de moivres theorem (1 Viewer)

kractus · Jan 20, 2023

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im not sure how you would do c given b. I initially thought just base off of inspection that there would be a substitution involving cis or cos theta but its hard for me to tell what to substitute. Or there's just no substirution

member 6003 · Jan 20, 2023

i think you substitute z=cistheta and divide by z^3 on both sides. then like using z^n+z^-n=2cosntheta to get the rest of it.

kractus · Jan 20, 2023

tywebb said:
Looks like a cambridge question.

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ya it is cambridhe. I also have another question:

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Just iii.

5uckerberg · Jan 20, 2023

kractus said:
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im not sure how you would do c given b. I initially thought just base off of inspection that there would be a substitution involving cis or cos theta but its hard for me to tell what to substitute. Or there's just no substirution

To start we will first divide $z^{6}+z^{3}+1$ with $z^{3}$ which gives us $z^{3}+\frac{1}{z^{3}}+1=2\cos{3\theta}+1$
With the RHS since you got three multiples a division of $z^{3}$ is just simply dividing $z$ for each quadratic inside the bracket. This in turn will give us $\left(z-2\cos\frac{2\pi}{9}+\frac{1}{z}\right)\left(z-2\cos\frac{4\pi}{9}+\frac{1}{z}\right)\left(z-2\cos\frac{8\pi}{9}+\frac{1}{z}\right)$

Using our knowledge of conjugates $\left(z-2\cos\frac{2\pi}{9}+\frac{1}{z}\right)\left(z-2\cos\frac{4\pi}{9}+\frac{1}{z}\right)\left(z-2\cos\frac{8\pi}{9}+\frac{1}{z}\right)$ is just $\left(2\cos{\theta}-2\cos\frac{2\pi}{9}\right)\left(2\cos{\theta}-2\cos\frac{4\pi}{9}\right) \left(2\cos{\theta}-2\cos\frac{8\pi}{9}\right)$ .

Take the two out of everyone and multiply it two times it will become 8. This leads to $8\left(\cos{\theta}-\cos\frac{2\pi}{9}\right)\left(\cos{\theta}-\cos\frac{4\pi}{9}\right) \left(\cos{\theta}-\cos\frac{8\pi}{9}\right)$

5uckerberg · Jan 20, 2023

kractus said:
ya it is cambridhe. I also have another question:
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Just iii.

Help with Cambridge Extension 2 Enrichment Question - Complex Numbers

Hello, I am having trouble doing q a(iii) and b. Could someone help me please? Thanks.

boredofstudies.org

Have a look at this forum.

5uckerberg · Jan 20, 2023

tywebb said:
Looks like a cambridge question.

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Where is the 8 in front of the whole solution?

de moivres theorem (1 Viewer)

kractus

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member 6003

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kractus

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5uckerberg

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5uckerberg

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Help with Cambridge Extension 2 Enrichment Question - Complex Numbers

5uckerberg

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