Question 2:
2 | (a) | DR
− 2 4 + 7 i = ( − 2 4 ) 2 + 7 2 or
7
a r g ( − 2 4 + 7 i ) = t a n − 1 oe
− 2 4 | M1 | 1.1 | Attempt to find modulus or
argument using a correct formula
(values must be real) | If 7/24 allow only if supported by
explanation, further working or clear
diagram.
may use alternative trig function
−24+7i =25 and awrt –0.284 or 2.86 | A1 | 1.1 | Condone use of degrees for this
mark (–16.3 or 163.7) | accept arctan -7/24
–24 + 7i = 25(cos2.86 + isin2.86) | A1 | 1.1 | Final answer | Accept equivalent notation. E.g cis,
(r,θ) or exponential form, not (-C +iS)
nor rC + riS.
[3]
(b) | DR
6 i z + 1 8 w = − 4 2 i
− 6 i z − 5 w = 3 − i 1 3 | *M1 | 1.1 | Scaling both equations (using
i2 = –1) so that the coefficient of z
or w is the same in magnitude. | − 5 z + 1 5 i w = 3 5
− 1 8 z + 1 5 i w = 9 + 3 9 i
13w=−13−39i so w=−1−3i | A1 | 1.1 | 13𝑧 = 26−39iso 𝑧 = 2−3i
i𝑧+3(−1−3i) = −7i or −6𝑧+5i(−1−3i) =
3+13i
i𝑧 = 3+2i or −6𝑧 = −12+18i | dep*M1 | 1.1 | Substituting back into one equation
and attempt to solve by collecting
real and imaginary parts. | i(2−3i)+3𝑤 = −7i
or −6(2−3i)+5i𝑤 = 3+13i
i.e. reaches 𝑘𝑧 = 𝑎+𝑏ifor real 𝑎,𝑏
z = 2 − 3 i | A1 | 1.1
Alternative method:
−7i−i𝑧 3+13i+6𝑧
𝑤 = or 𝑤 =
3 5i | M1 | Using one equation to express one
unknown in terms of the other.
−7i−3w 3+13i−5i𝑤
𝑧 = or 𝑧 =
i −6
−7i−i𝑧
−6𝑧+5i( ) = 3+13i
3 | M1 | Substituting into the other equation
and using i2 = –1 at least once
∴ −18𝑧+35+5𝑧 = 9+39i
∴ (−13𝑧 = −26+39i so) 𝑧 = 2−3i | A1
A1 | A1
−7i−i(2−3i)
𝑤 = = −1−3i
3
Alternative method:
−7i−i𝑧 3+13i+6𝑧
𝑤 = or 𝑤 =
3 5i
M1
Using one equation to express one
unknown in terms of the other.
Substituting into the other equation
and using i2 = –1 at least once
Alternative method 2: | M1 | replaces z and w with two
Cartesian forms in both equations
i(𝑎+𝑏i)+3(𝑐+𝑑i) = −7i
−6(𝑎+𝑏i)+5i(𝑐+𝑑i) = 3+13i
Re:−𝑏+3𝑐 = 0,−6𝑎−5𝑑 = 3 | M1 | Takes real and imaginary parts | condone i’s left in
Im:𝑎+3𝑑 = −7,−6𝑏+5𝑐 = 13 | from both complex equations
𝑧 = 2−3i | A1 | Could be BC | a = 2, b = –3 sufficient
A1 | A1 | c = –1, d = –3 sufficient | c = –1, d = –3 sufficient
𝑤 = −1−3i
Alternative method 3: | M1 | Writes in matrix form and derives
𝑧
an equation for ( )
𝑤 | Must left-multiply by their inverse
matrix
Type equation here.
 i 3   z   − 7 i 
=
− 6 5 i w 3 + 1 3 i
− 1
 z   i 3   − 7 i 
=
w − 6 5 i 3 + 1 3 i
− 1
 i 3  1  5 i − 3 
=
where − 6 5 i 1 3 6 i | A1 | …with correct inverse matrix
1  5 i − 3   − 7 i  1  3 5 − 9 − 3 9 i 
=
1 3 6 i 3 + 1 3 i 1 3 − 4 2 i + 3 i − 1 3 | M1 | Expands…
 z = 2 − 3 i 𝑤 = −1−3i , | A1 | … to correct simplified solution | can be in vector form
[4]
replaces z and w with two
Cartesian forms in both equations
Writes in matrix form and derives
𝑧
an equation for ( )
𝑤
Must left-multiply by their inverse
matrix