Date | May Example questions | Marks available | 2 | Reference code | EXM.3.AHL.TZ0.2 |
Level | Additional Higher Level | Paper | Paper 3 | Time zone | Time zone 0 |
Command term | Show that and Hence | Question number | 2 | Adapted from | N/A |
Question
This question will explore connections between complex numbers and regular polygons.
The diagram below shows a sector of a circle of radius 1, with the angle subtended at the centre O being α,0<α<π2. A perpendicular is drawn from point P to intersect the x-axis at Q. The tangent to the circle at P intersects the x-axis at R.
By considering the area of two triangles and the area of the sector show that cosαsinα<α<sinαcosα.
Hence show that limα→0αsinα=1.
Let zn=1,z∈C,n∈N,n⩾5. Working in modulus/argument form find the n solutions to this equation.
Represent these n solutions on an Argand diagram. Let their positions be denoted by P0,P1,P2,…Pn−1 placed in order in an anticlockwise direction round the circle, starting on the positive x-axis. Show the positions of P0,P1,P2 and Pn−1.
Show that the length of the line segment P0P1 is 2sinπn.
Hence, write down the total length of the perimeter of the regular n sided polygon P0P1P2…Pn−1P0.
Using part (b) find the limit of this perimeter as n→∞.
Find the total area of this n sided polygon.
Using part (b) find the limit of this area as n→∞.
Markscheme
Area triangle OPQ=12cosαsinα A1
Area sector =1212α A1
Area triangle OPR=121tanα A1
So looking at the diagram 12cosαsinα<12α<12sinαcosα M1
⇒cosαsinα<α<sinαcosα AG
[5 marks]
Hence cosα<αsinα<1cosα and as α→0,cosα→1 we have M1R1
limα→0αsinα=1 AG
[2 marks]
(rcisθ)n=1cis0⇒rncisnθ=1cisθ M1A1M1A1
rn=1⇒r=1 nθ=0+2πk,k∈Z A1A1
θ=2πkn,0⩽k⩽n−1 A1
z=cis2πkn,0⩽k⩽n−1 A1
[8 marks]
A1
[1 mark]
Bisecting the triangle OP0P1 to form two right angle triangles M1
Length of P0P1=2t where t=sin(2πn2) M1A1A1
So length is 2sinπn AG
[4 marks]
Length of perimeter is 2nsinπn A1
[1 mark]
2nsinπn=2πnπsinπn→2π as n→∞ M1A1
[2 marks]
Area of OP0P1=121×1sin2πn so total area is n2sin2πn. M1A1A1
[3 marks]
n2sin2πn=πn2πsin2πn→π as n→∞ M1A1
[2 marks]