How do you find all complex solutions to the equation z^4 = 2/(1+i)

write your answers in the form re^(i*theta) , where -pi< theta<=pi
can you please show all workings?

thanks

Hello,

z⁴ = 2/(1 + i)
z⁴ = 2(1 - i) / (1² - i²) →→→ Multiply denominator and numerator by 1-i
z⁴ = 2(1 - i) / 2 →→→ i²=-1
z⁴ = 1 - i
z⁴ = (√2) × [(√2)/2 - i(√2)/2] →→→ Factoring by √2
z⁴ = (√2) × [cos(-π/4) + i.sin(-π/4)] →→→ Showing the cos and sin
z⁴ = (√2) × exp(-iπ/4) →→→ Obtain polar expression

Since we want z = r.exp(iθ):
[r.exp(iθ)]⁴ = (√2) × exp(-iπ/4)
r⁴.exp(4iθ) = (√2) × exp(-iπ/4)

Two complexes are equal iff they have the same modulus and the same argument:
{ r⁴ = √2
{ 4θ = -π/4 + 2kπ (with k in Z)

Hence:
{ r = 2^⅛
{ θ = -π/16 + kπ/2 (with k in Z)

Thus the four solutions with θ in (-π; π] :
If k=-1, z₋₁ = (2^⅛).exp(-9π/16)
If k=0, z₀ = (2^⅛).exp(-π/16)
If k=1, z₁ = (2^⅛).exp(7π/16)
If k=2, z₂ = (2^⅛).exp(15π/16)

Methodically,
Dragon.Jade :-)