b) will be done later

Alternative to a)

iw + z = 5 ① , w² + (4i - 1)z = -11 + 18i ②

Let w = a + bi, z = c + di, where a,b,c and d are real numbers.

From ①,

i(a + bi) + c + di = 5
ai + bi² + c + di = 5
c - b + (a + d)i = 5

Comparing real and imaginary parts on both sides,

c - b = 5 → c = 5 + b ③
a + d = 0 → d = -a ④

From ②,

(a + bi)² + (4i - 1)(c + di) = -11 + 18i
a² + 2abi + b²i² + 4ci + 4di² - c - di = -11 + 18i
a² + 2abi - b² + 4ci - 4d - c - di = -11 + 18i
a² - b² - c - 4d + (2ab + 4c - d)i = -11 + 18i

Comparing real and imaginary parts,

a² - b² - c - 4d = -11 ⑤
2ab + 4c - d = 18 ⑥

Sub ③ and ④ into ⑥,

2ab + 4(5 + b) - (-a) = 18
2ab + 20 + 4b + a = 18
2ab + 2 + 4b + a = 0
2b(a + 2) + a + 2 = 0
(2b + 1)(a + 2) = 0
b = -½ → c = 9/2 or a = -2 → d = 2

Sub ③ and ④ into ⑤,
a² - b² - 5 - b + 4a = -11

When a = -2,
(-2)² - b² - 5 - b + 4(-2) = -11
4 - b² - 5 - b - 8 = -11
b² + b - 2 = 0
(b + 2)(b - 1) = 0
b = -2 or b = 1

So c = 5 - 2 = 3 or c = 5 + 1 = 6

∴ w = -2 - 2i, z = 3 + 2i or w = -2 + i, z = 6 + 2i


(You should reject the following solution of a,b,c,d as a and d are complex numbers here but we have defined them to be real at the start .

The working is just done to show you that the final result of w and z is the same)

When b = -½,
a² - (-½)² - 5 - (-½) + 4a = -11
a² - ¼ - 5 + ½ + 4a = -11
a² + 4a + 25/4 = 0
a = [-4 ± √(4² - 4(25/4))] / 2
a = [-4 ± √-9] / 2
a = -2 ± 3i/2 → d = 2 ± 3i/2

w = -2 + 3i/2 - ½i or w = -2 - 3i/2 - ½i
w = -2 + i or w = -2 - 2i

z = 9/2 + 3/2 + 2i or z = 9/2 - 3/2 + 2i
z = 6 + 2i or z = 3 + 2i

bi)

iz = (1 + i)⁶ / (-1 - i√3)⁴


|(1 + i)| = √(1² + 1²) = √2
arg (1 + i) = tan-¹ (1/1) = π/4

So in exponential form,
1 + i = √2 e^(π/4 i)
(1 + i)⁶ = (√2 e^(π/4 i) )⁶ = 8 e^(3π/2 i)


|(-1 - i√3)| = √((-1)² + (-√3)²) = √4 = 2
arg (-1 - i√3) = tan-¹ ( (-√3)/(-1) ) = tan-¹(√3)
= π/3

So in exponential form,
(-1 - i√3) = 2 e^(π/3 i)
(-1 - i√3)⁴ = (2 e^(π/3 i))⁴ = 16 e^(4π/3 i)


So,

iz = 8e^(3π/2 i) / 16e^(4π/3 i)
iz = ½ e^(π/6 i)

But i = e^(π/2 i)

So,

e^(π/2 i) z = ½ e^(π/6 i)
z = ½ e^(π/6 i)/e^(π/2 i)

= ½ e^(-π/3 i)
= ½ e^(-iπ/3)

bii)

z = ½ e^(-iπ/3)
zⁿ = (½)ⁿ e^(-niπ/3)

|zⁿ|= (½)ⁿ and arg zⁿ = -nπ/3 = -n/3 π

For z to be a real number, -n/3 π has to be a integer multiple of π .

(Because the real numbers only lie on the real axis, where argument/angle is always ....-2π,-π,0,π,2π,3π,.... basically = kπ, k∈Z)


This means that -n/3 is an integer.
(-n/3 ∈ Z)

And for -n/3 to be an integer, n has to be a multiple of 3, i.e n = 3m , m ∈ Z

Smallest positive integer n will make -n/3 an integer = 3, such that -n/3 = -3/3 = 1