# De Moivre Theorem: Solve (1+z)^5 = (1-z)^5

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),

Last edited:

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hunt_mat
Homework Helper
I think you're making this far too hard for yourself. Clearly z=0 is a solution, so expand using the binomial theorem.

vela
Staff Emeritus
Homework Helper
Or try using

$$\frac{(1-z)^5}{(1+z)^5} = \left(\frac{1-z}{1+z}\right)^5 = 1$$

Revised:
Divide both side by (1+z)^5, gives (1-z)^5 / (1+z)^5 = [(1-z) / (1+z)]^5=1
Let capital Z = [(1-z) /(1+z )], thus Z^5 = [(1-z) /(1+z)]^5 =1
Z^5 = 1 = (x + i*y) is (1+i*0y),
r = (x^2 +y^2)^1/2 = (1^2 + 0)^1/2 = 1
(theta) = tan^(-1) |y/x| =0.

z = r* e^(i*(theta))

[Z^5]^1/5 = 1^1/5 = 1*(cos(theta) + i*sin(theta))^1/5
[Z^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
= = = = = = = = = = = = = = = = = = = = = =
When:
k=0, Z(1) = (cos(2*pi*0/5) + i sin(2*pi*0/5) = e^(i*2*pi*0/5) = e^(0) = 1
k=1, Z(2) = (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, Z(3) = (cos(2*pi*2/5) + i sin(2*pi*2/5) = e^(i*2*pi*2/5) = e^(i*4*pi/5) = w^2
k=3, Z(4) = (cos(2*pi*3/5) + i sin(2*pi*3/5) = e^(i*(2*pi*3/5) = e^(i*6*pi/5) = w^3
k=4, Z(5) = (cos(2*pi*4/5) + i sin(2*pi*4/5) = e^(i*(2*pi*4/5) = e^(i*8*pi/5) = w^4

= = = = = = = = = = = = = = = = = = = = = =
Back substitute to original equation:
So, Z(1) = 1 Since: [Z] = [(1-z) /(1+z)= 1] = 1, thus = 1-1 = 0
Z(2) = w Since: [Z] = [(1-z) /(1+z)= w] = 1, thus =?????

I don’t seem to get the same answers: 0, (w-1)/(w+1), (w^2-1)/(w^2+1), (w^3-1)/(w^3+1), (w^4-1)/(w^4+1). At all possible would you please guide me through the substitution part. Thanks.

= = = = = = = = = = = = = = = = = = = = = =
I also tried binomial formula expansion way, add/subtract/factor 2*z out (z=0). Then, used the quadratic equation to solve; answers were (+/-3.077, +/- 0.7280, 0), but not sure this was what professor was looking for.

vela
Staff Emeritus
Homework Helper
One suggestion: When you're trying to solve Z5=1, do it like this:
$$Z^5 = 1 = e^{i2\pi k}$$
where k is an integer, so
$$Z = (e^{i2\pi k})^{1/5} = e^{i(2\pi/5)k} = w^k$$
where $w=e^{i2\pi/5}$. There's no need to break it into real and imaginary parts to solve for Z.

So now you have

$$\frac{1-z}{1+z} = w^k$$

Just solve for z using regular algebra.

Thank you very much Vela. However, there is one small step left, would you please guide me. Thanks.

(1-z)/(1+z) = 1 => (1-z)= (1+z) => 1-1= 0 <=> 2z=z+z, thus z=0 true.
Similiar technique:
(1-z)/(1+z)=w => (1-z)= w(1+z) => 0=w+zw-1+z => 0= (w-1)+z(w+1)
Solve for z =-(w-1)/(w+1) => z= (1-w)/(w+1)
However, z= (1-w)/(w+1) is NOT the same as (w-1)/(w+1)
...
...
z= (1-w^4)/(w^4+1) is NOT the same as (w^4-1)/(w^4+1)

Best,

vela
Staff Emeritus