Date | May 2012 | Marks available | 3 | Reference code | 12M.1.hl.TZ2.6 |
Level | HL only | Paper | 1 | Time zone | TZ2 |
Command term | Find | Question number | 6 | Adapted from | N/A |
Question
Given that (4−5i)m+4n=16+15i(4−5i)m+4n=16+15i , where i2=−1i2=−1, find m and n if
m and n are real numbers;
m and n are conjugate complex numbers.
Markscheme
attempt to equate real and imaginary parts M1
equate real parts: 4m+4n=164m+4n=16; equate imaginary parts: −5m=15−5m=15 A1
⇒m=−3, n=7⇒m=−3, n=7 A1
[3 marks]
let m=x+iy, n=x−iym=x+iy, n=x−iy M1
⇒(4−5i)(x+iy)+4(x−iy)=16+15i⇒(4−5i)(x+iy)+4(x−iy)=16+15i
⇒4x−5ix+4iy+5y+4x−4iy=16+15i⇒4x−5ix+4iy+5y+4x−4iy=16+15i
attempt to equate real and imaginary parts M1
8x+5y=16, −5x=158x+5y=16, −5x=15 A1
⇒x=−3, y=8⇒x=−3, y=8 A1
(⇒m=−3+8i, n=−3−8i)(⇒m=−3+8i, n=−3−8i)
[4 marks]
Examiners report
Part (a) was generally well answered. In (b), however, some candidates put m=a+ib and n=c+id which gave four equations for two unknowns so that no further progress could be made.
Part (a) was generally well answered. In (b), however, some candidates put m=a+ib and n=c+id which gave four equations for two unknowns so that no further progress could be made.