Verified answer

z⁵ = (1+i) = ... r⁵ e^(i ¼𝝅), r⁵ e^(i ⁹∕₄𝝅), r⁵ e^(i ¹⁷∕₄𝝅), r⁵ e^(i ²⁵∕₄𝝅), r⁵ e^(i ³³∕₄𝝅), ...

z = ... r e^(i ¹∕₂₀𝝅), r e^(i ⁹∕₂₀𝝅), r e^(i ¹⁷∕₂₀𝝅), r e^(i ⁵∕₄𝝅), r⁵ e^(i ³³∕₂₀𝝅), ...

θ = ¹∕₂₀𝝅, ⁹∕₂₀𝝅, ¹⁷∕₂₀𝝅, ⁵∕₄𝝅, and ³³∕₂₀𝝅 for 0 ≤ θ < 2𝝅

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

z⁵ = 1 + i

The modulus is of z⁵ is: m = √[(1)² + (1)²] → m = √2 = 2^(1/2)

The argument is of z⁵ is such as: tan(α) = 1/1 = 1 → α = π/4

According this result, you can deduce that the modulus of z is: → m^(1/5) = [2^(1/2)]^(1/5) = 2^(1/10)

According this result, you can deduce that the argument of z is: θ = α/5 = (π/4)/5 = π/20

…and you can write:

z₁ = 2^(1/10) * [cos(π/20) + i.sin(π/20)] → to get the second root, you add (2π/5)

z₂ = 2^(1/10) * [cos{(π/20) + (2π/5)} + i.sin{(π/20) + (2π/5)}] → to get the third root, you add (2π/5)

z₃ = 2^(1/10) * [cos{(π/20) + (4π/5)} + i.sin{(π/20) + (4π/5)}] → to get the fourth root, you add (2π/5)

z₄ = 2^(1/10) * [cos{(π/20) + (6π/5)} + i.sin{(π/20) + (6π/5)}] → to get the fifth root, you add (2π/5)

z₅ = 2^(1/10) * [cos{(π/20) + (8π/5)} + i.sin{(π/20) + (8π/5)}]

Resume:

z₁ = 2^(1/10) * [cos(π/20) + i.sin(π/20)]

z₂ = 2^(1/10) * [cos(9π/20) + i.sin(9π/20)]

z₃ = 2^(1/10) * [cos(17π/20) + i.sin(17π/20)]

z₄ = 2^(1/10) * [cos(25π/20) + i.sin(25π/20)]

z₄ = 2^(1/10) * [cos(5π/4) + i.sin(5π/4)]

z₅ = 2^(1/10) * [cos(33π/20) + i.sin(33π/20)]

z₅ = 2^(1/10) * [cos{(20π + 13π)/20} + i.sin{(20π + 13π)/20)}]

z₅ = 2^(1/10) * [cos{π + (13π/20)} + i.sin{π + (13π/20)}]

z₅ = 2^(1/10) * [- cos(13π/20) - i.sin(13π/20)]

z₅ = - 2^(1/10) * [cos(13π/20) + i.sin(13π/20)]

The answer as: z = r.e^(iθ)

r = 2^(1/10) ← this is its modulus of z

θ = π/20 ← this is its argument of z

z₁ = 2^(1/10) * [cos(π/20) + i.sin(π/20)] = 2^(1/10) * e^(i.π/20)

z₂ = 2^(1/10) * [cos(9π/20) + i.sin(9π/20)] = 2^(1/10) * e^(i.9π/20)

z₃ = 2^(1/10) * [cos(17π/20) + i.sin(17π/20)] = 2^(1/10) * e^(i.17π/20)

z₄ = 2^(1/10) * [cos(5π/4) + i.sin(5π/4)] = 2^(1/10) * e^(i.5π/40)

z₅ = - 2^(1/10) * [cos(13π/20) + i.sin(13π/20)] = 2^(1/10) * e^(i.13π/40)

Let a + ib = 1 + i

We know that:

• IF to> 0 THEN θ = arctg (b / a)

• IF to <0 THEN θ = π + arctg (b / a)

In our case θ = arctg (1/2) = π/4

We are interested in the fifth root then

θk=θ/5+2kπ/5 for k=0,1,2,3,4.

So we get

for k=0 ⇒ θ₁=π/20

for k=1 ⇒ θ₂=π/20+2π/5=9π/20

for k=2 ⇒ θ₃=π/20+4π/5=17π/20

for k=3 ⇒ θ₄=π/20+6π/5=25π/20=5π/4

for k=4 ⇒ θ₅=π/20+8π/5=33π/20