Answer:
2/7.
Step--step explanation:
1/2 * 4/7.
Multiply the numerators and the denominators:
= 1*4 / 2*7
= 4 / 14 Now divide numerator and denominator by 2:
= 2/7.
If <em>z</em> ⁷ = 128<em>i</em>, then there are 7 complex numbers <em>z</em> that satisfy this equation.

![\implies z=\sqrt[7]{2^7} e^{i\frac17\left(\frac\pi2+2n\pi\right)}](https://tex.z-dn.net/?f=%5Cimplies%20z%3D%5Csqrt%5B7%5D%7B2%5E7%7D%20e%5E%7Bi%5Cfrac17%5Cleft%28%5Cfrac%5Cpi2%2B2n%5Cpi%5Cright%29%7D)
(where <em>n</em> = 0, 1, 2, …, 6)


I am putting this into rows. Ex: row one has 0. row two has blank and seven.
1: 0
2: 9 7
3: - 9 6
4: 1 3
5: - 1 2
6: 1
So the answer on top would be 8 1 r 1.