An arithmetic series has first term $a$ and common difference $d$, where $a > 0$ and $d \ne 0$.
The first, sixth and ninth terms of the arithmetic series are consecutive terms of a geometric series.

1. Express the 1st, 6th, and 9th terms of the arithmetic series in terms of $a$ and $d$.

Let $u_n$ be the $n$-th term of the arithmetic series.
$u_1 = a$
$u_6 = a + 5d$
$u_9 = a + 8d$

2. Set up an equation using the property of consecutive terms in a geometric series.

Since $u_1, u_6$, and $u_9$ are consecutive terms of a geometric series, they share a common ratio:
$\frac{u_6}{u_1} = \frac{u_9}{u_6}$
$\frac{a + 5d}{a} = \frac{a + 8d}{a + 5d}$

3. Expand, simplify, and solve to show the required relationship.

$(a + 5d)^2 = a(a + 8d)$
$a^2 + 10ad + 25d^2 = a^2 + 8ad$
$2ad + 25d^2 = 0$
$d(2a + 25d) = 0$
Since we are given $d \ne 0$, we can divide by $d$:
$2a + 25d = 0$
$25d = -2a \quad \text{(shown)}$

1. Set up the inequality using the sum of an arithmetic series formula and substitute $d = -\frac{2}{25}a$.

Using the sum of an arithmetic progression formula, $S = \frac{n}{2}(2a + (n-1)d)$, and the condition $S > 6a$:
$\frac{n}{2}(2a + (n-1)d) > 6a$
From part (a), we know $25d = -2a \implies d = -\frac{2}{25}a$. Substituting this into the inequality:
$\frac{n}{2}\left(2a + (n-1)\left(-\frac{2}{25}a\right)\right) > 6a$

2. Simplify the inequality into a standard quadratic form.

Since it is given that $a > 0$, we can divide both sides by $a$ without reversing the inequality sign:
$\frac{n}{2}\left(2 - \frac{2(n-1)}{25}\right) > 6$
$n\left(1 - \frac{n-1}{25}\right) > 6$
$n\left(\frac{25 - (n-1)}{25}\right) > 6$
$n(26 - n) > 150$
$26n - n^2 > 150$
$n^2 - 26n + 150 < 0$

3. Solve the quadratic inequality to find the exact range, and determine the valid integer values for $n$.

First, find the roots of the corresponding quadratic equation $n^2 - 26n + 150 = 0$:
$n = \frac{26 \pm \sqrt{(-26)^2 - 4(1)(150)}}{2} = \frac{26 \pm \sqrt{76}}{2} = 13 \pm \sqrt{19}$
The critical values are approximately $n \approx 8.641$ and $n \approx 17.359$.
For $n^2 - 26n + 150 < 0$, the range of $n$ is between the roots:
$8.641 < n < 17.359$
Since $n$ must be a positive integer (the number of terms), the possible values are:
$9 \le n \le 17$

1. Find the common ratio of the geometric series using the relationship from part (a).

The common ratio $r$ can be found using the first and second terms of the geometric series, which correspond to the first and sixth terms of the arithmetic series.
$r = \frac{u_6}{u_1} = \frac{a + 5d}{a}$
From part (a), we have $25d = -2a \implies d = -\frac{2}{25}a$.
Substitute this into the expression for $r$:
$r = \frac{a + 5\left(-\frac{2}{25}a\right)}{a} = \frac{a - \frac{2}{5}a}{a} = \frac{3}{5}$

2. Deduce the convergence of the geometric series.

For a geometric series to be convergent, its common ratio $r$ must satisfy $|r| < 1$.
Since $r = \frac{3}{5}$, we have $|r| = \frac{3}{5} < 1$.
Therefore, the geometric series is convergent.

1. Determine the common ratio $r$ of the geometric series based on earlier parts.

From part (a), $25d = -2a \implies d = -0.08a$.
The first two terms of the geometric series are the first and sixth terms of the arithmetic series, which are $a$ and $a+5d$.
The common ratio $r$ is:
$r = \frac{a+5d}{a} = 1 + \frac{5(-0.08a)}{a} = 1 - 0.4 = 0.6$

2. Formulate the inequality for the sum after the $m$th term.

Let the first term of the geometric series be $u_1$. The sum to infinity is $S_\infty = \frac{u_1}{1-r}$.
The terms after the $m$th term form an infinite geometric series starting with the $(m+1)$th term, $u_{m+1} = u_1 r^m$.
The sum after the $m$th term is $\frac{u_1 r^m}{1-r}$.
We require this sum to be less than $1\%$ of $S_\infty$:
$\frac{u_1 r^m}{1-r} < 0.01 \left(\frac{u_1}{1-r}\right)$

3. Simplify and solve the exponential inequality for $m$.

Since $u_1 = a > 0$ and $1-r = 0.4 > 0$, we can safely divide both sides by $\frac{u_1}{1-r}$ without changing the inequality sign:
$r^m < 0.01$
Substitute $r = 0.6$:
$0.6^m < 0.01$
Taking the natural logarithm on both sides:
$m \ln(0.6) < \ln(0.01)$
Because $\ln(0.6)$ is negative, dividing by it reverses the inequality sign:
$m > \frac{\ln(0.01)}{\ln(0.6)}$
$m > 9.0152$

4. Determine the smallest integer value of $m$.

Since $m$ must be an integer indicating the term position, the smallest integer strictly greater than $9.0152$ is:
$m = 10$

1. Relate tan(π/12) to the cartesian components of z

From the earlier part, $\arg z = \dfrac{\pi}{12}$. Writing $z = x + \mathrm{i}y$ (the cartesian form found in (ii)), the argument of a complex number satisfies $\tan(\arg z) = \frac{\text{Im}\,z}{\text{Re}\,z} = \frac{y}{x}.$ Hence $\tan\frac{\pi}{12} = \frac{y}{x}.$

2. Substitute the components and rationalise the surd quotient

Substituting the real and imaginary parts of $z$ from (ii) gives the ratio $\tan\frac{\pi}{12} = \frac{\sqrt{3}-1}{\sqrt{3}+1}.$ Rationalise by multiplying numerator and denominator by $\sqrt{3}-1$: $\frac{(\sqrt{3}-1)^2}{(\sqrt{3}+1)(\sqrt{3}-1)} = \frac{3 - 2\sqrt{3} + 1}{3 - 1} = \frac{4 - 2\sqrt{3}}{2}.$

3. Simplify to the required form and identify c and d

$\tan\frac{\pi}{12} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3}.$ Comparing with $c + d\sqrt{3}$ gives $c = 2$ and $d = -1$.