theta-6cos^2\theta sin^2\theta+sin^4\theta C os 4 θ = co s 4 θ − 6 co s 2 θ s i n 2 θ + s i n 4 θ
sin 5 θ 5\theta 5 θ
By De Moivre's Theorem;
C o s 5 θ + i s i n 5 θ = ( c o s θ + i s i n θ ) 5 Cos5\theta+isin5\theta=(cos\theta+isin\theta)^5 C os 5 θ + i s in 5 θ = ( cos θ + i s in θ ) 5
C o s 5 θ Cos5\theta C os 5 θ i s i n 5 θ isin5\theta i s in 5 θ
C o s 5 θ + i s i n 5 θ = c o s 5 θ + 5 c o s 4 θ i s i n θ + 10 c o s 3 θ i 2 s i n 2 θ + 10 c o s 2 θ i 3 s i n 3 θ + 5 c o s θ i 4 s i n 4 θ + i 5 s i n 5 θ 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 C os 5 θ + i s in 5 θ = co s 5 θ + 5 co s 4 θ i s in θ + 10 co s 3 θ i 2 s i n 2 θ + 10 co s 2 θ i 3 s i n 3 θ + 5 cos θ i 4 s i n 4 θ + i 5 s i n 5 θ
Group real parts and imaginary parts together;
C o s 5 θ + i s i n 5 θ = ( c o s 5 θ − 10 c o s 3 θ s i n 2 θ + 5 c o s θ s i n 4 θ ) + i ( 5 c o s 4 θ s i n θ − 10 c o s 2 θ s i n 3 θ + s i n 5 θ ) 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) C os 5 θ + i s in 5 θ = ( co s 5 θ − 10 co s 3 θ s i n 2 θ + 5 cos θ s i n 4 θ ) + i ( 5 co s 4 θ s in θ − 10 co s 2 θ s i n 3 θ + s i n 5 θ )
Equate the imaginary parts of both sides together;
i s i n 5 θ = i ( 5 c o s 4 θ s i n θ − 10 c o s 2 θ s i n 3 θ + s i n 5 θ ) isin5\theta=i(5cos^4\theta sin\theta-10cos^2\theta sin^3\theta+sin^5\theta) i s in 5 θ = i ( 5 co s 4 θ s in θ − 10 co s 2 θ s i n 3 θ + s i n 5 θ )
Divide out i
Answer
S i n 5 θ = 5 c o s 4 θ s i n θ − 10 c o s 2 θ s i n 3 θ + s i n 5 θ ) Sin5\theta=5cos^4\theta sin\theta-10cos^2\theta sin^3\theta+sin^5\theta) S in 5 θ = 5 co s 4 θ s in θ − 10 co s 2 θ s i n 3 θ + s i n 5 θ )
7.3
Solution:
By De Moivre's Theorem;
zn +z-n =2cosnθ \theta θ
Take n=1
2cosθ \theta θ 1 +z-1
(2cosθ \theta θ 6 =(z1 +z-1 )6
(2cosθ \theta θ 6 =z6 +6z4 +15z2 +20+15z-2 +6z-4 +z-6
64cos6 θ \theta θ 6 +z-6 )+6(z4 +z-4 )+15(z2 +z-2 )+20
64 c o s 6 θ = 2 c o s 6 θ + 12 c o s 4 θ + 30 c o s 2 θ + 20 64cos^6\theta=2cos6\theta+12cos4\theta+30cos2\theta+20 64 co s 6 θ = 2 cos 6 θ + 12 cos 4 θ + 30 cos 2 θ + 20
Divide by 64;
Answer
C o s 6 θ = 2 32 c o s 6 θ + 3 16 c o s 4 θ + 15 32 c o s 2 θ + 5 16 Cos^6\theta=\frac 2{32}cos6\theta+\frac3{16}cos4\theta+\frac{15}{32}cos2\theta+\frac5{16} C o s 6 θ = 32 2 cos 6 θ + 16 3 cos 4 θ + 32 15 cos 2 θ + 16 5
7.4
Solution;
Find C o s 3 θ Cos^3\theta C o s 3 θ
(2cosθ \theta θ 3 =(z1 +z-1 )3 =z3 +3z1 +3z-1 +z-3
8cos3 θ \theta θ 3 +z-3 )+3(z1 +z-1 )
=2c o s 3 θ + 6 c o s θ cos3\theta+6cos\theta cos 3 θ + 6 cos θ
cos3 θ = 1 4 c o s 3 θ + 3 4 c o s θ ^3\theta=\frac14cos3\theta+\frac34cos\theta 3 θ = 4 1 cos 3 θ + 4 3 cos θ
Find sin4 θ \theta θ
(2isinθ \theta θ 4 =(z4 -z-4 )4 =z4 -4z4 +6+4z-2 +z-4
16sin4 θ ^4\theta 4 θ 2 c o s 4 θ − 6 c o s 2 θ + 6 2cos4\theta-6cos2\theta+6 2 cos 4 θ − 6 cos 2 θ + 6
S i n 4 θ = 1 8 c o s 4 θ − 1 2 c o s 2 θ + 3 8 Sin^4\theta=\frac18cos4\theta-\frac12cos2\theta+\frac38 S i n 4 θ = 8 1 cos 4 θ − 2 1 cos 2 θ + 8 3
Therefore;
c o s 2 θ s i n 4 θ = cos^2\theta sin^4\theta= co s 2 θ s i n 4 θ = 1 4 c o s 3 θ + 3 8 c o s θ ) ( 1 8 c o s 4 θ − 1 2 c o s 2 θ + 3 8 ) \frac14cos3\theta+\frac38cos\theta)(\frac18cos4\theta-\frac12cos2\theta+\frac38) 4 1 cos 3 θ + 8 3 cos θ ) ( 8 1 cos 4 θ − 2 1 cos 2 θ + 8 3 )
Using the identity ;
CosAcosB=1 2 \frac12 2 1
Simplify.
Answer;
1 64 ( c o s 7 θ − c o s 5 θ − 3 c o s 3 θ + 3 c o s θ ) \frac1{64}(cos7\theta-cos5\theta-3cos3\theta+3cos\theta) 64 1 ( cos 7 θ − cos 5 θ − 3 cos 3 θ + 3 cos θ )
#340153 on Dec 2023
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