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# Complex Number Arithmetic...

+5
99
4
+251

Simplify $$(1 + i\sqrt{3})^6.$$

Feb 1, 2022

#1
+117100
+2

$$(1 + i\sqrt{3})^6\\ =\binom{6}{0}(i\sqrt3)^0+\binom{6}{1}(i\sqrt3)^1+\binom{6}{2}(i\sqrt3)^2 +\binom{6}{3}(i\sqrt3)^3+\binom{6}{4}(i\sqrt3)^4+\binom{6}{5}(i\sqrt3)^5+\binom{6}{6}(i\sqrt3)^6\\ =1+6*(i\sqrt3)^1+15(i\sqrt3)^2+20(i\sqrt3)^3+15(i\sqrt3)^4+6(i\sqrt3)^5+(i\sqrt3)^6\\ =1+(6\sqrt3)i-15(\sqrt3)^2-20(\sqrt3)^3i+15(\sqrt3)^4+6(\sqrt3)^5i-(\sqrt3)^6\\ =1+(6\sqrt3)i-15*3-60(\sqrt3)i+15*9+6*9\sqrt3 i-27\\ =1+6\sqrt3i-45-60\sqrt3i+135+54\sqrt3 i-27\\ =1-45+135-27+(6\sqrt3-60\sqrt3+54\sqrt3) i\\ =1-45+135-27+(6\sqrt3-60\sqrt3+54\sqrt3) i\\ =64$$

You need to check my working.   there may also have been a simpler path.

LaTex

(1 + i\sqrt{3})^6\\
=\binom{6}{0}(i\sqrt3)^0+\binom{6}{1}(i\sqrt3)^1+\binom{6}{2}(i\sqrt3)^2
+\binom{6}{3}(i\sqrt3)^3+\binom{6}{4}(i\sqrt3)^4+\binom{6}{5}(i\sqrt3)^5+\binom{6}{6}(i\sqrt3)^6\\
=1+6*(i\sqrt3)^1+15(i\sqrt3)^2+20(i\sqrt3)^3+15(i\sqrt3)^4+6(i\sqrt3)^5+(i\sqrt3)^6\\
=1+(6\sqrt3)i-15(\sqrt3)^2-20(\sqrt3)^3i+15(\sqrt3)^4+6(\sqrt3)^5i-(\sqrt3)^6\\
=1+(6\sqrt3)i-15*3-60(\sqrt3)i+15*9+6*9\sqrt3 i-27\\
=1+6\sqrt3i-45-60\sqrt3i+135+54\sqrt3 i-27\\
=1-45+135-27+(6\sqrt3-60\sqrt3+54\sqrt3) i\\
=1-45+135-27+(6\sqrt3-60\sqrt3+54\sqrt3) i\\
=64

Feb 1, 2022
#4
+251
+1

Ohh got it. Thank you Melody!

dolphinia  Feb 10, 2022
#2
+1

Put the number into polar form,

$$\displaystyle r(\cos\theta + i\sin\theta) .$$

$$\displaystyle r^{2}=1^{2}+(\sqrt{3})^{2}=4, \text{ }r=2.$$

$$\displaystyle \tan\theta = \sqrt{3}/1, \text{ }\theta= \pi/3.$$

$$\displaystyle (1 + i\sqrt{3})^{6}\\=[2(\cos(\pi/3)+i\sin(\pi/3))]^{6} \\= 2^{6}(\cos(2\pi)+i\sin(2\pi))\\=64.$$

Feb 1, 2022
#3
+117100
0

That is much nicer, thanks for reminding me :))

Melody  Feb 1, 2022