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#