Question

Question #6ec53

Answer 1

`z=1-i`

And it checks out when you plug this value of `z` in the original equation

Explanation:

We start with

`1)(z-1)^3=i`

If we multiply both sides by `i^3` we get

`2)i^3*(z-1)^3=i*i^3`

And we see these three identities:

`a^c*b^c=(a*b)^c`, so `i^3*(z-1)^3` becomes `(i*(z-1))^3`

`a^b*a^c=a^(b+c)`, so `i*i^3` becomes `i^4`

and `i^4=1`,

so

`3)(i*(z-1))^3=1`

Distributing the `i` we get

`4)(z*i-i)^3=1`

We take the cube root of both sides to eliminate the power of `3`

`5)z*i-i=1`

Factor out the `i` again

`6)i*(z-1)=1`

Multiply both sides by `i^3` to get

`7)1*(z-1)=-i`

Add `1` to both sides

`:. z=1-i`

Alternatively you could multiply by `i` after step `6` instead of `i^3`, but the result is the same

`8)-1*(z-1)=i`

Multiply (or divide) by `-1`

`z-1=-i`

Add `1` to both sides

`:. z=1-i`