复数 $z_{1}$ 和 $z_{2}$ 分别为 $z_{1}=\frac{1+i}{1-i} \quad\text{and}\quad z_{2}=\cos\left(\frac{\pi}{4}\right)+i\sin\left(\frac{\pi}{4}\right).$ 求 $z_{1}+z_{2}$，并表示为 $re^{i\theta}$ 的形式，其中 $r$ 是以三角函数形式表示的精确实常数且 $r>0$，$\theta$ 的形式为 $k\pi$，其中 $k$ 是一个精确的实常数，满足 $-1<k\le 1$。 同样求出 $z_{1}+z_{2}$ 并表示为 $x+iy$ 的形式，其中 $x$ 和 $y$ 是精确的实常数。由此证明 $\tan\left(\frac{3\pi}{8}\right)=1+\sqrt{2}$。 复数 $w$ 为 $w=\cos\theta + i\sin\theta$，其中 $0<\theta<\frac{\pi}{2}$。 证明 $1-w^{2}=-2i w \sin\theta$。 由此求出 $1-w^{2}$ 的模和辐角（用 $\theta$ 表示）。 已知 $\left(\dfrac{1-w^{2}}{iw^{*}}\right)^{n}$ 是负实数，且 $\theta=\dfrac{\pi}{5}$，求 $n$ 的前三个最小正整数值。

1. Simplify $z_1$ and express both $z_1$ and $z_2$ in exponential form.

$z_1 = \frac{1+i}{1-i} = \frac{(1+i)^2}{(1-i)(1+i)} = \frac{1+2i-1}{2} = i$
Since $i$ lies on the positive imaginary axis, its modulus is $1$ and argument is $\frac{\pi}{2}$. In exponential form:
$z_1 = e^{i\frac{\pi}{2}}$
The given $z_2$ can also be written in exponential form:
$z_2 = e^{i\frac{\pi}{4}}$

2. Add $z_1$ and $z_2$, and factor out the exponential of the average of their arguments.

$z_1 + z_2 = e^{i\frac{\pi}{2}} + e^{i\frac{\pi}{4}}$
The average of the arguments $\frac{\pi}{2}$ and $\frac{\pi}{4}$ is $\frac{\frac{\pi}{2} + \frac{\pi}{4}}{2} = \frac{3\pi}{8}$. Factoring out $e^{i\frac{3\pi}{8}}$ yields:
$z_1 + z_2 = e^{i\frac{3\pi}{8}} \left( e^{i\left(\frac{\pi}{2} - \frac{3\pi}{8}\right)} + e^{i\left(\frac{\pi}{4} - \frac{3\pi}{8}\right)} \right)$
$z_1 + z_2 = e^{i\frac{3\pi}{8}} \left( e^{i\frac{\pi}{8}} + e^{-i\frac{\pi}{8}} \right)$

3. Use Euler's identity to simplify the expression into the $re^{i\theta}$ form.

Using the identity $e^{i\alpha} + e^{-i\alpha} = 2\cos\alpha$, we have:
$e^{i\frac{\pi}{8}} + e^{-i\frac{\pi}{8}} = 2\cos\left(\frac{\pi}{8}\right)$
Substituting this back gives:
$z_1 + z_2 = 2\cos\left(\frac{\pi}{8}\right) e^{i\frac{3\pi}{8}}$
Here, $r = 2\cos\left(\frac{\pi}{8}\right)$, which is an exact real constant in trigonometric form. Since $0 < \frac{\pi}{8} < \frac{\pi}{2}$, $r > 0$ as required. Also, $\theta = \frac{3\pi}{8}$ meets the required form $k\pi$ with $k = \frac{3}{8}$, satisfying $-1 < k \le 1$.

1. Convert $z_1$ and $z_2$ to Cartesian form $x+iy$ and sum them.

Rationalising the denominator for $z_1$:

Evaluating trigonometric ratios for $z_2$:

Summing $z_1$ and $z_2$:
$z_1 + z_2 = \frac{\sqrt{2}}{2} + i\left(1 + \frac{\sqrt{2}}{2}\right)$

2. Use the argument of $z_1+z_2$ from part (a)(i) to prove the given trigonometric identity.

From part (a)(i), we know that $\arg(z_1+z_2) = \frac{3\pi}{8}$.
Using the property $\tan(\arg(z)) = \frac{\text{Im}(z)}{\text{Re}(z)}$:

1. Substitute $w = \cos\theta + i\sin\theta$ into the left-hand side and expand the square.

2. Simplify using trigonometric identities and factor out $-2i\sin\theta$ to reach the right-hand side.

1. Find the modulus of $1-w^2$ using the properties of modulus.

From (b)(i), we have $1-w^2 = -2iw\sin\theta$.
Taking the modulus on both sides:
$|1-w^2| = |-2i w \sin\theta|$
$|1-w^2| = |-2i| \cdot |w| \cdot |\sin\theta|$
Since $w = \cos\theta + i\sin\theta$, we have $|w| = 1$.
Given $0 < \theta < \frac{\pi}{2}$, we know $\sin\theta > 0$, so $|\sin\theta| = \sin\theta$.
Also, $|-2i| = 2$.
Therefore,
$|1-w^2| = 2(1)(\sin\theta) = 2\sin\theta$

2. Find the argument of $1-w^2$ using the properties of argument.

Taking the argument on both sides:
$\arg(1-w^2) = \arg(-2iw\sin\theta)$
$\arg(1-w^2) = \arg(-2i) + \arg(w) + \arg(\sin\theta)$
We know:
$\arg(-2i) = -\frac{\pi}{2}$
$\arg(w) = \theta$
Since $\sin\theta > 0$ for $0 < \theta < \frac{\pi}{2}$, it is a positive real number, so $\arg(\sin\theta) = 0$.
Therefore,
$\arg(1-w^2) = -\frac{\pi}{2} + \theta + 0 = \theta - \frac{\pi}{2}$
(Note: Since $0 < \theta < \frac{\pi}{2}$, $-\frac{\pi}{2} < \theta - \frac{\pi}{2} < 0$, which is within the principal argument range of $(-\pi, \pi]$.)

1. Substitute the identity from (b)(i) and simplify the algebraic fraction.

Using the result $1-w^2 = -2iw\sin\theta$ from (b)(i), substitute this into the expression:
$\frac{1-w^2}{iw^*} = \frac{-2iw\sin\theta}{iw^*} = -2\sin\theta \left( \frac{w}{w^*} \right)$

2. Express the simplified expression in polar/exponential form using $\theta = \frac{\pi}{5}$.

Substitute $\theta = \frac{\pi}{5}$. We have $w = e^{i\frac{\pi}{5}}$, which implies its conjugate is $w^* = e^{-i\frac{\pi}{5}}$.
$-2\sin\left(\frac{\pi}{5}\right) \frac{e^{i\frac{\pi}{5}}}{e^{-i\frac{\pi}{5}}} = -2\sin\left(\frac{\pi}{5}\right) e^{i\frac{2\pi}{5}}$

3. Apply the power of $n$ and set the condition for the expression to be real.

$\left( \frac{1-w^2}{iw^*} \right)^n = \left( -2\sin\frac{\pi}{5} e^{i\frac{2\pi}{5}} \right)^n = (-1)^n \left(2\sin\frac{\pi}{5}\right)^n e^{i\frac{2n\pi}{5}}$
For this expression to be real, the imaginary part must be zero. This requires the argument $\frac{2n\pi}{5}$ to be an integer multiple of $\pi$:
$\frac{2n\pi}{5} = p\pi \implies 2n = 5p$
Since $n$ and $p$ are integers, $n$ must be a multiple of $5$.
Let $n = 5q$ for some positive integer $q$. Then $e^{i\frac{2(5q)\pi}{5}} = e^{i2q\pi} = 1$.

4. Apply the condition for the expression to be negative and determine the smallest positive values of $n$.

Substitute $n = 5q$ into the expression:
$(-1)^{5q} \left(2\sin\frac{\pi}{5}\right)^{5q} (1) = (-1)^q \left(2\sin\frac{\pi}{5}\right)^{5q}$
Since $0 < \frac{\pi}{5} < \frac{\pi}{2}$, $\sin\frac{\pi}{5} > 0$, so the magnitude $\left(2\sin\frac{\pi}{5}\right)^{5q}$ is strictly positive.
For the entire result to be negative, we must have $(-1)^q = -1$, which means $q$ must be an odd integer.
The three smallest positive odd integer values for $q$ are $1, 3,$ and $5$.
Therefore, the three smallest positive integer values of $n$ are:
$n = 5(1) = 5$
$n = 5(3) = 15$
$n = 5(5) = 25$