Math Ia

Math IA Math Internal Assessment EF world(prenominal) Academy NY Student Name Joo Hwan Kim Teacher Ms. Gueye Date March sixteenth 2012 Contents Introduction Part A Part B purpose Introduction The aim of this IA is to note tabu the pattern of the equivalences with complex poetry by using our know conductge. I used de Moivres theorem and binomial expansion, to pay off forth the specific pattern and make conjecture ab discover it. I basically used property of binominal theory with the relationship among the duration of the identify constituents and the ascendents. Part ATo obtain the solutions to the comparability ) Moivres theorem, ( equation, we entrust involve , I used de Moivres theorem. According to de . So if we apply this theorem in to the ( ) ( ( ) ) ( ) If we rewrite the equation with the found take to be of , it shows ( ( ( ( ( ) )) )) Let k be 0, 1, and 2. When k is 0, ( ) ( ) v v Now I know that if I apply this equation with the roots of ( ) ( ) we can find the answers on the unit circle. I plotted these values in to the graphing softw ar, GeoGebra and then I got a graph as beneath control 1 The roots of z-1=0 I chose a root of and I tried to find place the length of cardinal sections from the point Z. I divided each trilateral in to two same beneficial angle triangles. By cunning that the radius of the unit circle is 1, with the knowledge of the length from D or Z to their mid-point C is length of the segment segment ) v , I found out . So I multiplied this answer by 2. And I got the v . I used same method to find out the length of the . (v v project 2 The graph of the equation z3-1=0 after determination out line segment Thus we can write that the three roots of , and we can as well as factorize the equation by long division.Since I know that cardinal of the roots is 1, I can divide the whole equation by (z-1). And then I got . So if we factorize the equation as ( )( ) As question asks I repeat the work above for the equations . Using De Moivres theorem, can be rewritten as ( ) call up So the roots of the equation are . As we can see the graph below, I draw a graph of the roots and affiliated two other from a point A. The question wants me to find out the length of the line segments which I connected from a single roots to two other roots, . Since are isosceles right-angle triangles with two sides of 1.With the basic knowledge of right triangle with two I found out that the length of the v v portend 3 Graph of z4-1=0 before finding out the line segment Figure 4 Graph of z4-1=0 after finding out the line segments Again I am finding out the roots of ( ( ( Suppose that the k is equal to 0,1,2,3 and 4. ) ) ) ( ( ( ( I plotted those roots of the equation ) ) ) ) ( ( ( ( ) ) ) ) ( ( ( ( ) ) ) ) in to GeoGebra and on an Argand Diagram. And as shown below I found out the length of the line segments Figure 5 Graph of z5-1=0 before finding out the line segments Figure 6 Graph of z5-1=0 after find ing out the line segmentsSo if I rewrite the lengths of line segments for each polar equations and , they are , ( ) ( ) , ( ( ( ( ) ) ( ) ) ( ) ) ( ) With my values of distance of the line segments between the chosen root and others, I made a conjecture that says ( ( ( ) ) ( ( ) ) I tried to see this conjecture. But as shown below, it is im practicable to prove due to unknown gist of multiple of the sin properties ( ) Then I tried to prove it by binominal expansion, which is totally different way. I drew a graph of an equation (shown below) and connected between a root to all the other roots.Figure 7 The graph of zn-1=0, with its roots connected As shown above, the graph has authoritative amount of roots, and they are connected to a root as told in the problems. And the lengths of those line segments are able to be written as ( So I rewrote the equation ( And with the knowledge of ( )( )( )( ) ( ) )( ) in the form of )( )( ) ( ) And since the angles , And I will have ( ) And then, with the binominal expansion, I folded it out, and got ( ( ( ( ) ( ) ) )( )( )( ) ( ) ) ( ) And I can find out that ( ) ( ( ( )( ) )( ) )( ( ) ( ) ) And I know that ( ) , so with this knowledge, I rewrote ( ( )( ) )( ( )( ( ) ) And all those ( to zero. So it finally has )and ( ) tinct ( ( ( ( ( ) ( )( )( )( ) ) ) ( ( ) ) ) ) And there are two condition where n can be even number or odd number, And according to this condition the value of n ( ) So the total reaping of the length of the line segment equal to the power of the equation Proved. And I factorized When I factorized ( ( ( )( )( , I got the answers alike )( ) ) ) And I to a fault tried to test my conjecture with some more values of For ( ) Suppose ( ) ( ) ( ( ) ) Figure 8 The graph of z6-1=0 with line segments The product of lengths of the line segments are v vFor ( Suppose ) Figure 9 The graph of z7-1=0 with its line segments Part B I am going to find the solutions of this equation for each Moivres the orem to obtain solutions to the equation . And I will use de . And I also drew diagrams for each roots of the equation s. I used Geo Gebra to exhibit each roots of the equation on the Argand Diagram. So, when ( ) ( ( ) ) ( ( ( ) ) ) ( ( ) ) ( ) ( ) v v ( ) ( ) Figure 10 The graph of roots of equation z3=i As shown above, the equation has three distinct roots. And the distance of arc between each neighboring roots are same with others.Roots of this equation development by are three roots on the unit circle. , so we can find that there When ( ( ( ) ) ) Suppose ( ) ( ( ( ) ) ) Figure 11 The graph of roots of equation z4=i When n=5, ( ( ) ) Suppose ( ( ) ( ( ( ) ) ) ) ( ) Figure 12 The graph of roots of the equation z5=i Basically all the roots we found are on the lane of the unit-circle, because we use the complex ( ) number whose modulus is 1. . So if I interpolate the equation of , I would get ( ( So for the equation like equation is Generalize the equations of , ) ) that satisfy this ( ). And I can should be (0+1i)= i.And the value of into , where n=3,4 and 5. rad. So we can change the equation ( ) ( ( ( ) ) ) With the knowledge of in the right triangle of a b So With the knowledge v It is possible state that This elicitation is proved naturally as we found out that the angle of the roots is . When But when under the condition of has a generalization of the generalization would change as Conclusion I found out some patterns about two different equation some conjectures that led me to find out and prove it. For of all length of the line segments connected form a root to others. . There were n is equal to the product