Date | May 2022 | Marks available | 3 | Reference code | 22M.1.AHL.TZ1.9 |
Level | Additional Higher Level | Paper | Paper 1 (without calculator) | Time zone | Time zone 1 |
Command term | Find | Question number | 9 | Adapted from | N/A |
Question
Consider the complex numbers z1=1+bi and z2=(1-b2)-2bi, where b∈ℝ, b≠0.
Find an expression for z1z2 in terms of b.
Hence, given that arg(z1z2)=π4, find the value of b.
Markscheme
z1z2=(1+bi)((1-b2)-(2b)i)
=(1-b2-2i2b2)+i(-2b+b-b3) M1
=(1+b2)+i(-b-b3) A1A1
Note: Award A1 for 1+b2 and A1 for -bi-b3i.
[3 marks]
arg(z1z2)=arctan(-b-b31+b2)=π4 (M1)
EITHER
arctan(-b)=π4 (since 1+b2≠0, for b∈ℝ) A1
OR
-b-b3=1+b2 (or equivalent) A1
THEN
b=-1 A1
[3 marks]
Examiners report
Part (a) was generally well done with many completely correct answers seen. Part (b) proved to be challenging with many candidates incorrectly equating the ratio of their imaginary and real parts to π4 instead of tanπ4. Stronger candidates realized that when θ=π4, it forms an isosceles right-angled triangle and equated the real and imaginary parts to obtain the value of b .