QCE Vault / Specialist Maths Complex numbers practice QCE Specialist Maths · Original practice questions with worked solutions
Browse all questions Complex numbers practice Vectors practice Original practice exam Practise Cartesian and polar forms, complex roots and geometric reasoning. Start by deciding which representation makes the operation simplest, then check the location of your result on an Argand diagram.
Key ideas In Cartesian form, multiply out and use i 2 = − 1 i^2=-1 i 2 = − 1 . For division, multiply numerator and denominator by the conjugate of the denominator. In polar form, multiply moduli and add arguments. De Moivre’s theorem gives ( r cis θ ) n = r n cis ( n θ ) (r\operatorname{cis}\theta)^n=r^n\operatorname{cis}(n\theta) ( r cis θ ) n = r n cis ( n θ ) . For z n = r cis θ z^n=r\operatorname{cis}\theta z n = r cis θ , find all n n n roots using arguments ( θ + 2 k π ) / n (\theta+2k\pi)/n ( θ + 2 k π ) / n , for k = 0 , … , n − 1 k=0,\ldots,n-1 k = 0 , … , n − 1 . Worked example For z 3 = 8 z^3=8 z 3 = 8 , the roots have modulus 2 2 2 and arguments 0 , 2 π / 3 , 4 π / 3 0,2\pi/3,4\pi/3 0 , 2 π /3 , 4 π /3 . They are 2 2 2 , − 1 + i 3 -1+i\sqrt3 − 1 + i 3 and − 1 − i 3 -1-i\sqrt3 − 1 − i 3 . A common mistake An inverse tangent alone does not identify the correct quadrant. Check the real and imaginary components before assigning the argument.
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Q2 · Practice question Simple familiar 1 mark
QUESTION 2 Which of the following is a root of z 3 = − 8 z^3=-8 z 3 = − 8
WORKED SOLUTION
Answer C 1 mark Sample response and mark allocation
− 8 = 8 cis ( π ) -8=8\operatorname{cis}(\pi) − 8 = 8 cis ( π ) , so the roots have modulus 2 and arguments
π / 3 , π , 5 π / 3 \pi/3,\pi,5\pi/3 π /3 , π , 5 π /3 . The first-quadrant root is
2 cis ( π / 3 ) = 1 + i 3 2\operatorname{cis}(\pi/3)=1+i\sqrt3 2 cis ( π /3 ) = 1 + i 3 .
Selects the correct option. [1 mark]
Practice question aligned to the current QCAA syllabus; review the worked solution and marking criteria.
View the QCAA syllabus Q7 · Practice question Complex familiar 5 marks
QUESTION 7 (5 marks) The complex number z z z
z 4 + 4 = 0. z^4+4=0. z 4 + 4 = 0. a) Determine the four values of z z z
[3 marks] b) Hence factorise z 4 + 4 z^4+4 z 4 + 4
[2 marks] WORKED SOLUTION
Practice marking scheme 5 marks ANSWER z = 1 + i , − 1 + i , − 1 − i , 1 − i ; z 4 + 4 = ( z 2 − 2 z + 2 ) ( z 2 + 2 z + 2 ) z=1+i,-1+i,-1-i,1-i;\quad z^4+4=(z^2-2z+2)(z^2+2z+2) z = 1 + i , − 1 + i , − 1 − i , 1 − i ; z 4 + 4 = ( z 2 − 2 z + 2 ) ( z 2 + 2 z + 2 ) Sample response and mark allocation
z 4 = − 4 = 4 cis ( π + 2 k π ) z^4=-4=4\operatorname{cis}(\pi+2k\pi) z 4 = − 4 = 4 cis ( π + 2 k π ) .
Expresses the equation in polar form. [1 mark]
z = 2 cis ( π / 4 + k π / 2 ) , k = 0 , 1 , 2 , 3 z=\sqrt2\operatorname{cis}(\pi/4+k\pi/2),\ k=0,1,2,3 z = 2 cis ( π /4 + k π /2 ) , k = 0 , 1 , 2 , 3 .
Determines the four polar-form roots. [1 mark]
z = 1 + i , − 1 + i , − 1 − i , 1 − i z=1+i,-1+i,-1-i,1-i z = 1 + i , − 1 + i , − 1 − i , 1 − i .
Expresses all roots in Cartesian form. [1 mark]
( z − ( 1 + i ) ) ( z − ( 1 − i ) ) = z 2 − 2 z + 2 (z-(1+i))(z-(1-i))=z^2-2z+2 ( z − ( 1 + i )) ( z − ( 1 − i )) = z 2 − 2 z + 2 and
( z − ( − 1 + i ) ) ( z − ( − 1 − i ) ) = z 2 + 2 z + 2 (z-(-1+i))(z-(-1-i))=z^2+2z+2 ( z − ( − 1 + i )) ( z − ( − 1 − i )) = z 2 + 2 z + 2 .
