Please help me with my Trigonometry?

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

When you multiply 2 complex numbers togethers, you obtain a new complex number.

The modulus of the new complex number is the product of each modulus.

The argument of the new complex number is the sum of each argument.

Now, let's take an example: z³

The modulus of z³ will be: m³

The argument of z³ will be: 3α

z = 2.[cos(π/20)) + i.sin(π/20)]

The modulus od z is 2 → the modulus of z⁵ is: 2⁵ = 32

The argument od z is (π/20) → the argument of z⁵ is: 5 * (π/20) = π/4

Then you can deduce that:

z⁵ = { 2.[cos(π/20)) + i.sin(π/20)] }⁵

z⁵ = 32.[cos(π/4)) + i.sin(π/4)]

z⁵ = 32.[(√2)/2 + i.(√2)/2]

z⁵ = 16.(√2 + i.√2)

z⁵ = 16√2.(1 + i)

z = √1.[cos(3π/14) + i.sin(3π/14)]

The modulus od z is √1, i.e. 1 → the modulus of z⁷ is: 1⁷ = 1

The argument od z is (3π/14) → the argument of z⁷ is: 7 * (3π/14) = 3π/2

Then you can deduce that:

z⁷ = { √1.[cos(3π/14) + i.sin(3π/14)] }⁷

z⁷ = cos(3π/2) + i.sin(3π/2)

z⁷ = - i

z = √5.[cos(5π/16) + i.sin(5π/16)]

The modulus od z is √5, i.e. 1 → the modulus of z⁴ is: (√5)⁴ = √5 * √5 * √5 * √5 = 25

The argument od z is (5π/16) → the argument of z⁴ is: 4 * (5π/16) = 5π/4

Then you can deduce that:

z⁴ = { √5.[cos(5π/16) + i.sin(5π/16)] }⁴

z⁴ = 25.[cos(5π/4) + i.sin(5π/4)]

z⁴ = 25.[cos{(4π + π)/4} + i.sin{(4π + π)/4}]

z⁴ = 25.[cos{π + (π/4)} + i.sin{π + (π/4)}]

z⁴ = 25.[- cos(π/4) - i.sin(π/4)]

z⁴ = - 25.[cos(π/4) + i.sin(π/4)]

z⁴ = - 25.[(√2)/2 + i.(√2)/2]

z⁴ = - [(25√2)/2].(1 + i)

z = (³√7).[cos(π/18) + i.sin(π/18)]

The modulus od z is (³√7), i.e. 7^(1/3) → the modulus of z⁶ is: [7^(1/3)]⁶ = 7^(6/3) = 7² = 49

The argument od z is (π/18) → the argument of z⁶ is: 6 * (π/18) = π/3

Then you can deduce that:

z⁶ = { (³√7).[cos(π/18) + i.sin(π/18)] }⁶

z⁶ = 49.[cos(π/3) + i.sin(π/3)]

z⁶ = 49.[(1/2) + i.(√3)/2]

z⁶ = (49/2).(1 + i)

z = 1 + i → let's try another method if you want it, nut you can apply the method above

z⁵ = (1 + i)⁵

z⁵ = (1 + i)².(1 + i)².(1 + i)

z⁵ = (1 + 2i + i²).(1 + 2i + i²).(1 + i) → where: i² = - 1

z⁵ = (1 + 2i - 1).(1 + 2i - 1).(1 + i)

z⁵ = (2i).(2i).(1 + i)

z⁵ = 4i².(1 + i) → where: i² = - 1

z⁵ = - 4.(1 + i)

z = - √3 + i

z⁵ = (- √3 + i)⁵

z⁵ = (- √3 + i)².(- √3 + i)².(- √3 + i)

z⁵ = (3 - 2i + i²).(3 - 2i + i²).(- √3 + i) → where: i² = - 1

z⁵ = (3 - 2i - 1).(3 - 2i - 1).(- √3 + i)

z⁵ = (2 - 2i).(2 - 2i).(- √3 + i)

z⁵ = 2.(1 - i) * 2.(1 - i).(- √3 + i)

z⁵ = 4.(1 - i)².(- √3 + i)

z⁵ = 4.(1 - 2i + i²).(- √3 + i) → where: i² = - 1

z⁵ = 4.(1 - 2i - 1).(- √3 + i)

z⁵ = - 8i.(- √3 + i)

z⁵ = 8i√3 - 8i² → where: i² = - 1

z⁵ = 8 + 8i√3

z⁵ = 8.(1 + i√3)

1/

2^5[(cos(5π/20))+isin(5π/20)] = 32(√2/2+i√2/2) = 16√2 + 16√2 i

2/

(cos(21π/14))+isin(21π/14)=cos(3π/2))+isin(3π/2) = 0 - i

3/ and 4/ similar to 1/2/

5/

1+i = √2((cos(π/4))+isin(π/4)) -->(1+i)^5 = [√2((cos(π/4))+isin(π/4))]^5 = 4√2((cos(5π/4))+isin(5π/4))

=4√2(-1/√2 - i/√2)=-4-4i

6/

(-√3+i) = 2((cos(5π/6))+isin(5π/6))

(-√3+i) = [2((cos(5π/6))+isin(5π/6))]^5 = 32((cos(25π/6))+isin(25π/6))=32((cos(π/6))+isin(π/6)) = ...