Question

What are the solutions of \[{(z - 1)^3} = 8i\]?

Answer 1

Hint: If we expand the term as per the identity, we may get confused about solving the problem. So, we need to just try to express RHS also as a cubic term. Finally, we need to equate the base terms as the powers are the same on both sides of the equation.

Complete Step by step answer:
Observe the given problem
LHS part is a cube so try to express RHS also as a cube
\[{(z - 1)^3} = {( - 2i)^3}\]
Let’s check how 8i will become\[{( - 2i)^3}\]
\[{( - 2i)^3} = {( - 2)^3}{(i)^3}\] , $8$can be written as the cube of $2$.
\[ = - 8(i){(i)^2}\], as we have the value of the square of the imaginary part as $ - 1$
\[ = - 8(i)( - 1)\], as in the problem it is 8i and the square of imaginary is $ - 1$.
\[ = 8i\]
\[{(z - 1)^3} = {( - 2i)^3}\] (When the powers are the same we can equate the bases, as per the polynomials rule i.e.\[{(a)^m} = {(b)^m}\] then \[a{\text{ }} = {\text{ }}b\]
We can equate the base term as the powers are the same.
\[(z - 1) = - 2i \]
\[z = - 2i + 1 \]

Additional information:
You can try doing this problem by expanding as a formula
But you may not get a proper solution as it contains imaginary parts in it.
For this problem, we will need to know how to find the nth roots of a complex number. To do this, we will use the identity
\[{e^{i\theta }} = cos\left( \theta \right) + isin\left( \theta \right)\]
Because of this identity, we can represent any complex number as
\[a + bi = R{e^{i\theta }}\;\]where \[R = \surd {a^2} + {b^2}\;\] and \[\theta = arctan\left( {b/a} \right)\]
Now we will go over the steps to find the 3rd root of a complex number \[a + bi\]. The steps for finding the nth roots are similar.
Given\[a + bi = R{e^{i\theta }}\;\] we are looking for all complex numbers $z$such that
\[{z^3} = R{e^{i\theta }}\]
As $z$is a complex number, there exist ${R_0}$and ${\theta _0}$ such that
\[z = {R_0}{e^{i{\theta _0}}}\]
Then
\[{z^3} = {R^3}_0{e^{3i{\theta _0}}} = {\operatorname{Re} ^{i\theta }}\]
From this, we immediately have \[{R_0} = {R^{1/3}}\]. We also may equate the exponents of$e$, but note that as sine and cosine are periodic with period$2\pi $, and then from the original identity,${e^{i\theta }}$ will be as well. Then we have
\[3i{\theta _0} = i\left( {\theta + 2\pi k} \right)\;\] , where \[k \in Z\] .
\[ \Rightarrow {\theta _0} = (\theta + 2\pi k)/3\;\] , where \[k \in Z\]
However, as if we keep adding 2π over and over, we will end up with the same values, we can ignore the redundant values by adding the restriction\[{\theta _0} \in \left[ {0,2\pi } \right)\], that is, \[k \in \left\{ {0,1,2} \right\}\]
Putting it all together, we get the solution set
\[z \in \left\{ {{R^{1/3}}{e^{3i\theta }},{R^{1/3}}{e^{i\left( {\theta + 2\pi } \right)3}},{R^{1/3}}{e^{i(\theta + 4\pi )3}}} \right\}\]
We may convert this back to \[a + bi\;\]form if desired using the identity
\[{e^{i\theta }} = cos\left( \theta \right) + isin\left( \theta \right)\]

Note:
Now, we can also apply the above to the problem at hand:
\[\begin{array}{*{20}{l}} {{{\left( {z - 1} \right)}^3} = 8i} \\ { \Rightarrow z - 1 = 2{i^{1/3}}} \\ { \Rightarrow z = 2{i^{1/3}} + 1} \end{array}\]