Processing math: 100%

User interface language: English | Español

Date November 2012 Marks available 6 Reference code 12N.1.hl.TZ0.10
Level HL only Paper 1 Time zone TZ0
Command term Hence, Solve, and Write Question number 10 Adapted from N/A

Question

Consider the complex numbers

z1=23cis3π2 and z2=1+3 .

(i)     Write down z1 in Cartesian form.

(ii)     Hence determine (z1+z2) in Cartesian form.

[3]
a.

(i)     Write z2 in modulus-argument form.

(ii)     Hence solve the equation z3=z2 .

[6]
b.

Let z=rcisθ , where rR+  and 0θ<2π . Find all possible values of and θ ,

(i)     if z2=(1+z2)2;

(ii)     if z=1z2.

[6]
c.

Find the smallest positive value of n for which (z1z2)nR+ .

[4]
d.

Markscheme

(i)     z1=23cis3π2z1=23i     A1

 

(ii)     z1+z2=23i1+3i=13i     A1

(z1+z2)=1+3i     A1

[3 marks]

a.
(i)     |z2|=2
tanθ=3     (M1)
z2 lies on the second quadrant
θ=argz2=2π3
z2=2cis2π3     A1A1
 
(ii)     attempt to use De Moivre’s theorem     M1
z=32cis2π3+2kπ3, k=0, 1 and 2
z=32cis2π9,32cis8π9,32cis14π9(=32cis(4π9))     A1A1
Note: Award A1 for modulus, A1 for arguments.
 
Note: Allow equivalent forms for z .
 
[6 marks]
b.

(i)     METHOD 1

z2=(11+3i)2=3(z=±3i)     M1

z=3cisπ2 or z1=3cis3π2(=3cis(π2))     A1A1

so r=3 and θ=π2 or θ=3π2(=π2) 

Note: Accept rcis(θ) form.

 

 METHOD 2

z2=(11+3i)2=3z2=3cis((2n+1)π)     M1

 r2=3r=3     A1

2θ=(2n+1)πθ=π2 or θ=3π2 (as 0θ<2π)     A1 

Note: Accept rcis(θ) form.

 

(ii)     METHOD 1

z=12cis2π3z=cisπ2cis2π3     M1

z=12cisπ3

so r=12 and θ=π3     A1A1

METHOD 2

z1=11+3iz1=13i(1+3i)(13i)     M1

z=1+3i4z=12cisπ3

so r=12 and θ=π3     A1A1

[6 marks]

c.

z1z2=3cis5π6     (A1)

(z1z2)n=3ncis5nπ6     A1

equating imaginary part to zero and attempting to solve     M1

obtain n = 12     A1

Note: Working which only includes the argument is valid.

 

[4 marks]

d.

Examiners report

Many candidates were perhaps fortunate in this question due to there being several follow through marks available. Part a) was often done correctly. In part b), incorrect answers of 2cis(π3) were common, though many of these candidates often applied De Moivre’s Theorem correctly to their answers. In c) the majority found z2=3 but could then get no further. The second part was often poorly done, with those rationalising the Cartesian form of z1 having the most success. Part d) posed problems for a great many, and correct solutions were rarely seen. cis(5nπ6) was often seen, but then finding n = 12 proved to be a step too far for many. In general, the manipulation of complex numbers in polar form is not well understood.

a.

Many candidates were perhaps fortunate in this question due to there being several follow through marks available. Part a) was often done correctly. In part b), incorrect answers of 2cis(π3) were common, though many of these candidates often applied De Moivre’s Theorem correctly to their answers. In c) the majority found z2=3 but could then get no further. The second part was often poorly done, with those rationalising the Cartesian form of z1 having the most success. Part d) posed problems for a great many, and correct solutions were rarely seen. cis(5nπ6) was often seen, but then finding n = 12 proved to be a step too far for many. In general, the manipulation of complex numbers in polar form is not well understood.

b.

Many candidates were perhaps fortunate in this question due to there being several follow through marks available. Part a) was often done correctly. In part b), incorrect answers of 2cis(π3) were common, though many of these candidates often applied De Moivre’s Theorem correctly to their answers. In c) the majority found z2=3 but could then get no further. The second part was often poorly done, with those rationalising the Cartesian form of z1 having the most success. Part d) posed problems for a great many, and correct solutions were rarely seen. cis(5nπ6) was often seen, but then finding n = 12 proved to be a step too far for many. In general, the manipulation of complex numbers in polar form is not well understood.

c.

Many candidates were perhaps fortunate in this question due to there being several follow through marks available. Part a) was often done correctly. In part b), incorrect answers of 2cis(π3) were common, though many of these candidates often applied De Moivre’s Theorem correctly to their answers. In c) the majority found z2=3 but could then get no further. The second part was often poorly done, with those rationalising the Cartesian form of z1 having the most success. Part d) posed problems for a great many, and correct solutions were rarely seen. cis(5nπ6) was often seen, but then finding n = 12 proved to be a step too far for many. In general, the manipulation of complex numbers in polar form is not well understood.

d.

Syllabus sections

Topic 1 - Core: Algebra » 1.6 » Modulus–argument (polar) form z=r(cosθ+isinθ)=rcisθ=reiθ
Show 22 related questions

View options