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

4 Answers
sankarankalyanam
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.

#"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)#

Ratnaker Mehta
Jul 21, 2018

# -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!

Kalyanam S.
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)#

Harish Chandra Rajpoot
Jul 21, 2018

#-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)#

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