Question #6ec53

1 Answer
Oct 25, 2017

z=1-iz=1i

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

Explanation:

We start with

1)(z-1)^3=i1)(z1)3=i

If we multiply both sides by i^3i3 we get

2)i^3*(z-1)^3=i*i^32)i3(z1)3=ii3

And we see these three identities:

a^c*b^c=(a*b)^cacbc=(ab)c, so i^3*(z-1)^3i3(z1)3 becomes (i*(z-1))^3(i(z1))3

a^b*a^c=a^(b+c)abac=ab+c, so i*i^3ii3 becomes i^4i4

and i^4=1i4=1,

so

3)(i*(z-1))^3=13)(i(z1))3=1

Distributing the ii we get

4)(z*i-i)^3=14)(zii)3=1

We take the cube root of both sides to eliminate the power of 33

5)z*i-i=15)zii=1

Factor out the ii again

6)i*(z-1)=16)i(z1)=1

Multiply both sides by i^3i3 to get

7)1*(z-1)=-i7)1(z1)=i

Add 11 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