Question

What are the roots of `((1+iz)/z)^3 = 8i` ?

Answer 1

The solutions are `S={1/sqrt3;-1/sqrt3 ;i/3 }`

Explanation:

Reminder : The Euler's Identity

`e^(itheta)=costheta+isintheta`

`e^(ipi/2)=cos(pi/2)+isin(pi/2)=0+i`

Therefore,

`((1+iz)/z)^3=8i=8e^(i(pi/2+2kpi))`, `AA k in {0,1,2}`

Taking the cube roots

`((1+iz)/z)=2e^(i(pi/6+2/3kpi))`

`1+iz=2ze^(i(pi/6+2/3kpi))`

`z(2e^(i(pi/6+2/3kpi))-i)=1`

`z=1/(2e^(i(pi/6+2/3kpi))-i)`

When `k=0`

`z_0=1/(2(cos(pi/6)+isin(pi/6))-i)`

`z_0=1/((sqrt3+i-i))=1/sqrt3`

`=(1-i(sqrt3-1))/(5-2sqrt3)`

When `k=1`

`z_1=1/(2(cos(pi/6+2/3pi)+isin(pi/6+2/3pi))-i)`

`z_1=1/((-sqrt3+i-i))=-1/sqrt3`

When `k=2`

`z_1=1/(2(cos(pi/6+4/3pi)+isin(pi/6+4/3pi))-i)`

`=1/(2(cos(3/2pi)+isin(3/2pi)-i)`

`=1/(2(0-i)-i)=1/(-3i)=i/3`

Answer 2

The roots are `1/3i` and `+-sqrt(3)/3`

Explanation:

Given:

`((1+iz)/z)^3 = 8i`

Since `(-i)^3 = i` and `2^3=8` this simplifies to:

`(1/z+i)^3 = (-2i)^3`

Hence:

`1/z+i = omega^k (-2i)" "` for `k = 0, 1, 2`

where `omega = -1/2+sqrt(3)/2i` is the primitive complex cube root of `1`.

Note that `omega^2 = bar(omega) = -1/2-sqrt(3)/2i`

So the three roots `z_k` are:

`z_k = 1/(omega^k(-2i)-i)" "k = 0, 1, 2`

Simplifying each in turn, we have:

`z_0 = 1/(-2i-i) = 1/(-3i) = 1/3i`

`z_1 = 1/((-1/2+sqrt(3)/2i)(-2i)-i) = 1/((color(red)(cancel(color(black)(i)))-sqrt(3))-color(red)(cancel(color(black)(i)))) = -1/sqrt(3) = -sqrt(3)/3`

`z_2 = 1/((-1/2-sqrt(3)/2i)(-2i)-i) = 1/((color(red)(cancel(color(black)(i)))+sqrt(3))-color(red)(cancel(color(black)(i)))) = 1/sqrt(3) = sqrt(3)/3`