Question

How do you find the fourth roots of $ - 8 + 8\sqrt {3i} $?

Answer 1

Hint: First we know that complex number, the complex numbers are $1, - 1,i, - i$. The complex number denoted by $z$. If you express your complex number in polar form as $r(\cos \theta + i\sin \theta )$.
We use the ${n^{th}}$ root theorem.
$\sqrt[n]{z} = \sqrt[n]{r}\left[ {\cos (\alpha ) + i\sin (\alpha )} \right]$
This is an extension of DeMoivre’s theorem.
Where, $\alpha = \dfrac{{\theta + 360k}}{n}$ ,$k = 0,1,2,3....,n - 1$

Complete step-by-step solution:
The first thing we should recognize is if we want the $4th$ roots, $n$ will be $4$.
$n = 4$
Next we need to write this complex number in complex form. So we need to find $r$ and $\theta $
Let $a + ib$
Let $z = - 8 + 8i\sqrt 3 $
$a = - 8$ and $b = 8\sqrt 3 $
Now $r$ formula is, $r = \sqrt {{a^2} + {b^2}} $ ,
Now apply the complex number in the $r$ equation, we get
$ \Rightarrow r = \sqrt {{{( - 8)}^2} + {{(8\sqrt 3 )}^2}} $
Square on two terms
$ \Rightarrow r = \sqrt {64 + 192} $
Now add the terms,
$ \Rightarrow r = \sqrt {256} = 16$
Next we need to find $\theta $. But before we do that let’s recognize that if $a$ is negative and $b$ is positive, $\theta $ will be in this second quadrant. And we know that tangent theta is equal to $\dfrac{b}{a}$
$ \Rightarrow \tan \theta = \dfrac{b}{a}$
Now substitute $a$ and $b$ in tangent
\[ \Rightarrow \tan \theta = \dfrac{{\not{8}\sqrt 3 }}{{ - \not{8}}} = \dfrac{{\sqrt 3 }}{{ - 1}}\]
Then the theta value is ${120^ \circ }$
$ \Rightarrow \theta = {120^ \circ }$
Now we find $\alpha $value,
$ \Rightarrow \alpha = \dfrac{{\theta + {{360}^ \circ }k}}{n}$, $k = 0,1,2,3....,n - 1$
\[ \Rightarrow \alpha = \dfrac{{{{120}^ \circ } + {{360}^ \circ }k}}{4}\]
Let’s see if we can simplify this expression, before we $120$ divide by $4$ would be $30$degrees and $360k$ divided by 4 would be $90$degrees.
$\alpha = {30^ \circ } + {90^ \circ }k$
Now we have,
$n = 4$
$r = 16$
$\alpha = {30^ \circ } + {90^ \circ }k$
$k = 0,1,2,3.$
If $k = 0$, then $\alpha = {30^ \circ } + {90^ \circ }(0) = {30^ \circ }$
If $k = 1$, then $\alpha = {30^ \circ } + {90^ \circ }(1) = {120^ \circ }$
If $k = 2$, then $\alpha = {30^ \circ } + {90^ \circ }(2) = {210^ \circ }$
If $k = 3$, then $\alpha = {30^ \circ } + {90^ \circ }(3) = {300^ \circ }$
Now these values are substitute in the equation of $\sqrt[n]{z} = \sqrt[n]{r}\left[ {\cos (\alpha ) + i\sin (\alpha )} \right]$
If $\theta = {30^ \circ }$
$\sqrt[4]{z} = \sqrt[4]{{16}}(\cos {30^ \circ } + i\sin {30^ \circ })$
$ \Rightarrow 2\left( {\dfrac{{\sqrt 3 }}{2} + i\left( {\dfrac{1}{2}} \right)} \right) = \sqrt 3 + i$
If $\theta = {120^ \circ }$
$\sqrt[4]{z} = \sqrt[4]{{16}}(\cos {120^ \circ } + i\sin {120^ \circ })$
$ \Rightarrow 2\left( {\dfrac{{ - 1}}{2} + i\left( {\dfrac{{\sqrt 3 }}{2}} \right)} \right) = - 1 + i\sqrt 3 $
If $\theta = {210^ \circ }$
$\sqrt[4]{z} = \sqrt[4]{{16}}(\cos {210^ \circ } + i\sin {210^ \circ })$
$ \Rightarrow 2\left( {\dfrac{{ - \sqrt 3 }}{2} + i\left( {\dfrac{{ - 1}}{2}} \right)} \right) = - \sqrt 3 - i$
If $\theta = {300^ \circ }$
$\sqrt[4]{z} = \sqrt[4]{{16}}(\cos {300^ \circ } + i\sin {300^ \circ })$
$ \Rightarrow 2\left( {\dfrac{1}{2} + i\left( {\dfrac{{ - \sqrt 3 }}{2}} \right)} \right) = 1 - i\sqrt 3$

Note: The applet below shows the complex $4^{th}$ roots of a complex number. DeMoivre’s Theorem shows that there are always fourth roots, spaced evenly around a circle. The segment of $z$ indicates the given number $z$, and the segments of fourth roots ${z_0},{z_1},{z_2}$ and ${z_3}$.