Complex number math?

z⁵ = 1

1 - z⁵ = 0

(1 - z)(1 + z + z² + z³ + z⁴) = 0

As z ≠ 1, 1 - z ≠ 0

Then, 1 + z + z² + z³ + z⁴ = 0

z + z² + z³ + z⁴ = -1

Completely

"Given that z is an imaginary root of the equation x⁵=1..."

x⁵=1 doesn't have any imaginary roots.

Solutions to z^5 - 1 = 0, known as the fifth roots of unity, are

e^i(2𝛑/5), e^i(4𝛑/5), e^i(6𝛑/5), e^i(8𝛑/5), and 1

These are at equiangular spacing arround the origin, so it is clear that none

of them are purely imaginary. Probably the question should have been:-

“Given that z is an COMPLEX root of the equation

z^5 = 1 and z ≠ 1, find the value of z + z^2 + z^3 + z^4"

z^5 – 1 = (z – 1)(z^4 + z^3 + z^2 + z + 1)

but if (z – 1) ≠ 0 then z^4 + z^3 + z^2 + z + 1 = 0

z + z^2 + z^3 + z^4 = -1

This is related to the interesting subject of cyclotomic polynomials.

If z isn't 1, then it's either cos(72) +/- i*sin(72) or else it's cos(144) +/- i*sin(144). Let's consider just one such solution, how about

z = cos(144) + i*sin(144).

Then z^2 = cos(288) + i*sin(288)

= cos(72) - i*sin(72);

z^3 = cos(432) + i*sin(432)

= cos(72) + i*sin(72);

z^4 = cos(576) + i*sin(576)

= cos(216) + i*sin(216)

= cos(144) - i*sin(144).

Add up these four numbers and you get

2*cos(144) + 2*cos(72)

= -1.

Each of the other three possible choices for z gives you the same set of four addends.

Recall:

z = a + ib ← this is a complex number

m = √(a² + b²) ← this is its modulus

tan(α) = b/a → then you can deduce α ← this is the argument

In your case:

x⁵ = 1 → you can see that the modulus is (1) and the argument is (0)

You can say that the modulus of x is: 1^(1/5) = 1

You can say that the argument of x is: 0/5 = 0

...and you can deduce that the first root of x⁵ is:

x₁ = 1.[cos(0) + i.sin(0)] → then you add (2π/5) to obtain the second root

x₂ = 1.[cos{0 + (2π/5)} + i.sin{0 + (2π/5)}] → you add (2π/5) to obtain the third root

x₃ = 1.[cos{0 + (4π/5)} + i.sin{0 + (4π/5)}] → you add another time (2π/5) to obtain the fourth root

x₄ = 1.[cos{0 + (6π/5)} + i.sin{0 + (6π/5)}] → and (2π/5) more to obtain the fifth root

x₅ = 1.[cos{0 + (8π/5)} + i.sin{0 + (8π/5)}]

Resume the roots of x⁵:

x₁ = cos(0) + i.sin(0)

x₂ = cos(2π/5) + i.sin(2π/5)

x₃ = cos(4π/5) + i.sin(4π/5)

x₄ = cos(6π/5) + i.sin(6π/5)

x₅ = cos(8π/5) + i.sin(8π/5)

...and after simplification

x₁ = 1

x₂ = cos(2π/5) + i.sin(2π/5)

x₃ = cos(4π/5) + i.sin(4π/5)

x₄ = cos(6π/5) + i.sin(6π/5)

x₅ = cos(8π/5) + i.sin(8π/5)

Given that z is an imaginary root of the equation x⁵ = 1 and z ≠ 1, so you can deduce that:

z = x₂ or x₃ or x₄ or x₅

..but as z is an imaginary root, it's no possible, because x₂, x₃, x₄ and x₅ have imaginary part and also a real part.

You meant z⁵=1?

z+z^2+z^3+z^4 is the sum of all roots apart from z = 1.

As the sum of all 5 roots is 0, z + z^2 + z^3 + z^4 = -1.