I don't think I can do it! 2

Algebra Level 2

( 3 + i 2 ) 12 = ? \left(\frac{\sqrt{3}+i}{2}\right)^{12}= \ ? \

Note : i = − 1 i=\sqrt{-1}

Try #1 here .


The answer is 1.

Akshat Sharda by Akshat Sharda , 21, India

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3 solutions

Aareyan Manzoor
Dec 5, 2015

this can be written as ( i ( − ω 2 ) ) 12 (i(-\omega^2))^{12} where ω \omega is a primitive 3rd root of unity. so, i 12 ω 24 = 1 ∗ 1 = 1 i^{12} \omega^{24}=1*1=1

6 Helpful 0 Interesting 0 Brilliant 0 Confused

I like your solution,because it is not common.. because we usually prefer euler's formula to solve this.

Akhil Bansal - 5 years, 6 months ago

( 3 2 + i 1 2 ) 12 = ( cos ⁡ π 6 + i sin ⁡ π 6 ) 12 By Euler’s identity: = ( e i π 6 ) 12 = e i 2 π = cos ⁡ ( 2 π ) + i sin ⁡ ( 2 π ) = 1 \begin{aligned} \left(\frac{\sqrt{3}}{2}+i\frac{1}{2}\right)^{12} & = \left(\color{#3D99F6}{\cos \frac{\pi}{6} + i \sin \frac {\pi}{6}} \right)^{12} \quad \quad \color{#3D99F6}{\text{By Euler's identity:}} \\ & = \left(\color{#3D99F6}{e^{i\frac{\pi}{6}}}\right)^{12} = e^{i2\pi} = \cos (2\pi) + i \sin (2\pi) = \boxed{1} \end{aligned}

4 Helpful 0 Interesting 0 Brilliant 0 Confused

( 3 2 + i 2 ) 12 = ( c i s ( π 6 ) ) 12 (\frac{\sqrt{3}}{2}+\frac{i}{2})^{12}=(cis(\frac{\pi}{6}))^{12}

According to de Moivre's theorem...

( c i s ( π 6 ) ) 12 = ( c i s ( 12 × π 6 ) ) = ( c i s ( 2 π ) ) = 1 (cis(\frac{\pi}{6}))^{12}=(cis(12\times\frac{\pi}{6}))=(cis(2\pi))=\boxed{1}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

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