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

• hugo28
In summary: This leads to the solutions you listed. In summary, by dividing both sides by (1+z)^5, we can rewrite the equation as [(1-z)/(1+z)]^5=1. Using the properties of complex numbers and the quadratic formula, we can solve for z and find the solutions to be 0, (w-1)/(w+1), (w^2-1)/(w^2+1), (w^3-1)/(w^3+1), (w^4-1)/(w^4+1), where w=e^(i*2*pi/5). Alternatively, we can also start with [(1+z)/(1-z)]^5=1, which leads to the solutions (
hugo28
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:
I think you're making this far too hard for yourself. Clearly z=0 is a solution, so expand using the binomial theorem.

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.

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)

If you calculate the solutions using both formulas, you'll find you get the same answers, just in a different order. They should also match the solutions you found by expanding the polynomials.

## 1. What is "De Moivre Theorem"?

The De Moivre Theorem is a mathematical formula that relates to complex numbers. It states that if a complex number is raised to a power, the result can be found by multiplying the modulus (magnitude) of the number by the angle (argument) raised to that same power.

## 2. How do you solve (1+z)^5 = (1-z)^5 using De Moivre Theorem?

To solve this equation, we can start by writing both sides in the polar form (r∠θ). We then use De Moivre Theorem to raise each side to the fifth power. This will give us two equations in the form of r^5∠5θ = r^5∠5θ. We can then equate the moduli (r) and the arguments (θ) to find the values of z that satisfy the equation.

## 3. What is the significance of the exponent being raised to the fifth power in this equation?

The fifth power is significant because it is the same as the number of roots of unity (5) in the equation. This means that there will be five solutions to this equation, which can be found using De Moivre Theorem.

## 4. Can De Moivre Theorem be applied to any power or just the fifth power?

De Moivre Theorem can be applied to any power, not just the fifth power. It is most commonly used for powers that are multiples of the number of roots of unity in the equation, as it simplifies the solution process.

## 5. What are some real-life applications of De Moivre Theorem?

De Moivre Theorem has many applications in fields such as engineering, physics, and economics. It is used to solve problems involving alternating currents, oscillatory motion, and complex numbers in general.

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