How do you simplify (3+3*sqrt3 i)^4(3+33i)4?

4 Answers
Jul 21, 2018

=> color(maroon)(-648 (1 + sqrt3 i)648(1+3i)

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)^4Given (3+33i)4

"can be writen as "((3 + 3 sqrt3 i)^2)^2can be writen as ((3+33i)2)2

=. (9 + 18 sqrt3 i + 27 i^2) ^2=.(9+183i+27i2)2

=> (9 + 18 sqrt3 i - 27 ) ^2(9+183i27)2

=> (-18 + 18 sqrt3 i)^2(18+183i)2

=>. 324 - -648 sqrt3 i + 972 i^2.3246483i+972i2

=> 324 - 648 sqrt3 i - 9723246483i972

=> -648 - 648 sqrt3 i6486483i

=> color(maroon)(-648 (1 + sqrt3 i)648(1+3i)

Jul 21, 2018

-648(1+sqrt3i)648(1+3i).

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!

Jul 21, 2018

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)

-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)