Groups conjugate roots into real quadratic factors. [1 mark]
z 4 + 4 = ( z 2 − 2 z + 2 ) ( z 2 + 2 z + 2 ) z^4+4=(z^2-2z+2)(z^2+2z+2) z 4 + 4 = ( z 2 − 2 z + 2 ) ( z 2 + 2 z + 2 ) .
States the required factorisation. [1 mark]
Practice question aligned to the current QCAA syllabus; review the worked solution and marking criteria.
View the QCAA syllabus How many marks did you earn? Compare your working with the guide above.
Q12 · Practice question Complex familiar 6 marks
QUESTION 12 (6 marks) Let z = − 3 + 3 3 i z=-3+3\sqrt3\,i z = − 3 + 3 3 i . a) Express z z z in the form r cis θ r\operatorname{cis}\theta r cis θ , where 0 ≤ θ < 2 π 0\le\theta<2\pi 0 ≤ θ < 2 π . [2 marks] b) Determine z 4 z^4 z 4 in Cartesian form. [2 marks] c) Determine all solutions of w 3 = z w^3=z w 3 = z in polar form. [2 marks] WORKED SOLUTION
Practice marking scheme 6 marks ANSWER (a) 6 cis ( 2 π / 3 ) 6\operatorname{cis}(2\pi/3) 6 cis ( 2 π /3 ) ; (b) − 648 + 648 3 i -648+648\sqrt3\,i − 648 + 648 3 i ; (c) 6 3 cis ( 2 π / 9 + 2 k π / 3 ) \sqrt[3]6\operatorname{cis}(2\pi/9+2k\pi/3) 3 6 cis ( 2 π /9 + 2 k π /3 ) , k = 0 , 1 , 2 k=0,1,2 k = 0 , 1 , 2 . Worked solution
(a) ∣ z ∣ = 9 + 27 = 6 |z|=\sqrt{9+27}=6 ∣ z ∣ = 9 + 27 = 6 and z z z is in quadrant II with reference angle π / 3 \pi/3 π /3 . Thus z = 6 cis ( 2 π / 3 ) z=6\operatorname{cis}(2\pi/3) z = 6 cis ( 2 π /3 ) . (b) By De Moivre’s theorem, z 4 = 6 4 cis ( 8 π / 3 ) = 1296 cis ( 2 π / 3 ) = − 648 + 648 3 i . z^4=6^4\operatorname{cis}(8\pi/3)=1296\operatorname{cis}(2\pi/3)=-648+648\sqrt3\,i. z 4 = 6 4 cis ( 8 π /3 ) = 1296 cis ( 2 π /3 ) = − 648 + 648 3 i . (c) The cube roots have modulus 6 3 \sqrt[3]6 3 6 and arguments 2 π / 3 + 2 k π 3 = 2 π 9 + 2 k π 3 , k = 0 , 1 , 2. \frac{2\pi/3+2k\pi}{3}=\frac{2\pi}{9}+\frac{2k\pi}{3},\qquad k=0,1,2. 3 2 π /3 + 2 k π = 9 2 π + 3 2 k π , k = 0 , 1 , 2. Therefore the roots are 6 3 cis ( 2 π / 9 ) \sqrt[3]6\operatorname{cis}(2\pi/9) 3 6 cis ( 2 π /9 ) , 6 3 cis ( 8 π / 9 ) \sqrt[3]6\operatorname{cis}(8\pi/9) 3 6 cis ( 8 π /9 ) and 6 3 cis ( 14 π / 9 ) \sqrt[3]6\operatorname{cis}(14\pi/9) 3 6 cis ( 14 π /9 ) . Correct modulus and argument.
[2 marks] Correct use of De Moivre’s theorem and Cartesian result.
[2 marks] All three cube roots with correct moduli and arguments.
[2 marks] Practice question aligned to the current QCAA syllabus; review the worked solution and marking criteria.
View the QCAA syllabus How many marks did you earn? Compare your working with the guide above.
