- #1

hugo28

- 7

- 0

Divide both side by (1+z)^5:

(1-z)^5 / (1+z)^5 = 1

Let Z1 = (1-z); (Z1)^5 = (1-z)^5

Z2 = (1+z); (Z2)^5 = (1+z)^5

Work numerator and denominator separately:

(a) Numerator: (Z1)^5 = 1 = (cos(theta) + I sin(theta))

since (Z1)^5 = 1 => (x + iy) is (1+i*0y), thus r = (x^2 +y^2)^1/2 = (1^2 + 0)^1/2 = 1

thus, theta = tan^-1 |y/x| = 0

(Z(1)^5)^1/5 = 1 = ((cos(theta)+2*pi*k) + (i sin(theta) +2*pi*k))^1/5

= ((cos(0)+2*pi*k) + (i sin(0) +2*pi*k))^1/5

= (cos(2*pi*k) + i sin(2*pi*k))^1/5

Z(1) = 1 = (cos(2*pi*k/5) + i sin(2*pi*k/5) = e^(i*(2*pi*k/5)

k=0, (cos(2*pi*0/5) + i sin(2*pi*0/5) = e^(i*2*pi*0/5) = e^(0) = 1

k=1, (cos(2*pi*1/5) + i sin(2*pi*1/5) = e^(i*2*pi*1/5) = e^(i*2*pi/5) = w(1)

Let set: e^(i*2*pi/5) = w, then:

k=2, (cos(2*pi*2/5) + i sin(2*pi*2/5) = e^(i*2*pi*2/5) = e^(i*4*pi/5) = w(1)^2

k=3, (cos(2*pi*3/5) + i sin(2*pi*3/5) = e^(i*(2*pi*3/5) = e^(i*6*pi/5) = w(1)^3

k=4, (cos(2*pi*4/5) + i sin(2*pi*4/5) = e^(i*(2*pi*4/5) = e^(i*8*pi/5) = w(1)^4

(b) Denominator: similar work. Z(2) = 1, w(2), w(2)^2, w(2)^3, w(2)^4

Substitute back to (Z1)^5 / (Z2)^5 = (1-z)^5 / (1+z)^5 =

= (1, w(1), w(1)^2, w(1)^3, w(1)^4) / (1, w(2), w(2)^2, w(3)^3, w(4)^4)

I don’t know what the necessary steps that need to take, at all possible would you please guide me through. Thanks.

Answers are: 0, (w-1)/(w+1), (w^2-1)/(w^2+1), (w^3-1)/(w^3+1), (w^4-1)/(w^4+1),

(1-z)^5 / (1+z)^5 = 1

Let Z1 = (1-z); (Z1)^5 = (1-z)^5

Z2 = (1+z); (Z2)^5 = (1+z)^5

Work numerator and denominator separately:

(a) Numerator: (Z1)^5 = 1 = (cos(theta) + I sin(theta))

since (Z1)^5 = 1 => (x + iy) is (1+i*0y), thus r = (x^2 +y^2)^1/2 = (1^2 + 0)^1/2 = 1

thus, theta = tan^-1 |y/x| = 0

(Z(1)^5)^1/5 = 1 = ((cos(theta)+2*pi*k) + (i sin(theta) +2*pi*k))^1/5

= ((cos(0)+2*pi*k) + (i sin(0) +2*pi*k))^1/5

= (cos(2*pi*k) + i sin(2*pi*k))^1/5

Z(1) = 1 = (cos(2*pi*k/5) + i sin(2*pi*k/5) = e^(i*(2*pi*k/5)

k=0, (cos(2*pi*0/5) + i sin(2*pi*0/5) = e^(i*2*pi*0/5) = e^(0) = 1

k=1, (cos(2*pi*1/5) + i sin(2*pi*1/5) = e^(i*2*pi*1/5) = e^(i*2*pi/5) = w(1)

Let set: e^(i*2*pi/5) = w, then:

k=2, (cos(2*pi*2/5) + i sin(2*pi*2/5) = e^(i*2*pi*2/5) = e^(i*4*pi/5) = w(1)^2

k=3, (cos(2*pi*3/5) + i sin(2*pi*3/5) = e^(i*(2*pi*3/5) = e^(i*6*pi/5) = w(1)^3

k=4, (cos(2*pi*4/5) + i sin(2*pi*4/5) = e^(i*(2*pi*4/5) = e^(i*8*pi/5) = w(1)^4

(b) Denominator: similar work. Z(2) = 1, w(2), w(2)^2, w(2)^3, w(2)^4

Substitute back to (Z1)^5 / (Z2)^5 = (1-z)^5 / (1+z)^5 =

= (1, w(1), w(1)^2, w(1)^3, w(1)^4) / (1, w(2), w(2)^2, w(3)^3, w(4)^4)

I don’t know what the necessary steps that need to take, at all possible would you please guide me through. Thanks.

Answers are: 0, (w-1)/(w+1), (w^2-1)/(w^2+1), (w^3-1)/(w^3+1), (w^4-1)/(w^4+1),

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