One of the roots of the equation
2z4−14z3+33z2−26z+p=0, where p is a constant2z^4 - 14z^3 + 33z^2 - 26z + p = 0\text{, where } p \text{ is a constant}
is 3+i3 + \mathrm{i}.

  1. (a)

    Based on the above information only, a student claims that the equation has a root 3−i3 - \mathrm{i}. State, with a reason, why the student's claim may not be true.

  2. (b)

    Show that p=10p = 10.

  3. (c)

    For the rest of this question, do not use a calculator.

    Find the roots of the equation 2z4−14z3+33z2−26z+10=02z^4 - 14z^3 + 33z^2 - 26z + 10 = 0 and mark them clearly on a single labelled Argand diagram.

  4. (d)

    The points of the Argand diagram in part (c) form the vertices of a quadrilateral. Identify the type of quadrilateral and determine its area.

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Worked solution

4(a)

1. State the condition required for roots to occur in complex conjugate pairs and identify why it is not met here.

For a polynomial equation to necessarily have complex roots in conjugate pairs, all of its coefficients must be real numbers. The given information only states that pp is a constant, but does not specify that pp is a real number. Therefore, the coefficients may not all be real, and the complex roots might not occur in conjugate pairs.

Answer

The theorem that complex roots occur in conjugate pairs applies only to equations with real coefficients. Since pp is only stated to be a constant and not specifically a real constant, the coefficients might not all be real.

4(b)

1. Calculate the powers of $z = 3 + \mathrm{i}$.

z=3+iz = 3 + \mathrm{i}
z2=(3+i)2=9+6i+i2=8+6iz^2 = (3 + \mathrm{i})^2 = 9 + 6\mathrm{i} + \mathrm{i}^2 = 8 + 6\mathrm{i}
z3=z⋅z2=(3+i)(8+6i)=24+18i+8i+6i2=18+26iz^3 = z \cdot z^2 = (3 + \mathrm{i})(8 + 6\mathrm{i}) = 24 + 18\mathrm{i} + 8\mathrm{i} + 6\mathrm{i}^2 = 18 + 26\mathrm{i}
z4=z2⋅z2=(8+6i)2=64+96i+36i2=28+96iz^4 = z^2 \cdot z^2 = (8 + 6\mathrm{i})^2 = 64 + 96\mathrm{i} + 36\mathrm{i}^2 = 28 + 96\mathrm{i}

2. Substitute $z$ and its powers into the polynomial equation.

2(28+96i)−14(18+26i)+33(8+6i)−26(3+i)+p=02(28 + 96\mathrm{i}) - 14(18 + 26\mathrm{i}) + 33(8 + 6\mathrm{i}) - 26(3 + \mathrm{i}) + p = 0
56+192i−252−364i+264+198i−78−26i+p=056 + 192\mathrm{i} - 252 - 364\mathrm{i} + 264 + 198\mathrm{i} - 78 - 26\mathrm{i} + p = 0

3. Group real and imaginary parts to solve for $p$.

(56−252+264−78+p)+(192−364+198−26)i=0(56 - 252 + 264 - 78 + p) + (192 - 364 + 198 - 26)\mathrm{i} = 0
(−10+p)+0i=0(-10 + p) + 0\mathrm{i} = 0
p=10p = 10 (shown)

Answer

p = 10

4(c)

1. State the complex conjugate root due to the coefficients being real.

Since the coefficients of the polynomial equation 2z4−14z3+33z2−26z+10=02z^4 - 14z^3 + 33z^2 - 26z + 10 = 0 are all real numbers, the complex roots must occur in conjugate pairs.
Therefore, the complex conjugate of 3+i3 + \mathrm{i}, which is z=3−iz = 3 - \mathrm{i}, is also a root.

