Date | May 2008 | Marks available | 14 | Reference code | 08M.2.hl.TZ1.14 |
Level | HL only | Paper | 2 | Time zone | TZ1 |
Command term | Find and Hence | Question number | 14 | Adapted from | N/A |
Question
z1=(1+i√3)m and z2=(1−i)nz1=(1+i√3)m and z2=(1−i)n .
(a) Find the modulus and argument of z1z1 and z2z2 in terms of m and n, respectively.
(b) Hence, find the smallest positive integers m and n such that z1=z2z1=z2 .
Markscheme
(a) |1+i√3|=2 or |1−i|=√2∣∣1+i√3∣∣=2 or |1−i|=√2 (A1)
arg(1+i√3)=π3 or arg(1−i)=−π4(accept 7π4)arg(1+i√3)=π3 or arg(1−i)=−π4(accept 7π4) (A1)
|z1|=2m|z1|=2m A1
|z2|=√2n|z2|=√2n A1
arg(z1)=marctan√3=mπ3arg(z1)=marctan√3=mπ3 A1
arg(z2)=narctan(−1)=n−π4(accept n7π4)arg(z2)=narctan(−1)=n−π4(accept n7π4) A1 N2
[6 marks]
(b) 2m=√2n⇒n=2m2m=√2n⇒n=2m (M1)A1
mπ3=n−π4+2πkmπ3=n−π4+2πk , where k is an integer M1A1
⇒mπ3+nπ4=2πk⇒mπ3+nπ4=2πk
⇒mπ3+2mπ4=2πk⇒mπ3+2mπ4=2πk (M1)
56mπ=2πk56mπ=2πk
⇒m=125k⇒m=125k A1
The smallest value of k such that m is an integer is 5, hence
m =12 A1
n = 24. A1 N2
[8 marks]
Total [14 marks]
Examiners report
Part (a) of this question was answered fairly well by candidates who attempted this question. The main error was the sign of the argument of z2z2. Few candidates attempted part (b), and of those who did, most scored the first two marks for equating the modulii. Only a very small number equated the arguments correctly using 2πk2πk.