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Date May 2019 Marks available 3 Reference code 19M.2.AHL.TZ1.H_2
Level Additional Higher Level Paper Paper 2 Time zone Time zone 1
Command term Solve Question number H_2 Adapted from N/A

Question

Solve z 2 = 4 e π 2 i , giving your answers in the form

r e i θ where r θ R r > 0 .

[3]
a.

a + i b where a , b R .

[2]
b.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

z = 2 e π 4 i ( = 2 e 0.785 i )       A1

Note: Accept all answers in the form  2 e ( π 4 + 2 π n ) i .

z = 2 e 5 π 4 i ( = 2 e 3.93 i )   OR   z = 2 e 3 π 4 i ( = 2 e 2.36 i )        (M1)A1

Note: Accept all answers in the form 2 e ( 3 π 4 + 2 π n ) i .

Note: Award M1A0 for correct answers in the incorrect form, eg  2 e π 4 i .

[3 marks]

a.

z = 1.41 + 1.41 i z = 1.41 1.41 i        A1A1

( z = 2 + 2 i , z = 2 2 i )

[2 marks]

b.

Examiners report

[N/A]
a.
[N/A]
b.

Syllabus sections

Topic 1—Number and algebra » AHL 1.14—Complex roots of polynomials, conjugate roots, De Moivre’s, powers & roots of complex numbers
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Topic 1—Number and algebra

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