Question
(a) Show that the complex number i is a root of the equation\[{x^4} – 5{x^3} + 7{x^2} – 5x + 6 = 0{\text{ }}.\](b) Find the other roots of this equation.
▶️Answer/Explanation
Markscheme
(a) \({{\text{i}}^4} – 5{{\text{i}}^3} + 7{{\text{i}}^2} – 5{\text{i}} + 6 = 1 + 5{\text{i}} – 7 – 5{\text{i}} + 6\) M1A1
\( = 0\) AG N0
(b) \({\text{i}}\) root \( \Rightarrow \) \( – {\text{i}}\) is second root (M1)A1
moreover, \({x^4} – 5{x^3} + 7{x^2} – 5x + 6 = \left( {x – {\text{i}}} \right)\left( {x + {\text{i}}} \right)q(x)\)
where \(q(x) = {x^2} – 5x + 6\)
finding roots of \(q(x)\)
the other two roots are \(2\) and \(3\) A1A1
Note: Final A1A1 is independent of previous work.
[6 marks]
Examiners report
A surprising number of candidates solved the question by dividing the expression by \(1 – i\) rather than substituting \(l\) into the expression. Many students were not aware that complex roots occur in conjugate pairs, and many did not appreciate the difference between a factor and a root. Generally the question was well done.
Question
Consider the polynomial \(p(x) = {x^4} + a{x^3} + b{x^2} + cx + d\), where a, b, c, d \( \in \mathbb{R}\).
Given that 1 + i and 1 − 2i are zeros of \(p(x)\), find the values of a, b, c and d.
▶️Answer/Explanation
Markscheme
METHOD 1
1 + i is a zero \( \Rightarrow \) 1 – i is a zero (A1)
1 – 2i is a zero \( \Rightarrow \) 1 + 2i is a zero (A1)
\(\left( {x – (1 – {\text{i}})} \right)\left( {x – (1 + {\text{i}})} \right) = ({x^2} – 2x + 2)\) (M1)A1
\(\left( {x – (1 – 2{\text{i}})} \right)\left( {x – (1 + 2{\text{i}})} \right) = ({x^2} – 2x + 5)\) A1
\(p(x) = ({x^2} – 2x + 2)({x^2} – 2x + 5)\) M1
\( = {x^4} – 4{x^3} + 11{x^2} – 14x + 10\) A1
\(a = – 4,{\text{ }}b = 11,{\text{ }}c = – 14,{\text{ }}d = 10\)
[7 marks]
METHOD 2
\(p(1 + {\text{i}}) = – 4 + ( – 2 + 2{\text{i}})a + (2{\text{i}})b + (1 + {\text{i}})c + d\) M1
\(p(1 + {\text{i}}) = 0 \Rightarrow \left\{ {\begin{array}{*{20}{c}}
{ – 4 – 2a + c + d = 0} \\
{2a + 2b + c = 0}
\end{array}} \right.\) M1A1A1
\(p(1 – 2{\text{i}}) = – 7 + 24{\text{i}} + ( – 11 + 2{\text{i}})a + ( – 3 – 4{\text{i}})b + (1 – 2{\text{i}})c + d\)
\(p(1 – 2{\text{i}}) = 0 \Rightarrow \left\{ {\begin{array}{*{20}{c}}
{ – 7 – 11a – 3b + c + d = 0} \\
{24 + 2a – 4b – 2c = 0}
\end{array}} \right.\) A1
\(\left( {\begin{array}{*{20}{c}}
a \\
b \\
c \\
d
\end{array}} \right) = {\left( {\begin{array}{*{20}{c}}
{ – 2}&0&1&1 \\
2&2&1&0 \\
{ – 11}&{ – 3}&1&1 \\
2&{ – 4}&{ – 2}&0
\end{array}} \right)^{ – 1}}\left( {\begin{array}{*{20}{c}}
4 \\
0 \\
7 \\
{ – 24}
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
{ – 4} \\
{11} \\
{ – 14} \\
{10}
\end{array}} \right)\) M1A1
\(a = – 4,{\text{ }}b = 11,{\text{ }}c = – 14,{\text{ }}d = 10\)
[7 marks]
Examiners report
Most candidates attempted this question, using different approaches. The most successful approach was the method of complex conjugates and the product of linear factors. Candidates who used this method were in general successful whereas candidates who attempted direct substitution and separation of real and imaginary parts to obtain four equations in four unknowns were less successful because either they left the work incomplete or made algebraic errors that led to incorrect answers.
Question
One root of the equation \({x^2} + ax + b = 0\) is \(2 + 3{\text{i}}\) where \(a,{\text{ }}b \in \mathbb{R}\). Find the value of \(a\) and the value of \(b\).
▶️Answer/Explanation
Markscheme
METHOD 1
substituting
\( – 5 + 12{\text{i}} + a(2 + 3{\text{i}}) + b = 0\) (A1)
equating real or imaginary parts (M1)
\(12 + 3a = 0 \Rightarrow a = – 4\) A1
\( – 5 + 2a + b = 0 \Rightarrow b = 13\) A1
METHOD 2
other root is \(2 – 3{\text{i}}\) (A1)
considering either the sum or product of roots or multiplying factors (M1)
\(4 = – a\) (sum of roots) so \(a = – 4\) A1
\(13 = b\) (product of roots) A1
[4 marks]
Examiners report
Question
The complex numbers \(u\) and \(v\) are represented by point A and point B respectively on an Argand diagram.
