EDEXCEL AS LEVEL FURTHER MATHEMATICS PAPER 21: FURTHER PURE MATHEMATICS 1 NEW UPDATE WITH ALL COMPREHENSIVE REAL EXAM QUESTIONS AND A DETAILED BREAKDOWN OF ALL CORRECT VERIFIED ANSWERS EXAM PLUS CERTIFIED RATIONALES | COMPLETE TEST SOLUTION | PA

EDEXCEL AS LEVEL FURTHER MATHEMATICS
PAPER 21: FURTHER PURE MATHEMATICS 1 NEW
2026-2027 UPDATE WITH ALL COMPREHENSIVE
REAL EXAM QUESTIONS AND A DETAILED
BREAKDOWN OF ALL CORRECT VERIFIED
ANSWERS EXAM PLUS CERTIFIED RATIONALES |
COMPLETE TEST SOLUTION | PASSED & REWARDED
WITH DISTINCTION FOR ORIGINALLY GRADE A+
BRAND NEW!!!




Section A: Complex Numbers
Question 1
Express (3+2i) (4−5i) (3+2i) (4−5i) in the
form a+bia+bi, where aa and bb are real numbers.
Correct Answer: 22−7i22−7i
Rationale: Expanding the
brackets: 12−15i+8i−10i2=12−7i+10=22−7i12−15i+
8i−10i2=12−7i+10=22−7i.
Question 2

,Find the modulus and argument of the complex
number z=−1+3iz=−1+3i, giving the argument in
radians to three decimal places.
Correct
Answer: ∣z∣=2∣z∣=2, arg⁡(z)=2.094arg(z)=2.094
Rationale: ∣z∣= (−1)2+(3)2=1+3=2∣z∣= (−1)2+(3)2
=1+3=2. The argument is in the second
quadrant: π−arctan⁡ (3) =π−π3=2π3≈2.094π−arctan
(3) =π−3π=32π≈2.094.
Question 3
Write z=4(cos⁡π6+isin⁡π6) z=4(cos6π+isin6π) in
Cartesian form.
Correct Answer: 23+2i23+2i
Rationale: 4cos⁡π6=4×32=234cos6π=4×23=23
and 4sin⁡π6=4×12=24sin6π=4×21=2.
Question 4
Given that z=2+3iz=2+3i, find zzˉzzˉ.
Correct Answer: 1313

,Rationale: zˉ=2−3izˉ=2−3i, so zzˉ=(2+3i) (2−3i)
=4−6i+6i−9i2=4+9=13zzˉ=(2+3i) (2−3i)
=4−6i+6i−9i2=4+9=13.
Question 5
Solve the equation z2+4z+13=0z2+4z+13=0, giving
your answers in the form a±bia±bi.
Correct Answer: z=−2±3iz=−2±3i
Rationale: Using the quadratic
formula: z=−4±16−522=−4±−362=−4±6i2=−2±3iz=
2−4±16−52=2−4±−36=2−4±6i=−2±3i.
Question 6
Express 12−i2−i1 in the form a+bia+bi.
Correct Answer: 25+15i52+51i
Rationale: Multiplying numerator and denominator
by the conjugate: 2+i(2−i)(2+i) =2+i4+1=2+i5(2−i)
(2+i)2+i=4+12+i=52+i.
Question 7
Find the values of xx and yy such that (x+iy) (2−i)
=5+5i(x+iy) (2−i) =5+5i.

, Correct Answer: x=1,y=3x=1,y=3
Rationale: Expanding: 2x−xi+2yi−yi2=(2x+y)+(2y
−x) i=5+5i2x−xi+2yi−yi2=(2x+y) +(2y−x) i=5+5i.
So 2x+y=52x+y=5 and 2y−x=52y−x=5. Solving
gives x=1, y=3x=1, y=3.
Question 8
Given z=3+4iz=3+4i, find z2z2 in the
form a+bia+bi.
Correct Answer: −7+24i−7+24i
Rationale: z2=(3+4i)2=9+24i+16i2=9+24i−16=−7+
24iz2=(3+4i)2=9+24i+16i2=9+24i−16=−7+24i.
Question 9
Find the complex number zz such
that z+2zˉ=6+3iz+2zˉ=6+3i.
Correct Answer: z=2+3iz=2+3i
Rationale: Let z=a+biz=a+bi, so zˉ=a−bizˉ=a−bi.
Then a+bi+2a−2bi=3a−bi=6+3ia+bi+2a−2bi=3a−bi
=6+3i.