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@@ What is Happening?  [Wrap4]   Thurs 25/12/15   Sydney,  Australia Time 11:59 am   ♪ ♫

Enjoy your Chrismas Eve everyone, don't forget to leave out milk and cookies for Santa !    

Hi all,

It has been a week since my last wrap.  That is largely because I was away on holidays. This also means that there are a number of threads and answers that I have not seen :(  Still I have collected many answerer names: Credits go to Alan, CPhill, anonymous4338, Complex, Heureka, Olivia 3.141592654, Cookiesandcream, Alleycat2, gibsonj338 and Dragonlance.  Thank you    

Thanks for a appreciating my colour scheme too Hayley  LOL   

Stranger danger reminder

Hi Complex,

I plotted the 3 points and decided that a cubic could go through them if there was a horizontal point of inflection at (2,4)

so the graph goes though (1,1). and (3,7) and (2,4)

also     y'(2)=0   and y''(2)=0

so I started with

\(y=ax^3+bx^2+cx+d\\ y'=3ax^2+2bx+c\\ y''=6ax+2b \)

\(0=6a(2) + 2b\\ 0=12a+2b\\ 0=6a+b\\ b=-6a\\~\\ 0=3a(2^2)+2b(2)+c\\ 0=12a+4b+c\\ 0=12a+4(-6a)+c\\ 0=12a-24a+c\\ 0=-12a+c\\ c=12a\\~\\ y=ax^3+bx^2+cx+d\\ y=ax^3-6ax^2+12ax+d\\\)
 

\(y=ax^3-6ax^2+12ax+d\\ though (1,1)\\ 1=a-6a+12a+d\\ 1=7a+d\\ d=1-7a\\ y=ax^3-6ax^2+12ax+1-7a\\~\\ \)
\(through (2,4)\\ 4=a*2^3-6a*2^2+12a*2+1-7a\\ 4=8a-24a+24a+1-7a\\ 4=a+1\\ a=3\)

\(y=ax^3-6ax^2+12ax+1-7a\\ y=3x^3-18x^2+36x+1-21\\ y=3x^3-18x^2+36x-20\\\)

OR using your letters

\(\boxed{n=3s^3-18s^2+36s-20}\)

Chris, I do not think that yours is a new formula, it is just a mor complicated way of writing exactly the same formula. 

Hi, How is the ventriloquism going ??

You have had a fair bit of time to practice now so I expect that you are getting good   

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Find all 4th roots of z=5000+i.  Show all work and draw a graph to reprecipent the answers.

I think I have worked out how to do this quite neatly I think but I do not know the correct terminology. 

I am not familiar with the correct terms so i used this page to help me there.

http://www.sparknotes.com/math/precalc/complexnumbers/terms.html

The distance (modulus) of this point from 0 is   \(\sqrt{5000^2+1^1}=\sqrt{25000001}\)

The distance (modulus) from 0 that the roots will be is  \((\sqrt{25000001})^{1/4}=25000001^{1/8}=\sqrt[8]{25000001}\)

(25000001)^(1/8) = 8.4089641945819655     approx 8.4090

First, 2pi/4 = pi/2  so the roots will be approx 8.409  units from 0 and pi/2 radians apart.

z=5000+i

I have just drawn a right angle triangle on a scrap of paper to work this out.

\(z=\sqrt{25000001}*(\frac{5000}{\sqrt{25000001}}+\frac{1i}{\sqrt{25000001}})\)

\(cos\theta=\frac{5000}{\sqrt{25000001}}\qquad and \qquad sin\theta=\frac{1}{\sqrt{25000001}}\)

argument (angle) of the first root is     acos(5000/sqrt(25000001) = 0.000199999998

asin(1/sqrt(25000001)) = 0.000199999997

They had to be the same, I was just showing you. :)

So the angle (argument) is very close to 0.0002 radians

The first 4th root angle is \(acos(\frac{5000}{\sqrt{25000001}})\div4\approx 0.0002\div4 = 0.00005 \;radians\)

So the angles of the 4 roots will be

0.00005, 0.00005+pi/2, 0.00005+pi, 0.0005+3pi/2

0.00005+pi/2 = 1.5708463267948966                    approx 1.5708 radians     

0.00005+pi/2+pi/2 = 3.1416426535897932            approx 3.1416  radians

0.00005+pi/2+pi/2+pi/2 = 4.7124389803846899    approx  4.7124  radians

So the 4th roots of  z=5000+i    are

8.409e^(0.00005i), 8.409e^(1.5708i), 8.409e^(3.1416i), 8.409e^(4.712i)

Check the first one.     

(8.409e^(0.00005i))^4 = 5000.08516065925605241+1.000017045882085632i   near enough

Hi Amazinggirl, welcome to the forum :)

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We'd like you to cut out your swearing though.

I think if your teacher approved your writing then it woud not be full of bad language :)

Hi Gibsonj338   

Find all cubic roots of

 \(z=-1+i \mbox{ distance of z to the origin }=\sqrt{1^2+1^2}=\sqrt2\\ \mbox{cubic toots so there are 3 of them so they are } \frac{2\pi}{3} radians\; apart.\\ z=\sqrt2(cis\frac{3\pi}{4})\\ \sqrt[3]{z}=\sqrt[6]{2}(cis(\frac{3\pi}{4}\div3)),\quad\sqrt[6]{2}(cis((\frac{3\pi}{4}+2\pi)\div3)),\quad \sqrt[6]{2}(cis((\frac{3\pi}{4}+4\pi)\div3))\\ \sqrt[3]{z}=\sqrt[6]{2}(cis(\frac{3\pi}{4}\div3)),\quad\sqrt[6]{2}(cis(\frac{11\pi}{4}\div3)),\quad \sqrt[6]{2}(cis(\frac{19\pi}{4}\div3))\\ \sqrt[3]{z}=\sqrt[6]{2}(cis(\frac{3\pi}{12})),\quad\sqrt[6]{2}(cis(\frac{11\pi}{12})),\quad \sqrt[6]{2}(cis(\frac{19\pi}{12}))\\ \sqrt[3]{z}=\sqrt[6]{2}(cis(\frac{\pi}{4})),\quad\sqrt[6]{2}(cis(\frac{11\pi}{12})),\quad \sqrt[6]{2}(cis(\frac{19\pi}{12}))\\ or\ \)

\(\sqrt[3]{z}=\frac{1+i}{\sqrt[3]{2}},\quad\sqrt[6]{2}(cos(\frac{11\pi}{12})+i\;sin(\frac{11\pi}{12})),\quad\sqrt[6]{2}(cos(\frac{19\pi}{12})+i\;sin(\frac{19\pi}{12}))\\ \sqrt[3]{z}=\frac{1+i}{\sqrt[3]{2}},\quad\sqrt[6]{2}(-cos(\frac{\pi}{12})+i\;sin(\frac{\pi}{12})),\quad\sqrt[6]{2}(cos(\frac{5\pi}{12})-i\;sin(\frac{5\pi}{12})) \)

Thanks Chris, I thought it would be something like that.    LOL     :)