Point A is rotated through \(\frac{\pi }{2}\) in the anticlockwise direction about the origin O to become point \({\text{A}}’\). Point B is rotated through \(\frac{\pi }{2}\) in the clockwise direction about O to become point \({\text{B}}’\).
a.Consider \(z = r(\cos \theta + {\text{i}}\sin \theta ),{\text{ }}z \in \mathbb{C}\).
Use mathematical induction to prove that \({z^n} = {r^n}(\cos n\theta + {\text{i}}\sin n\theta ),{\text{ }}n \in {\mathbb{Z}^ + }\).[7]
b.Given \(u = 1 + \sqrt 3 {\text{i}}\) and \(v = 1 – {\text{i}}\),
(i) express \(u\) and \(v\) in modulus-argument form;
(ii) hence find \({u^3}{v^4}\).[4]
c.Plot point A and point B on the Argand diagram.[1]
d.Find the area of triangle O\({\text{A}}’\)\({\text{B}}’\).[3]
e.Given that \(u\) and \(v\) are roots of the equation \({z^4} + b{z^3} + c{z^2} + dz + e = 0\), where \(b,{\text{ }}c,{\text{ }}d,{\text{ }}e \in \mathbb{R}\),
find the values of \(b,{\text{ }}c,{\text{ }}d\) and \(e\).[5]
▶️Answer/Explanation
Markscheme
let \({\text{P}}(n)\) be the proposition \({z^n} = {r^n}(\cos n\theta + {\rm{i}}\sin n\theta ),n \in {¢^ + }\)
let \(n = 1 \Rightarrow \)
\({\text{LHS}} = r(\cos \theta + {\text{i}}\sin \theta )\)
\({\text{RHS}} = r(\cos \theta + {\text{i}}\sin \theta ),{\text{ }}\therefore {\text{P}}(1)\) is true R1
assume true for \(n = k \Rightarrow {r^k}{(\cos \theta + {\text{i}}\sin \theta )^k} = {r^k}\left( {\cos (k\theta ) + {\text{i}}\sin (k\theta )} \right)\) M1
Note: Only award the M1 if truth is assumed.
now show \(n = k\) true implies \(n = k + 1\) also true
\({r^{k + 1}}{(\cos \theta + {\text{i}}\sin \theta )^{k + 1}} = {r^{k + 1}}{(\cos \theta + {\text{i}}\sin \theta )^k}(\cos \theta + {\text{i}}\sin \theta )\) M1
\( = {r^{k + 1}}\left( {\cos (k\theta ) + {\text{i}}\sin (k\theta )} \right)(\cos \theta + {\text{i}}\sin \theta )\)
\( = {r^{k + 1}}\left( {\cos (k\theta )\cos \theta – \sin (k\theta )\sin \theta + {\text{i}}\left( {\sin (k\theta )\cos \theta + \cos (k\theta )\sin \theta } \right)} \right)\) A1
\( = {r^{k + 1}}\left( {\cos (k\theta + \theta ) + {\text{i}}\sin (k\theta + \theta )} \right)\) A1
\( = {r^{k + 1}}\left( {\cos (k + 1)\theta + {\text{i}}\sin (k + 1)\theta } \right) \Rightarrow n = k + 1\) is true A1
\({\text{P}}(k)\) true implies \({\text{P}}(k + 1)\) true and \({\text{P}}(1)\) is true, therefore by mathematical induction statement is true for \(n \geqslant 1\) R1
Note: Only award the final R1 if the first 4 marks have been awarded.
[7 marks]
(i) \(u = 2{\text{cis}}\left( {\frac{\pi }{3}} \right)\) A1
\(v = \sqrt 2 {\text{cis}}\left( { – \frac{\pi }{4}} \right)\) A1
Notes: Accept 3 sf answers only. Accept equivalent forms.
Accept \(2{e^{\frac{\pi }{3}i}}\) and \(\sqrt 2 {e^{ – \frac{\pi }{4}i}}\).
(ii) \({u^3} = {2^3}{\text{cis}}(\pi ) = – 8\)
\({v^4} = 4{\text{cis}}( – \pi ) = – 4\) (M1)
\({u^3}{v^4} = 32\) A1
Notes: Award (M1) for an attempt to find \({u^3}\) and \({v^4}\).
Accept equivalent forms.
[4 marks]
A1
Note: Award A1 if A or \({\text{1 + }}\sqrt 3 i\) and B or \(1 – i\) are in their correct quadrants, are aligned vertically and it is clear that \(\left| u \right| > \left| v \right|\).
[1 mark]
Area \( = \frac{1}{2} \times 2 \times \sqrt 2 \times \sin \left( {\frac{{5\pi }}{{12}}} \right)\) M1A1
\( = 1.37{\text{ }}\left( { = \frac{{\sqrt 2 }}{4}\left( {\sqrt 6 + \sqrt 2 } \right)} \right)\) A1
Notes: Award M1A0A0 for using \(\frac{{7\pi }}{{12}}\).
[3 marks]
\((z – 1 + {\text{i}})(z – 1 – {\text{i}}) = {z^2} – 2z + 2\) M1A1
Note: Award M1 for recognition that a complex conjugate is also a root.
\(\left( {z – 1 – \sqrt 3 {\text{i}}} \right)\left( {z – 1 + \sqrt 3 {\text{i}}} \right) = {z^2} – 2z + 4\) A1
\(\left( {{z^2} – 2z + 2} \right)\left( {{z^2} – 2z + 4} \right) = {z^4} – 4{z^3} + 10{z^2} – 12z + 8\) M1A1
Note: Award M1 for an attempt to expand two quadratics.
[5 marks]