方程 $2z^{4} - 14z^{3} + 33z^{2} - 26z + p = 0\text{，其中 } p \text{ 是常数}$ 的一个根为 $3 + \mathrm{i}$。

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, $p$ 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.

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

$z = 3 + \mathrm{i}$
$z^2 = (3 + \mathrm{i})^2 = 9 + 6\mathrm{i} + \mathrm{i}^2 = 8 + 6\mathrm{i}$
$z^3 = z \cdot z^2 = (3 + \mathrm{i})(8 + 6\mathrm{i}) = 24 + 18\mathrm{i} + 8\mathrm{i} + 6\mathrm{i}^2 = 18 + 26\mathrm{i}$
$z^4 = z^2 \cdot z^2 = (8 + 6\mathrm{i})^2 = 64 + 96\mathrm{i} + 36\mathrm{i}^2 = 28 + 96\mathrm{i}$

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

Substitute into $2z^4 - 14z^3 + 33z^2 - 26z + p = 0$:
$2(28 + 96\mathrm{i}) - 14(18 + 26\mathrm{i}) + 33(8 + 6\mathrm{i}) - 26(3 + \mathrm{i}) + p = 0$
$56 + 192\mathrm{i} - 252 - 364\mathrm{i} + 264 + 198\mathrm{i} - 78 - 26\mathrm{i} + p = 0$

Grouping the real and imaginary parts:
$(56 - 252 + 264 - 78 + p) + (192 - 364 + 198 - 26)\mathrm{i} = 0$
$(-10 + p) + 0\mathrm{i} = 0$

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

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

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

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

2. Find the other quadratic factor by comparing coefficients.

Let the other quadratic factor be $az^2 + bz + c$.
$2z^4 - 14z^3 + 33z^2 - 26z + 10 = (z^2 - 6z + 10)(az^2 + bz + c)$
By comparing the coefficient of $z^4$, $a = 2$.
By comparing the constant terms, $10c = 10 \implies c = 1$.
So the factor takes the form $2z^2 + bz + 1$.

Comparing the coefficients of $z^3$:
$b + (-6)(2) = -14$
$b - 12 = -14 \implies b = -2$

The other quadratic factor is $2z^2 - 2z + 1$.

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

Set $2z^2 - 2z + 1 = 0$.
Using the quadratic formula without a calculator:
$z = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(2)(1)}}{2(2)}$
$z = \frac{2 \pm \sqrt{4 - 8}}{4} = \frac{2 \pm \sqrt{-4}}{4}$
$z = \frac{2 \pm 2\mathrm{i}}{4} = \frac{1}{2} \pm \frac{1}{2}\mathrm{i}$

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

The roots of the equation are $3+\mathrm{i}$, $3-\mathrm{i}$, $\frac{1}{2}+\frac{1}{2}\mathrm{i}$, and $\frac{1}{2}-\frac{1}{2}\mathrm{i}$.

To mark them on a single labelled Argand diagram, construct a coordinate system with the horizontal axis labelled $\mathrm{Re}$ and the vertical axis labelled $\mathrm{Im}$. Plot the four points corresponding to the cartesian coordinates $(3, 1)$, $(3, -1)$, $(0.5, 0.5)$, and $(0.5, -0.5)$, and clearly label each point with its respective complex root.

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

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

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

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

Solving $2z^2 - 2z + 1 = 0$:
$z = \frac{2 \pm \sqrt{(-2)^2 - 4(2)(1)}}{2(2)} = \frac{2 \pm \sqrt{-4}}{4} = \frac{1}{2} \pm \frac{1}{2}\mathrm{i}$

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

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

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

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

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