Given that $z_1 = r_1(\cos\theta_1 + \mathrm{i}\sin\theta_1)$ and $z_2 = r_2(\cos\theta_2 + \mathrm{i}\sin\theta_2)$, prove, using a trigonometric method, that $\frac{z_1}{z_2} = \frac{r_1}{r_2}[\cos(\theta_1 - \theta_2) + \mathrm{i}\sin(\theta_1 - \theta_2)]$. Hence state $\arg\left(\frac{z_1}{z_2}\right)$ in terms of $\theta_1$ and $\theta_2$. Let $w_r = r + \mathrm{i}$, where $r$ is a real constant, $r \ge 0$. Use the method of differences to express $\sum_{r=1}^{n} (\arg(w_{r-1}) - \arg(w_r))$ in the form $k - \arg(n + \mathrm{i})$, where $k$ is an exact real constant to be determined. Explain why $\sum_{r=1}^{\infty} (\arg(w_{r-1}) - \arg(w_r))$ converges and state its limit. Hence, or otherwise, find the exact value of
$\arg\left(\frac{1+\mathrm{i}}{2+\mathrm{i}}\right)^3 + \arg\left(\frac{2+\mathrm{i}}{3+\mathrm{i}}\right)^3 + \arg\left(\frac{3+\mathrm{i}}{4+\mathrm{i}}\right)^3 + \dots$

1. Set up the quotient and rationalise the denominator by multiplying the numerator and denominator by the conjugate of $\cos\theta_2 + \mathrm{i}\sin\theta_2$.

$\frac{z_1}{z_2} = \frac{r_1(\cos\theta_1 + \mathrm{i}\sin\theta_1)}{r_2(\cos\theta_2 + \mathrm{i}\sin\theta_2)}$
$\frac{z_1}{z_2} = \frac{r_1}{r_2} \left[ \frac{(\cos\theta_1 + \mathrm{i}\sin\theta_1)(\cos\theta_2 - \mathrm{i}\sin\theta_2)}{(\cos\theta_2 + \mathrm{i}\sin\theta_2)(\cos\theta_2 - \mathrm{i}\sin\theta_2)} \right]$

2. Expand the numerator and simplify the denominator using the Pythagorean identity.

$\frac{z_1}{z_2} = \frac{r_1}{r_2} \left[ \frac{(\cos\theta_1\cos\theta_2 + \sin\theta_1\sin\theta_2) + \mathrm{i}(\sin\theta_1\cos\theta_2 - \cos\theta_1\sin\theta_2)}{\cos^2\theta_2 + \sin^2\theta_2} \right]$
Since $\cos^2\theta_2 + \sin^2\theta_2 = 1$,
$\frac{z_1}{z_2} = \frac{r_1}{r_2} [(\cos\theta_1\cos\theta_2 + \sin\theta_1\sin\theta_2) + \mathrm{i}(\sin\theta_1\cos\theta_2 - \cos\theta_1\sin\theta_2)]$

3. Apply trigonometric compound angle identities to complete the proof.

Using the compound angle identities $\cos(A-B) = \cos A\cos B + \sin A\sin B$ and $\sin(A-B) = \sin A\cos B - \cos A\sin B$,
$\frac{z_1}{z_2} = \frac{r_1}{r_2} [\cos(\theta_1 - \theta_2) + \mathrm{i}\sin(\theta_1 - \theta_2)] \quad \text{(shown)}$

4. State the argument of the quotient.

From the polar form shown above, the argument is the angle of the complex number.
$\arg\left(\frac{z_1}{z_2}\right) = \theta_1 - \theta_2$

1. List the first few terms and the last term of the series to apply the method of differences.

$\sum_{r=1}^{n} (\arg(w_{r-1}) - \arg(w_r))$\n$= (\arg(w_0) - \arg(w_1))$\n$\phantom{= } + (\arg(w_1) - \arg(w_2))$\n$\phantom{= } + (\arg(w_2) - \arg(w_3))$\n$\phantom{= } + \dots$\n$\phantom{= } + (\arg(w_{n-1}) - \arg(w_n))$

