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Date November 2015 Marks available 6 Reference code 15N.1.hl.TZ0.11
Level HL only Paper 1 Time zone TZ0
Command term Solve Question number 11 Adapted from N/A

Question

Solve the equation z3=8i, zC giving your answers in the form z=r(cosθ+isinθ) and in the form z=a+bi where a, bR.

[6]
a.

Consider the complex numbers z1=1+i and z2=2(cos(π2)+isin(π6)).

(i)     Write z1 in the form r(cosθ+isinθ).

(ii)     Calculate z1z2 and write in the form z=a+bi where a, bR.

(iii)     Hence find the value of tan5π12 in the form c+d3, where c, dZ.

(iv)     Find the smallest value p>0 such that (z2)p is a positive real number.

[11]
b.

Markscheme

Note:     Accept answers and working in degrees, throughout.

 

z3=8(cos(π2+2πk)+isin(π2+2πk))     (A1)

attempt the use of De Moivre’s Theorem in reverse     M1

z=2(cos(π6)+isin(π6)); 2(cos(5π6)+isin(5π6));

2(cos(9π6)+isin(9π6))     A2

 

Note:     Accept cis form.

 

z=±3+i, 2i     A2

 

Note:     Award A1 for two correct solutions in each of the two lines above.

[6 marks]

a.

Note:     Accept answers and working in degrees, throughout.

 

(i)     z1=2(cos(π4)+isin(π4))     A1A1

(ii)     (z2=(3+i))

z1z2=(1+i)(3+i)     M1

=(31)+i(1+3)     A1

(iii)     z1z2=22(cos(π6+π4)+isin(π6+π4))     M1A1

 

Note:     Interpret “hence” as “hence or otherwise”.

 

tan5π12=3+131     A1

=2+3     M1A1

 

Note:     Award final M1 for an attempt to rationalise the fraction.

 

(iv)     z2p=2p(cis(pπ6))     (M1)

z2p is a positive real number when p=12     A1

 

Note:     Accept a solution based on part (a).

[11 marks]

Total [17 marks]

b.

Examiners report

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

Syllabus sections

Topic 1 - Core: Algebra » 1.6 » Modulus–argument (polar) form z=r(cosθ+isinθ)=rcisθ=reiθ
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