Question #6ec53

1 Answer
Oct 25, 2017

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