2. Cancel out the intermediate terms to simplify the sum.

$= \arg(w_0) - \arg(w_n)$

3. Evaluate the remaining terms using the given definition of $w_r$ to find the constant $k$.

Given $w_r = r + \mathrm{i}$.\nWhen $r = 0$, $w_0 = 0 + \mathrm{i} = \mathrm{i}$.\n$\arg(w_0) = \arg(\mathrm{i}) = \frac{\pi}{2}$\nWhen $r = n$, $w_n = n + \mathrm{i}$.\nSubstituting these back into the simplified sum:\n$\sum_{r=1}^{n} (\arg(w_{r-1}) - \arg(w_r)) = \frac{\pi}{2} - \arg(n + \mathrm{i})$\nComparing this to $k - \arg(n + \mathrm{i})$, we find:\n$k = \frac{\pi}{2}$

1. Deduce the expression for the partial sum from part (b).

From the method of differences in part (b), the sum telescopes to $\arg(w_0) - \arg(w_n)$.
Since $w_0 = 0 + \mathrm{i} = \mathrm{i}$, its argument is $\arg(\mathrm{i}) = \frac{\pi}{2}$.
Thus, the partial sum to $n$ terms is $S_n = \frac{\pi}{2} - \arg(n+\mathrm{i})$.

2. Set up the limit of the partial sum and explain the behaviour of $\arg(n+\mathrm{i})$ as $n \to \infty$.

To find the sum to infinity, we consider $\lim_{n \to \infty} S_n = \lim_{n \to \infty} \left( \frac{\pi}{2} - \arg(n+\mathrm{i}) \right)$.
For $n > 0$, the complex number $n+\mathrm{i}$ is in the first quadrant, so we can express its argument as $\arg(n+\mathrm{i}) = \tan^{-1}\left(\frac{1}{n}\right)$.
As $n \to \infty$, $\frac{1}{n} \to 0$, which means $\arg(n+\mathrm{i}) \to \tan^{-1}(0) = 0$.
Since the limit of the sequence of partial sums exists and is finite, the infinite series converges.

3. State the limit of the infinite series.

Evaluating the limit gives:
$\lim_{n \to \infty} S_n = \frac{\pi}{2} - 0 = \frac{\pi}{2}$

1. Express the given series in sigma notation and simplify using properties of arguments.

Let the given sum be $S$. We can write $S$ as:
$S = \sum_{r=1}^{\infty} \arg\left( \left(\frac{r+\mathrm{i}}{r+1+\mathrm{i}}\right)^3 \right)$
Using the property $\arg(z^3) = 3\arg(z)$ and the result from part (a) that $\arg\left(\frac{z_1}{z_2}\right) = \arg(z_1) - \arg(z_2)$:
$S = 3 \sum_{r=1}^{\infty} \arg\left( \frac{r+\mathrm{i}}{r+1+\mathrm{i}} \right)$
$S = 3 \sum_{r=1}^{\infty} \left[ \arg(r+\mathrm{i}) - \arg(r+1+\mathrm{i}) \right]$
Using the definition $w_r = r + \mathrm{i}$ from part (b):
$S = 3 \sum_{r=1}^{\infty} \left[ \arg(w_r) - \arg(w_{r+1}) \right]$

2. Use the method of differences to evaluate the infinite sum and apply the limit.

Consider the partial sum for the first $n$ terms:
$\sum_{r=1}^{n} \left[ \arg(w_r) - \arg(w_{r+1}) \right]$
$= (\arg(w_1) - \arg(w_2)) + (\arg(w_2) - \arg(w_3)) + \dots + (\arg(w_n) - \arg(w_{n+1}))$
$= \arg(w_1) - \arg(w_{n+1})$
Taking the limit as $n \to \infty$:
$\lim_{n \to \infty} \sum_{r=1}^{n} \left[ \arg(w_r) - \arg(w_{r+1}) \right] = \arg(w_1) - \lim_{n \to \infty} \arg(w_{n+1})$
Similar to part (c), as $n \to \infty$, $w_{n+1} = n+1+\mathrm{i} \to \infty$ along the real axis, so $\lim_{n \to \infty} \arg(w_{n+1}) = 0$.

3. Substitute the argument of the first term to find the exact value of the series.

We need to evaluate $\arg(w_1)$. Since $w_1 = 1+\mathrm{i}$, its argument is:
$\arg(1+\mathrm{i}) = \frac{\pi}{4}$
Substituting this back into our expression for $S$:
$S = 3 \left( \frac{\pi}{4} - 0 \right) = \frac{3\pi}{4}$