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, z∈C giving your answers in the form z=r(cosθ+isinθ) and in the form z=a+bi where a, b∈R.
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, b∈R.
(iii) Hence find the value of tan5π12 in the form c+d√3, where c, d∈Z.
(iv) Find the smallest value p>0 such that (z2)p is a positive real number.
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]
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
=(√3−1)+i(1+√3) A1
(iii) z1z2=2√2(cos(π6+π4)+isin(π6+π4)) M1A1
Note: Interpret “hence” as “hence or otherwise”.
tan5π12=√3+1√3−1 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]