IB Mathematics AHL 1.12 Complex numbers AA HL Paper 3
Question
This question asks you to explore cubic polynomials of the form $(x-r)\left(x^2-2 a x+a^2+b^2\right)$ for $x \in \mathbb{R}$ and corresponding cubic equations with one real root and two complex roots of the form $(z-r)\left(z^2-2 a z+a^2+b^2\right)=0$ for $z \in \mathrm{C}$.
In parts (a), (b) and (c), let $r=1, a=4$ and $b=1$.
Consider the equation $(z-1)\left(z^2-8 z+17\right)=0$ for $z \in \mathbb{C}$.
Consider the function $f(x)=(x-1)\left(x^2-8 x+17\right)$ for $x \in \mathbb{R}$.
Consider the function $g(x)=(x-r)\left(x^2-2 a x+a^2+b^2\right)$ for $x \in \mathbb{R}$ where $r, a \in \mathbb{R}$ and $b \in \mathbb{R}, b>0$.
The equation $(z-r)\left(z^2-2 a z+a^2+b^2\right)=0$ for $z \in \mathbb{C}$ has roots $r$ and $a \pm b$ i where $r, a \in \mathbb{R}$ and $b \in \mathbb{R}, b>0$.
On the Cartesian plane, the points $\mathrm{C}_1\left(a, \sqrt{g^{\prime}(a)}\right)$ and $\mathrm{C}_2\left(a,-\sqrt{g^{\prime}(a)}\right)$ represent the real and imaginary parts of the complex roots of the equation $(z-r)\left(z^2-2 a z+a^2+b^2\right)=0$
The following diagram shows a particular curve of the form $y=(x-r)\left(x^2-2 a x+a^2+16\right)$ and the tangent to the curve at the point $\mathrm{A}(a, 80)$. The curve and the tangent both intersect the $x$-axis at the point $\mathrm{R}(-2,0)$. The points $\mathrm{C}_1$ and $\mathrm{C}_2$ are also shown.
Consider the curve $y=(x-r)\left(x^2-2 a x+a^2+b^2\right)$ for $a \neq r, b>0$. The points $\mathrm{A}(a, g(a))$ and $\mathrm{R}(r, 0)$ are as defined in part (d)(ii). The curve has a point of inflexion at point $P$.
Consider the special case where $a=r$ and $b>0$.
a.i. Given that 1 and $4+\mathrm{i}$ are roots of the equation, write down the third root.
a.ii.Verify that the mean of the two complex roots is 4 .
b. Show that the line $y=x-1$ is tangent to the curve $y=f(x)$ at the point $\mathrm{A}(4,3)$.
[1]
[1]
c. Sketch the curve $y=f(x)$ and the tangent to the curve at point A, clearly showing where the tangent crosses the $x$-axis.
d.i. Show that $g^{\prime}(x)=2(x-r)(x-a)+x^2-2 a x+a^2+b^2$.
[2]
d.ii.Hence, or otherwise, prove that the tangent to the curve $y=g(x)$ at the point $\mathrm{A}(a, g(a))$ intersects the $x$-axis at the point $\mathrm{R}(r, 0)$.
e. Deduce from part (d)(i) that the complex roots of the equation $(z-r)\left(z^2-2 a z+a^2+b^2\right)=0$ can be expressed as $a \pm \mathrm{i} \sqrt{g^{\prime}(a)}$.
f.i. Use this diagram to determine the roots of the corresponding equation of the form $(z-r)\left(z^2-2 a z+a^2+16\right)=0$ for $z \in \mathbb{C}$.
f.ii. State the coordinates of $\mathrm{C}_2$.
g.i. Show that the $x$-coordinate of $\mathrm{P}$ is $\frac{1}{3}(2 a+r)$.
You are not required to demonstrate a change in concavity.
g.iiHence describe numerically the horizontal position of point $\mathrm{P}$ relative to the horizontal positions of the points $\mathrm{R}$ and $\mathrm{A}$.
[1]
h.i. Sketch the curve $y=(x-r)\left(x^2-2 a x+a^2+b^2\right)$ for $a=r=1$ and $b=2$.
h.ii.For $a=r$ and $b>0$, state in terms of $r$, the coordinates of points P and A.
[1]
▶️Answer/Explanation
Markscheme
a.i. 4 – i
A1
[1 mark]
a.iimean $=\frac{1}{2}(4+\mathrm{i}+4-\mathrm{i}) \quad \boldsymbol{A 1}$
$=4 \quad \boldsymbol{A G}$
[1 mark]
b. METHOD 1
attempts product rule differentiation
(M1)
Note: Award (M1) for attempting to express $f(x)$ as $f(x)=x^3-9 x^2+25 x-17$
$f(x)=(x-1)(2 x-8)+x^2-8 x+17\left(f(x)=3 x^2-18 x+25\right) \quad$ A1
$f(4)=1$
A1
Note: Where $f(x)$ is correct, award $\boldsymbol{A} 1$ for solving $f(x)=1$ and obtaining $x=4$.
