complex number stuffs (1 Viewer)

synthesisFR · Aug 9, 2023

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could someone please help show me how to do these ( the ones in yellow) in a logical way? thank you so much!
i was trying doing them in my maths class today given as revision questions but couldnt figure them out...

SadCeliac · Aug 9, 2023

surely for 13c you can do difference of two squares over and over?? wait let me try

SadCeliac · Aug 9, 2023

Could you even just sub in z = [cis(theta)]^2 ???

carrotsss · Aug 9, 2023

First one, can do the rest tomorrow if u need

synthesisFR · Aug 9, 2023

SadCeliac said:
Could you even just sub in z = [cis(theta)]^2 ???

the solution he wrote was subbing in z = (cos2(theta)-1)/2

but it makes more sense to sub in sin^2theta instead

ik its the same but like skipping that one step confsued me so much rip

tywebb · Aug 10, 2023

These are Cambridge Ext 2 Ex 3C Q13-15

Here the solutions to 13b,c,14c,15a. I included 13b too because the solution to 13c refers to part of the solution to 13b .

synthesisFR · Aug 10, 2023

Could you explain 15a please. I got lost after they started doing cis2a

tywebb · Aug 10, 2023

synthesisFR said:
Could you explain 15a please. I got lost after they started doing cis2a

Rather than explain the rather long solution to 15a, here is a shorter solution instead:

$(z+1)^{2n}+(z-1)^{2n}=0\Rightarrow(z+1)^{2n}=-(z-1)^{2n}\text{ and since }z\ne1,$
$\textstyle\left(\frac{z+1}{z-1}\right)^{2n}=-1=e^{i(2k+1)\pi}\text{ for integers }k.\ \therefore\frac{z+1}{z-1}=e^{i\theta}\text{ where }\theta=\frac{(2k+1)\pi}{2n}\text{ and so}$
$\begin{aligned}z&=\textstyle\frac{e^{i\theta}+1}{e^{i\theta}-1}\\&=\textstyle\frac{\cos\theta+i\sin\theta+1}{\cos\theta+i\sin\theta-1}\\&=\textstyle\frac{2\cos^2\frac{\theta}{2}+2i\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{-2\sin^2\frac{\theta}{2}+2i\sin\frac{\theta}{2}\cos\frac{\theta}{2}}\\&=\textstyle\frac{\cos\frac{\theta}{2}}{\sin\frac{\theta}{2}}\cdot\frac{\cos\frac{\theta}{2}+i\sin\frac{\theta}{2}}{-\sin\frac{\theta}{2}+i\cos\frac{\theta}{2}}\\&=\textstyle-i\cot\frac{\theta}{2}\cdot\frac{\cos\frac{\theta}{2}+i\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}+i\sin\frac{\theta}{2}}\\&=\textstyle-i\cot\frac{\theta}{2}\\&=\textstyle-i\cot\frac{(2k+1)\pi}{4n}\end{aligned}$

$\text{For }2n\text{ distinct values of }z\text{ we may choose }k=-n,-n+1,\dots,n-1$

tywebb · Aug 11, 2023

There is yet a third even quicker way using the hyperbolic cotangent method, which is outside the syllabus, but actually much easier:

$(z+1)^{2n}+(z-1)^{2n}=0\Rightarrow(z+1)^{2n}=-(z-1)^{2n}\text{ and since }z\ne1,$
$\textstyle\left(\frac{z+1}{z-1}\right)^{2n}=-1=e^{i(2k+1)\pi}\text{ for integers }k.\ \therefore\frac{z+1}{z-1}=e^{i\theta}\text{ where }\theta=\frac{(2k+1)\pi}{2n}\text{ and so}$
$\begin{aligned}z&=\textstyle\frac{e^{i\theta}+1}{e^{i\theta}-1}\\&=\textstyle\frac{e^{i\theta}+1}{e^{i\theta}-1}\cdot\frac{e^{-\frac{i\theta}{2}}}{e^{-\frac{i\theta}{2}}}\\&=\textstyle\frac{e^{\frac{i\theta}{2}}+e^{-\frac{i\theta}{2}}}{e^{\frac{i\theta}{2}}-e^{-\frac{i\theta}{2}}}\\&=\textstyle\coth(\frac{i\theta}{2})\\&=\textstyle i\cot (i\cdot\frac{i\theta}{2})\text{ since }\coth(x)=i\cot(ix)\\&=\textstyle-i\cot\frac{\theta}{2}\\&=\textstyle-i\cot\frac{(2k+1)\pi}{4n}\end{aligned}$

$\text{For }2n\text{ distinct values of }z\text{ we may choose }k=-n,-n+1,\dots,n-1$

complex number stuffs (1 Viewer)

synthesisFR

afterhscivemostlybeentrollingdonttakeitsrsly

SadCeliac

done hsc yay

SadCeliac

done hsc yay

carrotsss

New Member

synthesisFR

afterhscivemostlybeentrollingdonttakeitsrsly

tywebb

dangerman

synthesisFR

afterhscivemostlybeentrollingdonttakeitsrsly

tywebb

dangerman

tywebb

dangerman

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