Question

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

Answer 1

`=> 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.

`"Given " (3 + 3 sqrt3 i)^4`

`"can be writen as "((3 + 3 sqrt3 i)^2)^2`

`=. (9 + 18 sqrt3 i + 27 i^2) ^2`

`=> (9 + 18 sqrt3 i - 27 ) ^2`

`=> (-18 + 18 sqrt3 i)^2`

`=>. 324 - -648 sqrt3 i + 972 i^2`

`=> 324 - 648 sqrt3 i - 972`

`=> -648 - 648 sqrt3 i`

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

Answer 2

`-648(1+sqrt3i)`.

Explanation:

Prerequisite : D'Moivre's Theorem :

`r(costheta+isintheta)^n=r^n(cos(ntheta)+isin(ntheta)).......(star)`.

Let, `z=(3+3sqrt3i)^4={3(1+sqrt3i)}^4=3^4(1+sqrt3i)^4`.

Now, `(1+sqrt3i)=2(1/2+sqrt3/2i)=2(cos(pi/3)+isin(pi/3))`.

Using `(star), {(1+sqrt3i)}^4={2(cos(pi/3)+isin(pi/3))}^4`,

`=2^4(cos(4xxpi/3)+isin(4xxpi/3))`,

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

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

`=16{-cos(pi/3)-isin(pi/3)}`,

`=-16(1/2+isqrt3/2)`,

`=-8(1+sqrt3i)`.

`rArr z=-648(1+sqrt3i)`, as Respected Sankarankalyanam

has readily derived!

Answer 3

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

Explanation:

`(3 + 3 sqrt3 i)^4`

`=> 3^4 (1 + sqrt3 i)^4`

`=>3^4 ((1+sqrt 3 i)^2)^2`

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

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

`=> 3^4 (-2 + 2 sqrt3 i)^2`

`=> 324 (-1 + sqrt3 i)^2`

`=> 324 (1 -2 sqrt 3 i + 3 i^2)`

`=> 324 (-2 - 2 sqrt3 i)`

`=> -648 * (1 + sqrt3 i)`

Answer 4

`-648(1+i\sqrt3)`

Explanation:

`(3+3\sqrt3i)^4`

`=(6e^{i\pi/3})^4`

`=6^4(e^{i\pi/3})^4`

`=1296(e^{i4\cdot \pi/3})`

`=1296e^{i{4\pi}/3}`

`=1296(\cos({4\pi}/3)+i\sin({4\pi}/3))`

`=1296(-\cos({\pi}/3)-i\sin({\pi}/3))`

`=1296(-1/2-i\sqrt3/2)`

`=-648(1+i\sqrt3)`