DeMoivre's Theorem Year 2014

Algebra Level 1

Given

z = [ sin ⁡ ( π 2014 ) + i cos ⁡ ( π 2014 ) ] 2014 , z = \left[ \sin\left(\frac{\pi}{2014}\right) + i \cos\left(\frac{\pi}{2014}\right)\right]^{2014},

find 1 z 2014 \dfrac1{z^{2014}} .


The answer is 1.

Hobart Pao by Hobart Pao , 23, USA

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2 solutions

z = ( sin ⁡ π 2014 + i cos ⁡ π 2014 ) 2014 z = \left(\sin\dfrac{\pi}{2014} + i\cos\dfrac{\pi}{2014} \right)^{2014}

z = ( − i ) 2014 ( cos ⁡ π 2014 + i sin ⁡ π 2014 ) 2014 = − 1 × ( e i π 2014 × 2014 ) = − 1 × − 1 = 1 z = (-i)^{2014} \left(\cos \dfrac{\pi}{2014} + i\sin \dfrac{\pi}{2014}\right)^{2014} = -1 \times \left(e^{i \frac{\pi}{2014} \times 2014}\right) = -1 \times -1 = 1

1 z 2014 = 1 \large \dfrac{1}{z^{2014}} = \boxed{1}

17 Helpful 1 Interesting 3 Brilliant 2 Confused

z = ( c i s ( π 2014 ) ) 2014 = c i s ( π ) = − 1 z=(cis(\frac{\pi}{2014}))^{2014}=cis(\pi)=-1 → 1 z 2014 = 1 ( − 1 ) 2014 = 1 \rightarrow \frac{1}{z^{2014}}=\frac{1}{(-1)^{2014}}=1

4 Helpful 0 Interesting 0 Brilliant 3 Confused

In the problem statement, it is not cis, but sic. Sloppy problem design that the answer comes out the same when it's misread this way, in my opinion.

Thomas Hayes - 3 years, 4 months ago

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Haha, didn't notice, thanks for it

Hjalmar Orellana Soto - 3 years, 3 months ago

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