Complex numb help (1 Viewer)

Run hard@thehsc · Dec 1, 2021

Pls help!

Drongoski · Dec 1, 2021

Run hard@thehsc said:
View attachment 34363
Pls help!

(a)
$z^{2n} - 1 = (z-1)(1+z+z^2 + \cdots + z^{2n-1})\\ \\ \therefore 1+z+z^2 + \cdots + z^{2n-1} = \frac{z^{2n}-1}{z-1} (z \neq 1))\\ \\ = \frac{(cis{\frac{\pi}{n}})^{2n}-1}{cis{\frac {\pi}{n}}-1} = \frac{cis{\frac {2n\pi}{n}}-1}{\cdots} = \frac {cis{2\pi}-1}{\cdots} = \frac{0}{\cdots} = 0$

(b) Similarly

$1+z+z^2 + \cdots +z^{n-1} = \frac {z^n -1}{z-1} = \frac{cis\pi -1}{cis{\frac{\pi}{n}}-1} = \frac {-1-1}{\cdots}\\ \\ = \frac{2}{1-cos\frac{\pi}{n} -isin\frac{\pi}{n}} = \frac{2\times(1-cos\frac{\pi}{n} + isin\frac{\pi}{n})}{(1-cos\frac{\pi}{n}-isin\frac{\pi}{n})\times(1-cos\frac{\pi}{n}+isin\frac{\pi}{n})}\\ \\ = \frac {\cdots}{(1-cos\frac{\pi}{n})^2 +sin^2\frac{\pi}{n}} = \frac {\cdots}{1-2cos\frac{\pi}{n} + cos^2 \frac{\pi}{n}+ sin^2 \frac{\pi}{n}}\\ \\ = \frac{2((1-cos\frac{\pi}{n})+isin\frac{\pi}{n})}{2(1-cos\frac{\pi}{n})} =\frac{(2sin^2\frac{\pi}{2n}) +2isin\frac{\pi}{2n}cos\frac{\pi}{2n}}{2sin^2\frac{\pi}{2n} }\\ \\ = 1 +i \frac{cos\frac{\pi}{2n}}{sin\frac{\pi}{2n}} = 1 + icot \frac{\pi}{2n}$

amdspotter · Dec 1, 2021

Drongoski said:
$z^{2n} - 1 = (z-1)(1+z+z^2 + \cdots + z^{2n-1})\\ \\ \therefore 1+z+z^2 + \cdots + z^{2n-1} = \frac{z^{2n}-1}{z-1}\\ \\ = \frac{(cis{\frac{\pi}{n}})^{2n}-1}{cis{\frac {\pi}{n}}-1} = \frac{cis{\frac {2n\pi}{n}}-1}{\cdots} = \frac {cis{2\pi}-1}{\cdots} = \frac{0}{\cdots} = 0$

Can U if possible explain a bit more like I don't rlly quite get the approach U took

Drongoski · Dec 1, 2021

amdspotter said:
Can U if possible explain a bit more like I don't rlly quite get the approach U took

OK. Sorry, I have not spelt out in more detail because I really struggle to LaTeX my solution; so have to skip intermediate steps.

Which part would you like me to explain more fully?

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Complex numb help (1 Viewer)

Run hard@thehsc

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Drongoski

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amdspotter

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Drongoski

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