2. Find the quadratic factor associated with these two known roots.

The quadratic factor corresponding to the roots 3±i3 \pm \mathrm{i} is:
(z−(3+i))(z−(3−i))=(z−3)2−(i)2(z - (3 + \mathrm{i}))(z - (3 - \mathrm{i})) = (z - 3)^2 - (\mathrm{i})^2
=(z2−6z+9)−(−1)= (z^2 - 6z + 9) - (-1)
=z2−6z+10= z^2 - 6z + 10

3. Determine the remaining quadratic factor by comparing coefficients.

Let the polynomial be expressed as the product of two quadratic factors:
2z4−14z3+33z2−26z+10=(z2−6z+10)(Az2+Bz+C)2z^4 - 14z^3 + 33z^2 - 26z + 10 = (z^2 - 6z + 10)(Az^2 + Bz + C)
By comparing the z4z^4 coefficient: A=2A = 2.
By comparing the constant term: 10C=10  ⟹  C=110C = 10 \implies C = 1.
So the second factor is (2z2+Bz+1)(2z^2 + Bz + 1).
Comparing the z3z^3 coefficient:
B−6(2)=−14  ⟹  B−12=−14  ⟹  B=−2B - 6(2) = -14 \implies B - 12 = -14 \implies B = -2
Thus, the second quadratic factor is 2z2−2z+12z^2 - 2z + 1.

4. Solve the second quadratic equation for the remaining roots.

Set the second factor to zero:
2z2−2z+1=02z^2 - 2z + 1 = 0
Using the quadratic formula without a calculator:
z=−(−2)±(−2)2−4(2)(1)2(2)z = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(2)(1)}}{2(2)}
z=2±4−84z = \frac{2 \pm \sqrt{4 - 8}}{4}
z=2±−44=2±2i4z = \frac{2 \pm \sqrt{-4}}{4} = \frac{2 \pm 2\mathrm{i}}{4}
z=12±12iz = \frac{1}{2} \pm \frac{1}{2}\mathrm{i}

5. Draw the Argand diagram plotting all four roots.

The four roots are z1=3+iz_1 = 3 + \mathrm{i}, z2=3−iz_2 = 3 - \mathrm{i}, z3=12+12iz_3 = \frac{1}{2} + \frac{1}{2}\mathrm{i}, and z4=12−12iz_4 = \frac{1}{2} - \frac{1}{2}\mathrm{i}.
On an Argand diagram, the horizontal axis is labelled Re(z)\text{Re}(z) and the vertical axis is labelled Im(z)\text{Im}(z).
The roots are plotted as points at coordinates (3,1)(3, 1), (3,−1)(3, -1), (0.5,0.5)(0.5, 0.5), and (0.5,−0.5)(0.5, -0.5).

Answer

The roots are 3+i3 + \mathrm{i}, 3−i3 - \mathrm{i}, 12+12i\frac{1}{2} + \frac{1}{2}\mathrm{i}, and 12−12i\frac{1}{2} - \frac{1}{2}\mathrm{i}.

[On an Argand diagram with axes labelled Re(z)\text{Re}(z) and Im(z)\text{Im}(z), four points should be marked clearly at coordinates (3,1)(3, 1), (3,−1)(3, -1), (0.5,0.5)(0.5, 0.5), and (0.5,−0.5)(0.5, -0.5).]

4(d)

1. State the coordinates of the roots on the Argand diagram and deduce the type of quadrilateral.

From part (c), the roots are 3±i3 \pm \mathrm{i} and 12±12i\frac{1}{2} \pm \frac{1}{2}\mathrm{i}.
The points on the Argand diagram are (3,1)(3, 1), (3,−1)(3, -1), (12,12)(\frac{1}{2}, \frac{1}{2}) and (12,−12)(\frac{1}{2}, -\frac{1}{2}).
Since the line segments connecting (3,1)(3, 1) to (3,−1)(3, -1) and (12,12)(\frac{1}{2}, \frac{1}{2}) to (12,−12)(\frac{1}{2}, -\frac{1}{2}) are both parallel to the imaginary axis, the quadrilateral has one pair of parallel sides.
Because the points are conjugate pairs, it is symmetrical about the real axis. Thus, it is an isosceles trapezium.

2. Calculate the area of the trapezium.

The lengths of the parallel sides are:
1−(−1)=21 - (-1) = 2
12−(−12)=1\frac{1}{2} - (-\frac{1}{2}) = 1

The perpendicular distance (height) between the parallel sides is:
3−12=523 - \frac{1}{2} = \frac{5}{2}

Area=12×(sum of parallel sides)×height\text{Area} = \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}
Area=12×(2+1)×52=154=3.75\text{Area} = \frac{1}{2} \times (2 + 1) \times \frac{5}{2} = \frac{15}{4} = 3.75

Answer

Isosceles trapezium, Area = 3.753.75

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