# How do you simplify  (3+3*sqrt3 i)^4?

Jul 21, 2018

=> color(maroon)(-648 (1 + sqrt3 i)

#### Explanation:

Here are the steps required for Multiplying Complex Numbers:

Step 1: Distribute (or FOIL) to remove the parenthesis.
Step 2 : Simplify the powers of i, specifically remember that i2 = –1.
Step 3 : Combine like terms, that is, combine real numbers with real numbers and imaginary numbers with imaginary numbers.

$\text{Given } {\left(3 + 3 \sqrt{3} i\right)}^{4}$

$\text{can be writen as } {\left({\left(3 + 3 \sqrt{3} i\right)}^{2}\right)}^{2}$

$= . {\left(9 + 18 \sqrt{3} i + 27 {i}^{2}\right)}^{2}$

$\implies {\left(9 + 18 \sqrt{3} i - 27\right)}^{2}$

$\implies {\left(- 18 + 18 \sqrt{3} i\right)}^{2}$

$\implies . 324 - - 648 \sqrt{3} i + 972 {i}^{2}$

$\implies 324 - 648 \sqrt{3} i - 972$

$\implies - 648 - 648 \sqrt{3} i$

=> color(maroon)(-648 (1 + sqrt3 i)

Jul 21, 2018

$- 648 \left(1 + \sqrt{3} i\right)$.

#### Explanation:

Prerequisite : D'Moivre's Theorem :

$r {\left(\cos \theta + i \sin \theta\right)}^{n} = {r}^{n} \left(\cos \left(n \theta\right) + i \sin \left(n \theta\right)\right) \ldots \ldots . \left(\star\right)$.

Let, $z = {\left(3 + 3 \sqrt{3} i\right)}^{4} = {\left\{3 \left(1 + \sqrt{3} i\right)\right\}}^{4} = {3}^{4} {\left(1 + \sqrt{3} i\right)}^{4}$.

Now, $\left(1 + \sqrt{3} i\right) = 2 \left(\frac{1}{2} + \frac{\sqrt{3}}{2} i\right) = 2 \left(\cos \left(\frac{\pi}{3}\right) + i \sin \left(\frac{\pi}{3}\right)\right)$.

Using $\left(\star\right) , {\left\{\left(1 + \sqrt{3} i\right)\right\}}^{4} = {\left\{2 \left(\cos \left(\frac{\pi}{3}\right) + i \sin \left(\frac{\pi}{3}\right)\right)\right\}}^{4}$,

$= {2}^{4} \left(\cos \left(4 \times \frac{\pi}{3}\right) + i \sin \left(4 \times \frac{\pi}{3}\right)\right)$,

=16{cos((4pi/3)+isin((4pi/3))},

$= 16 \left\{\cos \left(\pi + \frac{\pi}{3}\right) + i \sin \left(\pi + \frac{\pi}{3}\right)\right\}$,

$= 16 \left\{- \cos \left(\frac{\pi}{3}\right) - i \sin \left(\frac{\pi}{3}\right)\right\}$,

$= - 16 \left(\frac{1}{2} + i \frac{\sqrt{3}}{2}\right)$,

$= - 8 \left(1 + \sqrt{3} i\right)$.

$\Rightarrow z = - 648 \left(1 + \sqrt{3} i\right)$, as Respected Sankarankalyanam

Jul 21, 2018

color(purple)(=> -648 * (1 + sqrt3 i)

#### Explanation:

${\left(3 + 3 \sqrt{3} i\right)}^{4}$

$\implies {3}^{4} {\left(1 + \sqrt{3} i\right)}^{4}$

$\implies {3}^{4} {\left({\left(1 + \sqrt{3} i\right)}^{2}\right)}^{2}$

=>3^4 (1 + 2 sqrt3 i + 3 i^2)^2#

$\implies {3}^{4} {\left(1 + 2 \sqrt{3} i - 3\right)}^{2}$

$\implies {3}^{4} {\left(- 2 + 2 \sqrt{3} i\right)}^{2}$

$\implies 324 {\left(- 1 + \sqrt{3} i\right)}^{2}$

$\implies 324 \left(1 - 2 \sqrt{3} i + 3 {i}^{2}\right)$

$\implies 324 \left(- 2 - 2 \sqrt{3} i\right)$

$\implies - 648 \cdot \left(1 + \sqrt{3} i\right)$

$- 648 \left(1 + i \setminus \sqrt{3}\right)$

#### Explanation:

${\left(3 + 3 \setminus \sqrt{3} i\right)}^{4}$

$= {\left(6 {e}^{i \setminus \frac{\pi}{3}}\right)}^{4}$

$= {6}^{4} {\left({e}^{i \setminus \frac{\pi}{3}}\right)}^{4}$

$= 1296 \left({e}^{i 4 \setminus \cdot \setminus \frac{\pi}{3}}\right)$

$= 1296 {e}^{i \frac{4 \setminus \pi}{3}}$

$= 1296 \left(\setminus \cos \left(\frac{4 \setminus \pi}{3}\right) + i \setminus \sin \left(\frac{4 \setminus \pi}{3}\right)\right)$

$= 1296 \left(- \setminus \cos \left(\frac{\setminus \pi}{3}\right) - i \setminus \sin \left(\frac{\setminus \pi}{3}\right)\right)$

$= 1296 \left(- \frac{1}{2} - i \setminus \frac{\sqrt{3}}{2}\right)$

$= - 648 \left(1 + i \setminus \sqrt{3}\right)$