Author Topic: the hardest complex number question i've encountered... help!! (Read 3713 times) Tweet Share

annaconda · « **on:** January 31, 2018, 06:31:10 pm »

Hello all,

I have encountered this question, and no matter how hard I try, I just can't seem to get anywhere.
Can someone help me?
The downside is, there are no solutions...

RuiAce · « **Reply #1 on:** January 31, 2018, 06:53:39 pm »

$\text{Clearly }z=1\text{ is not a solution, so dividing}\\ \text{both sides by }(z-1)^{2n}\text{ gives}\\ \boxed{ \left( \frac{z+1}{z-1}\right)^{2n}+1 = 0}$
$\text{Using the formula provided, the solution to this equation is}\\ \boxed{\frac{z+1}{z-1} = \text{ cis }\left(\frac{(2k-1)\pi}{2n}\right)}$
with $k=1,2,\dots,2n$
_____________________
$\text{We attempt to make }w\text{ the subject in }\frac{z+1}{z-1} = w.\\ \text{Execute:}\\ \begin{align*}\frac{z+1}{z-1} &= w\\ z+1 &= wz - w\\ w+1 &= wz -z\\ z(w-1)&=w+1\\ z &= \frac{w+1}{w-1}\end{align*}$
$\text{So by letting }w = \text{ cis }\left(\frac{(2k-1)\pi}{2n}\right)\\ \text{we have } \boxed{z = \frac{ \text{ cis }\left(\frac{(2k-1)\pi}{2n}\right) + 1}{\text{ cis }\left(\frac{(2k-1)\pi}{2n}\right) - 1}}$
Edit: Typo
$\text{This becomes a hassle to write, so let }\theta_k = \frac{(2k-1)\pi}{2n} \text{ purely for convenience.}\\ \text{We will stop using the cis notation here and write }\boxed{z = \frac{\cos\theta_k + i\sin\theta_k +1}{\cos \theta_k +i\sin\theta_k - 1}}$
_____________________
$\text{Recall the double angle identities}\\ \begin{align*}\sin 2\alpha &= 2\sin\alpha\cos\alpha\\ \cos 2\alpha &= 2\cos^2\alpha-1\\ \cos 2\alpha &= 1-2\sin^2\alpha\end{align*}\\ \text{We will let }\alpha = \frac{\theta_k}{2}\text{ and use these identities to simplify things down.}$
\begin{align*}z &=\frac{\cos \theta_k + i\sin\theta_k + 1}{\cos\theta_k + i\sin\theta_k - 1}\\ &= \frac{\left(2\cos^2 \frac{\theta_k}{2} - 1\right) + 2i\sin\frac{\theta_k}{2}\cos \frac{\theta_k}{2} - 1}{\left(1-2\sin^2 \frac{\theta_k}{2}\right) + 2i \sin \frac{\theta_k}{2} \cos \frac{\theta_k}{2} - 1}\\ &= \frac{2\cos \frac{\theta_k}{2} \left(\cos \frac{\theta_k}{2} + i \sin \frac{\theta_k}{2}\right)}{-2\sin \frac{\theta_k}{2} \left(\sin \frac{\theta_k}{2} -i\cos \frac{\theta_k}{2}\right)}\\ &= -\cot \frac{\theta_k}{2}\times \frac{\cos \frac{\theta_k}{2} + i \sin \frac{\theta_k}{2}}{i \left(\cos \frac{\theta_k}{2} + i \sin \frac{\theta_k}{2}\right)}\\ &= i \cot \frac{\theta_k}{2} \end{align*}
$\text{Which is purely imaginary, regardless of the value of }k.\\ \text{Hence all of the roots are purely imaginary.}$

annaconda · « **Reply #2 on:** January 31, 2018, 10:24:49 pm »

thank you!!! such a legend!

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Author Topic: the hardest complex number question i've encountered... help!! (Read 3713 times) Tweet Share

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the hardest complex number question i've encountered... help!!

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