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# Math Portfolio Type 2 - PATTERNS FROM COMPLEX NUMBERS

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Introduction Erzurum İhsan Doğramacı Vakfı

Özel Bilkent Laboratuvar Lisesi

Mathematics Portfolio HL Type I

Topic

PATTERNS FROM COMPLEX NUMBERS

Name – Surname

Selim TEPELER

IB No:

DPR160 (006040-047)

Introduction

In this portfolio we are gonna use complex number patterns which are related to roots of complex number , and we connect them our knowledge about analytic geometry ( which is distance formula) . Finally we will try to formulate a conjecture. This conjecture will help us to finding distance between roots.

PART A

Use de Moivre’s theorem to obtain solutions to the equation Questions asks ; and firstly we should turn it into complex form; De Moivre’s theorem claims that ; where k = 0, 1, 2 we found ;

- Cis(0) -cis(120) -cis(240) (by the way cis(α) means ‘’ cos(α)+isin(α))

• Use graphing software to plot these roots on an Argand diagram as well as a unit circle with centre origin. • Choose a root and draw line segments from this root to the other two roots. • Measure these line segments and comment on your results. As we see that the length of line segments are equal to each other. (one segment is 0.01 unit bigger than the others because of accuracy of graphing software)

Repeat the above for the equations and •  Comment on your results and try to formulate a conjecture.

Middle

value cos(0) , is cos(72) value is sin(0) value is sin(72).

Now we will put them on equation .  1+1-2(cos72*cos 0 +sin72*sin0)

We will use this trigonometric identities ; and When we put our value at these formula ; =cos 72

Lastly our final formula is ; Factorize •  for n = 3, 4 and 5. -1= (z-1)(  -1= (z-1)(  -1= (z-1)( • Use graphing software to test your conjecture for some more values of n € Z+ and make modifications to your conjecture if necessary.

Lets try it for n=6 Testing for Aw distance ;

a=cis0 and w is cis120  =1.732050… ≅17.3 (because of accuracy of software)

or testing for Az distance ;  =1= 1 from graph .

PART B

Conclusion

=i where where k = 0, 1, 2,3,4

We found ;

Cis(18) cis (90) cis(162) cis(234) cis(306)

• Use graphing software to represent each of these solutions on an Argand diagram.

For =i For =i For =i Generalize and prove your results for i • =a + bi , where |a +bi |=1.

|a +bi |=1 so it means that ; + 1= As we can see r=1 so =a+bi= r cis(θ)

When we use De Moivre’s theorem, it states ; For our solution we will use reverse way of it ; r=1 so we dont need r at equation ;

z= let’s find the value with tangent ;

tan( =  = and = so z=cis( )

• What happens when |a +bi |≠1?

That doesn’t apply when when |a +bi |≠1 because r value should equal to 1.

Resources

http://en.wikipedia.org/wiki/Distance

http://en.wikipedia.org/wiki/List_of_trigonometric_identities

http://www.rapidlearningcenter.com/mathematics/trigonometry/18-Complex-Numbers-and-De-Moivre-Theorem.html

http://demonstrations.wolfram.com/PatternsFromMathRulesUsingComplexNumbers/

http://en.wikipedia.org/wiki/Imaginary_number

This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.

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