OCR Further Pure Core 1 2020 November — Question 4 4 marks

Exam BoardOCR
ModuleFurther Pure Core 1 (Further Pure Core 1)
Year2020
SessionNovember
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComplex numbers 2
TypeDirect nth roots: purely real or imaginary RHS
DifficultyModerate -0.3 This is a straightforward Further Maths question requiring standard technique: convert 25i to modulus-argument form (r=25, θ=π/2), apply the nth root formula to get two roots with half the argument and square root of modulus, then plot on an Argand diagram. While Further Maths content, it's a direct application of a well-practiced algorithm with no problem-solving or novel insight required, making it slightly easier than average overall.
Spec4.02h Square roots: of complex numbers4.02k Argand diagrams: geometric interpretation

4 In this question you must show detailed reasoning.
  1. Determine the square roots of 25 i in the form \(r \mathrm { e } ^ { \mathrm { i } \theta }\), where \(0 \leqslant \theta < 2 \pi\).
  2. Illustrate the number 25i and its square roots on an Argand diagram.

Question 4:
AnswerMarks Guidance
4(a) π
i
25i=25e2 oe
π
i
25i=25e2
π 5π
i i
⇒ 25i =5e4 and 5e 4
π 5π
i i
AnswerMarks
So the square roots are 5e4 and 5e 4B1
M1
AnswerMarks
A13.3
3.1a
AnswerMarks
3.2aConversion soi
Attempt to find at least one root by rooting the
modulus and halving the argument soi
Both
AnswerMarks
Alternate MethodB1
M1
AnswerMarks
A1Conversion and attempt to square
a and b. Ignore ±
Both
25i =a+bi
5
⇒a =b=± 2
2
5  1 1 
⇒ 25i =± 2 ( 1+i )=±5cos π+isin π 
2  4 4 
1 5
π π
⇒ 25i =5e4 and 5e4
[3]
AnswerMarks Guidance
(b)B1 1.1
if no scale then the lines representing the roots should
be at 45° to axis.
Accept points.
No extras
Line representing 25i must be at least two times as
long
[1]
B1
M1
A1
Conversion and attempt to square
a and b. Ignore ±
Both
Question 4:
4 | (a) | π
i
25i=25e2 oe
π
i
25i=25e2
π 5π
i i
⇒ 25i =5e4 and 5e 4
π 5π
i i
So the square roots are 5e4 and 5e 4 | B1
M1
A1 | 3.3
3.1a
3.2a | Conversion soi
Attempt to find at least one root by rooting the
modulus and halving the argument soi
Both
Alternate Method | B1
M1
A1 | Conversion and attempt to square
a and b. Ignore ±
Both
25i =a+bi
5
⇒a =b=± 2
2
5  1 1 
⇒ 25i =± 2 ( 1+i )=±5cos π+isin π 
2  4 4 
1 5
π π
⇒ 25i =5e4 and 5e4
[3]
(b) | B1 | 1.1 | All three but no extras. Scales etc are not required but
if no scale then the lines representing the roots should
be at 45° to axis.
Accept points.
No extras
Line representing 25i must be at least two times as
long
[1]
B1
M1
A1
Conversion and attempt to square
a and b. Ignore ±
Both
4 In this question you must show detailed reasoning.
\begin{enumerate}[label=(\alph*)]
\item Determine the square roots of 25 i in the form $r \mathrm { e } ^ { \mathrm { i } \theta }$, where $0 \leqslant \theta < 2 \pi$.
\item Illustrate the number 25i and its square roots on an Argand diagram.
\end{enumerate}

\hfill \mbox{\textit{OCR Further Pure Core 1 2020 Q4 [4]}}