# What are the solutions to (z-1)^3 =8i?

Dec 28, 2015

#### Answer:

$z \in \left\{\sqrt{3} + 1 + i , - \sqrt{3} + 1 + i , 1 - 2 i\right\}$

#### Explanation:

For this problem, we will need to know how to find the ${n}^{\text{th}}$ roots of a complex number. To do this, we will use the identity
${e}^{i \theta} = \cos \left(\theta\right) + i \sin \left(\theta\right)$

Because of this identity, we can represent any complex number as
$a + b i = R {e}^{i \theta}$ where $R = \sqrt{{a}^{2} + {b}^{2}}$ and $\theta = \arctan \left(\frac{b}{a}\right)$

Now we will go over the steps to find the ${3}^{\text{rd}}$ roots of a complex number $a + b i$. The steps for finding the ${n}^{\text{th}}$ roots are similar.

Given $a + b i = R {e}^{i \theta}$ 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
${z}^{3} = {\left({R}_{0} {e}^{i {\theta}_{0}}\right)}^{3} = {R}_{0}^{3} {e}^{3 i {\theta}_{0}} = R {e}^{i \theta}$

From this, we immediately have ${R}_{0} = {R}^{\frac{1}{3}}$. We also may equate the exponents of $e$, but noting that as sine and cosine are periodic with period $2 \pi$, then from the original identity, ${e}^{i \theta}$ will be as well. Then we have

$3 i {\theta}_{0} = i \left(\theta + 2 \pi k\right)$ where $k \in \mathbb{Z}$

$\implies {\theta}_{0} = \frac{\theta + 2 \pi k}{3}$ where $k \in \mathbb{Z}$

However, as if we keep adding $2 \pi$ 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
$z \in \left\{{R}^{\frac{1}{3}} {e}^{i \frac{\theta}{3}} , {R}^{\frac{1}{3}} {e}^{i \frac{\left(\theta + 2 \pi\right)}{3}} , {R}^{\frac{1}{3}} {e}^{i \frac{\theta + 4 \pi}{3}}\right\}$

We may convert this back to $a + b i$ form if desired using the identity
${e}^{i \theta} = \cos \left(\theta\right) + i \sin \left(\theta\right)$

Applying the above to the problem at hand:

${\left(z - 1\right)}^{3} = 8 i$

$\implies z - 1 = 2 {i}^{\frac{1}{3}}$

$\implies z = 2 {i}^{\frac{1}{3}} + 1$

Using the above process, we can find the ${3}^{\text{rd}}$ roots of $i$:

$i = {e}^{i \frac{\pi}{2}} \implies {i}^{\frac{1}{3}} \in \left\{{e}^{i \frac{\pi}{6}} , {e}^{i \frac{5 \pi}{6}} , {e}^{i \frac{3 \pi}{2}}\right\}$

Applying ${e}^{i \theta} = \cos \left(\theta\right) + i \sin \left(\theta\right)$ we have

${i}^{\frac{1}{3}} \in \left\{\frac{\sqrt{3}}{2} + \frac{i}{2} , - \frac{\sqrt{3}}{2} + \frac{i}{2} , - i\right\}$

Finally, we substitute in these values for $z = 2 {i}^{\frac{1}{3}} + 1$

$z \in \left\{2 \left(\frac{\sqrt{3}}{2} + \frac{i}{2}\right) + 1 , 2 \left(- \frac{\sqrt{3}}{2} + \frac{i}{2}\right) + 1 , 2 \left(- i\right) + 1\right\}$

$= \left\{\sqrt{3} + 1 + i , - \sqrt{3} + 1 + i , 1 - 2 i\right\}$