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. Identify that the coefficients are not guaranteed to be real and state the conjugate root theorem's condition.

The complex roots of a polynomial equation occur in conjugate pairs only if all the coefficients of the polynomial are real. In the given equation, pp is only stated to be a "constant" and is not explicitly stated to be a real number. Hence, we cannot assume the coefficients are all real.

Answer

The constant pp is not explicitly stated to be a real number. Complex roots are only guaranteed to occur in conjugate pairs if all coefficients of the polynomial are real.

4(b)

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

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 the evaluated powers into the equation and group the real and imaginary parts to show $p = 10$.

Substitute into 2z4−14z3+33z2−26z+p=02z^4 - 14z^3 + 33z^2 - 26z + p = 0:
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

Grouping the real and imaginary parts:
(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

By comparing the real parts:
−10+p=0-10 + p = 0
p=10 (shown)p = 10 \text{ (shown)}

Answer

p = 10 \text{ (shown)}

4(c)

1. Identify the complex conjugate root and form a quadratic factor.

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

The quadratic factor corresponding to these roots is:
(z−(3+i))(z−(3−i))=((z−3)−i)((z−3)+i)(z - (3+\mathrm{i}))(z - (3-\mathrm{i})) = ((z-3) - \mathrm{i})((z-3) + \mathrm{i})
=(z−3)2−i2= (z-3)^2 - \mathrm{i}^2
=z2−6z+9−(−1)= z^2 - 6z + 9 - (-1)
=z2−6z+10= z^2 - 6z + 10

2. Find the other quadratic factor by comparing coefficients.

Let the other quadratic factor be az2+bz+caz^2 + bz + c.
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 coefficient of z4z^4, a=2a = 2.
By comparing the constant terms, 10c=10  ⟹  c=110c = 10 \implies c = 1.
So the factor takes the form 2z2+bz+12z^2 + bz + 1.

Comparing the coefficients of z3z^3:
b+(−6)(2)=−14b + (-6)(2) = -14
b−12=−14  ⟹  b=−2b - 12 = -14 \implies b = -2

The other quadratic factor is 2z2−2z+12z^2 - 2z + 1.

3. Solve the second quadratic factor to find the remaining roots.

Set 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±4−84=2±−44z = \frac{2 \pm \sqrt{4 - 8}}{4} = \frac{2 \pm \sqrt{-4}}{4}
z=2±2i4=12±12iz = \frac{2 \pm 2\mathrm{i}}{4} = \frac{1}{2} \pm \frac{1}{2}\mathrm{i}

4. State all roots and describe their representation on an Argand diagram.

The roots of the equation 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}.

To mark them on a single labelled Argand diagram, construct a coordinate system with the horizontal axis labelled Re\mathrm{Re} and the vertical axis labelled Im\mathrm{Im}. Plot the four points corresponding to the cartesian 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), and clearly label each point with its respective complex root.

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}.

(These four points should be plotted and labelled on an Argand diagram with a real axis, Re\mathrm{Re}, and an imaginary axis, Im\mathrm{Im}.)

4(d)

1. Determine the roots of the equation from part (c) to locate the vertices of the quadrilateral.

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

The quadratic factor is (z−(3+i))(z−(3−i))=(z−3)2+1=z2−6z+10(z - (3+\mathrm{i}))(z - (3-\mathrm{i})) = (z-3)^2 + 1 = z^2 - 6z + 10.

By comparing coefficients, we have:
2z4−14z3+33z2−26z+10=(z2−6z+10)(2z2−2z+1)2z^4 - 14z^3 + 33z^2 - 26z + 10 = (z^2 - 6z + 10)(2z^2 - 2z + 1)

Solving 2z2−2z+1=02z^2 - 2z + 1 = 0:
z=2±(−2)2−4(2)(1)2(2)=2±−44=12±12iz = \frac{2 \pm \sqrt{(-2)^2 - 4(2)(1)}}{2(2)} = \frac{2 \pm \sqrt{-4}}{4} = \frac{1}{2} \pm \frac{1}{2}\mathrm{i}

The roots are 3±i3 \pm \mathrm{i} and 12±12i\frac{1}{2} \pm \frac{1}{2}\mathrm{i}, which correspond to the coordinates (3,1),(3,−1),(0.5,0.5)(3, 1), (3, -1), (0.5, 0.5) and (0.5,−0.5)(0.5, -0.5) on the Argand diagram.

2. Identify the type of quadrilateral formed by these points.

The points (3,1)(3, 1) and (3,−1)(3, -1) form a vertical line segment of length 1−(−1)=21 - (-1) = 2.
The points (0.5,0.5)(0.5, 0.5) and (0.5,−0.5)(0.5, -0.5) form another vertical line segment of length 0.5−(−0.5)=10.5 - (-0.5) = 1.

Since both line segments are parallel to the imaginary axis, they are parallel to each other. Furthermore, the pairs of points are reflections of each other across the real axis, meaning the non-parallel sides are of equal length.

Therefore, the quadrilateral is an isosceles trapezium.

3. Calculate the area of the isosceles trapezium.

The lengths of the parallel sides are a=2a = 2 and b=1b = 1.
The perpendicular (horizontal) distance between these parallel vertical sides is h=3−0.5=2.5h = 3 - 0.5 = 2.5.

Area=12(a+b)h=12(2+1)(2.5)=3.75 units2\text{Area} = \frac{1}{2}(a + b)h = \frac{1}{2}(2 + 1)(2.5) = 3.75 \text{ units}^2

Answer

Isosceles trapezium
Area = 3.75 units23.75 \text{ units}^2

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