All complex numbers practice questions 42 original questions · Page 1 of 2
Q2 · Original practice · 1 mark Original Specialist Mathematics practice question 2: roots of complex numbers
Complex numbers Q7 · Original practice · 5 marks Original Specialist Mathematics practice question 7: complex roots and factorisation
Complex numbers Q11 · Original practice · 1 mark Let z 1 = 2 cis ( π / 3 ) z_1=2\operatorname{cis}(\pi/3) z 1 = 2 cis ( π /3 ) and z 2 = 3 cis ( − π / 6 ) z_2=3\operatorname{cis}(-\pi/6) z 2 = 3 cis ( − π /6 ) . The product z 1 z 2 z_1z_2 z 1 z 2 is Complex numbers Q12 · Original practice · 6 marks Let z = − 3 + 3 3 i z=-3+3\sqrt3\,i z = − 3 + 3 3 i . Complex numbers Q13 · Original practice · 1 mark The complex number z = 1 − i 3 1 + i z=\dfrac{1-i\sqrt3}{1+i} z = 1 + i 1 − i 3 can be written in polar form as Complex numbers Q14 · Original practice · 5 marks The polynomial P ( z ) = z 4 − 4 z 3 + 6 z 2 − 4 z − 15 P(z)=z^4-4z^3+6z^2-4z-15 P ( z ) = z 4 − 4 z 3 + 6 z 2 − 4 z − 15 has real coefficients. Given that 1 + 2 i 1+2i 1 + 2 i is a root, factorise P ( z ) P(z) P ( z ) completely over C \mathbb C C . Complex numbers Q15 · Original practice · 1 mark The distance in the Argand plane between two adjacent sixth roots of unity is
Complex numbers Q16 · Original practice · 6 marks Let ω \omega ω be a non-real cube root of unity. Complex numbers Q17 · Original practice · 1 mark A monic cubic polynomial with real coefficients has roots 2 + i 2+i 2 + i and − 1 -1 − 1 . Its constant term is Complex numbers Q19 · Original practice · 1 mark Which of the following is a solution of z 3 = − 8 i z^3=-8i z 3 = − 8 i ? Complex numbers Q62 · Original practice · 6 marks The equation z 5 = 10 + 10 i z^5=10+10i z 5 = 10 + 10 i has five roots. Complex numbers Q111 · Original practice · 1 mark A non-real root of z 4 = 16 z^{4}=16 z 4 = 16 with positive imaginary part is Complex numbers Q118 · Original practice · 1 mark If z = 3 cis ( π 4 ) z=3\operatorname{cis}(\frac{\pi}{4}) z = 3 cis ( 4 π ) and w = 2 cis ( π 6 ) w=2\operatorname{cis}(\frac{\pi}{6}) w = 2 cis ( 6 π ) , then z w zw z w is Complex numbers Q119 · Original practice · 1 mark For z = x + i y , ∣ z − 2 ∣ = ∣ z + 2 ∣ z=x+iy, |z-2|=|z+2| z = x + i y , ∣ z − 2∣ = ∣ z + 2∣ describes Complex numbers Q121 · Original practice · 5 marks Let z = 1 − 3 i z=1-\sqrt{3}i z = 1 − 3 i and w = 2 + 2 i w=2+2i w = 2 + 2 i . Complex numbers Q129 · Original practice · 6 marks Points z = x + i y z=x+iy z = x + i y satisfy ∣ z ∣ = 5 |z|=5 ∣ z ∣ = 5 and are equidistant from 1 + 2 i 1+2i 1 + 2 i and 5 + 2 i 5+2i 5 + 2 i . Complex numbers Q134 · Original practice · 1 mark The polar point 4 cis ( π 2 ) 4\operatorname{cis}(\frac{\pi}{2}) 4 cis ( 2 π ) is Complex numbers Q137 · Original practice · 1 mark Under w = ( 2 − i ) z , z = 1 + i w=(2-i)z, z=1+i w = ( 2 − i ) z , z = 1 + i maps to Complex numbers Q144 · Original practice · 6 marks Let z z z satisfy z 3 = − 8 i z^{3}=-8i z 3 = − 8 i . Complex numbers Q148 · Original practice · 1 mark The four roots of z 4 = − 16 z^4=-16 z 4 = − 16 are shown on the Argand plane. For each root, a new complex number is defined by w = z + 4 z . w=z+\frac{4}{z}. w = z + z 4 . Which statement describes the set of distinct values of w w w ? Further complex numbers Q153 · Original practice · 1 mark The points A A A , B B B and C C C represent the three roots of z 3 = − 8 i z^3=-8i z 3 = − 8 i on the Argand plane. The point P P P represents z = 2 z=2 z = 2 .
What is the exact value of P A × P B × P C PA\times PB\times PC P A × P B × P C ? Further complex numbers Q173 · Original practice · 5 marks Four beacons in a light installation have complex coordinates given by the solutions of ( z − 1 − i ) 4 = 16 (z-1-i)^4=16 ( z − 1 − i ) 4 = 16 . They are represented on the Argand plane. The dashed circle has centre C = 1 + i C=1+i C = 1 + i . Further complex numbers Q174 · Original practice · 8 marks A nonzero complex number satisfies ∣ z ∣ = 1 |z|=1 ∣ z ∣ = 1 . Define w = z + 1 / z w=z+1/z w = z + 1/ z . Further complex numbers Q175 · Original practice · 1 mark The polynomial p ( z ) = z 2 − 2 i z − 2 p(z)=z^2-2iz-2 p ( z ) = z 2 − 2 i z − 2 has zero 1 + i 1+i 1 + i . Which is its other zero? Further complex numbers