Moderate -0.3 This is a straightforward application of the quadratic formula to find complex roots, followed by routine conversion to modulus-argument form. While it requires multiple steps (solving, simplifying complex numbers, finding modulus and argument), these are all standard techniques with no conceptual difficulty or novel insight required. The 'show detailed reasoning' instruction indicates it's testing method rather than problem-solving ability.
Question 1:
1 | DR
−−20± (−20)2 −4×4×169
z=
2×4
5±12i
z=
2
2 2
5 12 13
r = + = oe
2 2 2
6
θ=tan−1 oe
2.5
13
(cos1.18+isin1.18)
2
13
(cos(−1.18)+isin(−1.18))
2 | M1
A1
B1ft
M1
A1
[5] | 1.1
1.1
1.1
1.1
2.5 | Term by term substituting into
formula.
If formula quoted, allow one slip
...
Or correctly completes the square
Attempting to find argument using
trigonometry
Angle must be in radians.
Argument could be 5.11 but both
angles must be the same. | Condone anything correct of the
p± q
form
r
2
5 25
eg 4z− − +169=0
2 4
Ft workings from complex
conjugate distinct pair (with real
component)
2.5 6
θ=cos−1 , θ=sin−1
6.5 6.5
13 13
oe eg cis1.18 or , 1.18
2 2
Not 5.10 (rounding error)
Not e.g. cos(−1.18)+isin(5.11)
1 In this question you must show detailed reasoning.\\
Solve the equation $4 z ^ { 2 } - 20 z + 169 = 0$. Give your answers in modulus-argument form.
\hfill \mbox{\textit{OCR Further Pure Core 2 2020 Q1 [5]}}