EITHER
$$
y-3=1(x-4) \quad \text { A1 }
$$
OR
$$
\begin{aligned}
& y=x+c \\
& 3=4+c \Rightarrow c=-1
\end{aligned}
$$
A1
OR
states the gradient of $y=x-1$ is also 1 and verifies that $(4,3)$ lies on the line $y=x-1$
A1
THEN
so $y=x-1$ is the tangent to the curve at $\mathrm{A}(4,3) \quad \boldsymbol{A G}$
Note: Award a maximum of (M0)AOA1A1 to a candidate who does not attempt to find $f(x)$.
METHOD 2
sets $f(x)=x-1$ to form $x-1=(x-1)\left(x^2-8 x+17\right)$
(M1)
EITHER
$(x-1)\left(x^2-8 x+16\right)=0\left(x^3-9 x^2+24 x-16=0\right)$
attempts to solve a correct cubic equation
$$
(x-1)(x-4)^2=0 \Rightarrow x=1,4
$$
OR
recognises that $x \neq 1$ and forms $x^2-8 x+17=1\left(x^2-8 x+16=0\right)$
attempts to solve a correct quadratic equation
$$
(x-4)^2=0 \Rightarrow x=4
$$
THEN
$x=4$ is a double root
R1
A1
(M1)
(M1)
A1
so $y=x-1$ is the tangent to the curve at $\mathrm{A}(4,3) \quad \boldsymbol{A G}$
Note: Candidates using this method are not required to verify that $y=3$.
[4 marks]
c.
a positive cubic with an $x$-intercept $(x=1)$, and a local maximum and local minimum in the first quadrant both positioned to the left of A
$A 1$
Note: As the local minimum and point A are very close to each other, condone graphs that seem to show these points coinciding. For the point of tangency, accept labels such as A, $(4,3)$ or the point labelled from both axes. Coordinates are not required.
a correct sketch of the tangent passing through A and crossing the $x$-axis at the same point $(x=1)$ as the curve
A1
Note: Award $\boldsymbol{A 1 A O}$ if both graphs cross the $x$-axis at distinctly different points.
[2 marks]
d.i.EITHER
$$
g^{\prime}(x)=(x-r)(2 x-2 a)+x^2-2 a x+a^2+b^2
$$
(M1)A1
OR
$$
g(x)=x^3-(2 a+r) x^2+\left(a^2+b^2+2 a r\right) x-\left(a^2+b^2\right) r
$$
attempts to find $g^{\prime}(x) \quad$ M1
$$
\begin{aligned}
& g^{\prime}(x)=3 x^2-2(2 a+r) x+a^2+b^2+2 a r \\
& =2 x^2-2(a+r) x+2 a r+x^2-2 a x+a^2+b^2 \\
& \left(=2\left(x^2-a x-r x+a r\right)+x^2-2 a x+a^2+b^2\right)
\end{aligned}
$$
A1
THEN
$$
g^{\prime}(x)=2(x-r)(x-a)+x^2-2 a x+a^2+b^2 \quad \text { AG }
$$
[2 marks]
d.iiMETHOD 1
$$
\begin{aligned}
& g(a)=b^2(a-r) \\
& g^{\prime}(a)=b^2
\end{aligned}
$$
(A1)
(A1)
attempts to substitute their $g(a)$ and $g^{\prime}(a)$ into $y-g(a)=g^{\prime}(a)(x-a) \quad$ M1
$$
y-b^2(a-r)=b^2(x-a)
$$
EITHER
$$
\begin{aligned}
& y=b^2(x-r)\left(y=b^2 x-b^2 r\right) \quad \text { A1 } \\
& \text { sets } y=0 \text { so } b^2(x-r)=0 \quad \text { M1 } \\
& b>0 \Rightarrow x=r \text { OR } b \neq 0 \Rightarrow x=r \quad \text { R1 }
\end{aligned}
$$
OR
$$
\begin{array}{ll}
\text { sets } y=0 \text { so }-b^2(a-r)=b^2(x-a) & \text { M1 } \\
b>0 \text { OR } b \neq 0 \Rightarrow-(a-r)=x-a & \text { R1 } \\
x=r \quad \text { A1 } &
\end{array}
$$
THEN
so the tangent intersects the $x$-axis at the point $\mathrm{R}(r, 0) \quad \boldsymbol{A G}$
METHOD 2
$$
\begin{aligned}
& g^{\prime}(a)=b^2 \\
& g(a)=b^2(a-r)
\end{aligned}
$$
attempts to substitute their $g(a)$ and $g^{\prime}(a)$ into $y=g^{\prime}(a) x+c$ and attempts to find $c \quad \boldsymbol{M 1}$
