User interface language: English | Español

Date November 2011 Marks available 8 Reference code 11N.2.hl.TZ0.14
Level HL only Paper 2 Time zone TZ0
Command term Hence and Show that Question number 14 Adapted from N/A

Question

Show that |eiθ|=1eiθ=1.

[1]
a.

Consider the geometric series 1+13eiθ+19e2iθ+ .1+13eiθ+19e2iθ+ .

Write down the common ratio, z, of the series, and show that |z|=13|z|=13.

[2]
b.

Find an expression for the sum to infinity of this series.

[2]
c.

Hence, show that sinθ+13sin2θ+19sin3θ+=9sinθ106cosθsinθ+13sin2θ+19sin3θ+=9sinθ106cosθ.

[8]
d.

Markscheme

|eiθ| (=|cosθ+isinθ|)=cos2θ+sin2θ=1eiθ (=|cosθ+isinθ|)=cos2θ+sin2θ=1     M1AG

[1 mark]

a.

z=13eiθz=13eiθ     A1

|z|=|13eiθ|=13|z|=13eiθ=13     A1AG

[2 marks]

b.

S=a1r=1113eiθS=a1r=1113eiθ     (M1)A1

[2 marks]

c.

EITHER

S=1113cosθ13isinθS=1113cosθ13isinθ     A1

=113cosθ+13isinθ(113cosθ13isinθ)(113cosθ+13isinθ)=113cosθ+13isinθ(113cosθ13isinθ)(113cosθ+13isinθ)     M1A1

=113cosθ+13isinθ(113cosθ)2+19sin2θ=113cosθ+13isinθ(113cosθ)2+19sin2θ     A1

=113cosθ+13isinθ123cosθ+19=113cosθ+13isinθ123cosθ+19     A1

OR

S=1113eiθS=1113eiθ

=113eiθ(113eiθ)(113eiθ)=113eiθ(113eiθ)(113eiθ)     M1A1

=113eiθ113(eiθ+eiθ)+19=113eiθ113(eiθ+eiθ)+19     A1

=113eiθ10923cosθ=113eiθ10923cosθ     A1

=113(cosθisinθ)10923cosθ=113(cosθisinθ)10923cosθ     A1

THEN

taking imaginary parts on both sides

13sinθ+19sin2θ+=13sinθ10923cosθ13sinθ+19sin2θ+=13sinθ10923cosθ     M1A1A1

=sinθ10923cosθ=sinθ10923cosθ

sinθ+13sin2θ+=9sinθ106cosθsinθ+13sin2θ+=9sinθ106cosθ     AG

[8 marks]

d.

Examiners report

Parts (b) and (c) were answered fairly well by quite a few candidates. In (a) many candidates failed to write the formula for the modulus of a complex number. (c) proved inaccessible for a large number of candidates. The algebraic manipulation required and the recognition of the imaginary and real parts in order to arrive at the necessary relationship were challenging for many candidates.

a.

Parts (b) and (c) were answered fairly well by quite a few candidates. In (a) many candidates failed to write the formula for the modulus of a complex number. (c) proved inaccessible for a large number of candidates. The algebraic manipulation required and the recognition of the imaginary and real parts in order to arrive at the necessary relationship were challenging for many candidates.

b.

Parts (b) and (c) were answered fairly well by quite a few candidates. In (a) many candidates failed to write the formula for the modulus of a complex number. (c) proved inaccessible for a large number of candidates. The algebraic manipulation required and the recognition of the imaginary and real parts in order to arrive at the necessary relationship were challenging for many candidates.

c.

Parts (b) and (c) were answered fairly well by quite a few candidates. In (a) many candidates failed to write the formula for the modulus of a complex number. (c) proved inaccessible for a large number of candidates. The algebraic manipulation required and the recognition of the imaginary and real parts in order to arrive at the necessary relationship were challenging for many candidates.

d.

Syllabus sections

Topic 1 - Core: Algebra » 1.5 » Complex numbers: the number i=1i=1 ; the terms real part, imaginary part, conjugate, modulus and argument.
Show 33 related questions

View options