$$
c=-b^2 r
$$
EITHER
$$
\begin{aligned}
& y=b^2(x-r)\left(y=b^2 x-b^2 r\right) \quad \text { A1 } \\
& \text { sets } y=0 \text { so } b^2(x-r)=0 \quad \text { M1 } \\
& b>0 \Rightarrow x=r \text { OR } b \neq 0 \Rightarrow x=r
\end{aligned}
$$
R1
OR
$$
\begin{aligned}
& \text { sets } y=0 \text { so } b^2(x-r)=0 \quad \text { M1 } \\
& b>0 \text { OR } b \neq 0 \Rightarrow x-r=0 \quad \text { R1 } \\
& x=r \quad \text { A1 }
\end{aligned}
$$
METHOD 3
$$
g^{\prime}(a)=b^2
$$
the line through $R(r, 0)$ parallel to the tangent at $\mathrm{A}$ has equation $y=b^2(x-r) \quad \boldsymbol{A 1}$
sets $g(x)=b^2(x-r)$ to form $b^2(x-r)=(x-r)\left(x^2-2 a x+a^2+b^2\right) \quad$ M1
$b^2=x^2-2 a x+a^2+b^2,(x \neq r) \quad \boldsymbol{A} 1$
$(x-a)^2=0$
A1
since there is a double root $(x=a)$, this parallel line through $R(r, 0)$ is the required tangent at $\mathrm{A}$
R1
[6 marks]
e. EITHER
$g^{\prime}(a)=b^2 \Rightarrow b=\sqrt{g^{\prime}(a)}($ since $b>0) \quad$ R1
Note: Accept $b= \pm \sqrt{g^{\prime}(a)}$.
OR
$(a \pm b \mathrm{i}=) a \pm \mathrm{i} \sqrt{b^2}$ and $g^{\prime}(a)=b^2 \quad$ R1
THEN
hence the complex roots can be expressed as $a \pm \mathrm{i} \sqrt{g^{\prime}(a)} \quad \quad \boldsymbol{A G}$
[1 mark]
f.i. $b=4$ (seen anywhere)
A1
EITHER
attempts to find the gradient of the tangent in terms of $a$ and equates to 16
(M1)
OR
substitutes $r=-2, x=a$ and $y=80$ to form $80=(a-(-2))\left(a^2-2 a^2+a^2+16\right)$
(M1)
OR
substitutes $r=-2, x=a$ and $y=80$ into $y=16(x-r)$
(M1)
THEN
$\frac{80}{a+2}=16 \Rightarrow a=3$
roots are -2 (seen anywhere) and $3 \pm 4 \mathrm{i}$
A1A1
Note: Award $\boldsymbol{A} \boldsymbol{1}$ for -2 and $\boldsymbol{A} \boldsymbol{1}$ for $3 \pm 4$ i. Do not accept coordinates.
[4 marks]
f.ii. $(3,-4)$
A1
Note: Accept ” $x=3$ and $y=-4$ “.
Do not award $\boldsymbol{A 1 F T}$ for $(a,-4)$.
[1 mark]
g.i. $g^{\prime}(x)=2(x-r)(x-a)+x^2-2 a x+a^2+b^2$
attempts to find $g^{\prime \prime}(x) \quad M 1$
$g^{\prime \prime}(x)=2(x-a)+2(x-r)+2 x-2 a(=6 x-2 r-4 a)$
sets $g^{\prime \prime}(x)=0$ and correctly solves for $x \quad A 1$
for example, obtaining $x-r+2(x-a)=0$ leading to $3 x=2 a+r$
So $x=\frac{1}{3}(2 a+r) \quad$ AG
Note: Do not award $\boldsymbol{A} \boldsymbol{1}$ if the answer does not lead to the $\boldsymbol{A G}$.
[2 marks]
g.iipoint $\mathrm{P}$ is $\frac{2}{3}$ of the horizontal distance (way) from point $\mathrm{R}$ to point A
A1
Note: Accept equivalent numerical statements or a clearly labelled diagram displaying the numerical relationship. Award $\mathbf{A O}$ for non-numerical statements such as ” $\mathrm{P}$ is between $\mathrm{R}$ and $\mathrm{A}$, closer to $\mathrm{A}$ “.
[1 mark]
h.i. $y=(x-1)\left(x^2-2 x+5\right)$
(A1)
a positive cubic with no stationary points and a non-stationary point of inflexion at $x=1$
A1
Note: Graphs may appear approximately linear. Award this $\boldsymbol{A 1}$ if a change of concavity either side of $x=1$ is apparent. Coordinates are not required and the $y$-intercept need not be indicated.
[2 marks]
h.ii. $(r, 0) \quad \boldsymbol{A 1}$
[1 mark]