Solve the equation $z^{5}+32=0$, expressing your answers in the form $re^{i\theta}$, where $r>0$ and $-\pi<\theta\le\pi$.

1. Identify the roots $z_1$ and $z_2$ from the equation $z^5+32=0$.

$z^5 = -32 = 32e^{i\pi}$
$z = 2e^{i\frac{\pi + 2k\pi}{5}} \quad \text{for } k = -2, -1, 0, 1, 2$
Considering the roots with arguments in $(0, \pi]$:
For $k=0$, $\arg z = \frac{\pi}{5}$
For $k=1$, $\arg z = \frac{3\pi}{5}$
For $k=2$, $\arg z = \pi$
Since $0 < \arg z_1 < \arg z_2 < \arg z_3 \le \pi$, we can deduce:
$z_1 = 2e^{i\frac{\pi}{5}} \quad \text{and} \quad z_2 = 2e^{i\frac{3\pi}{5}}$

2. Find the ratio $\frac{z_1}{z_2}$ in exponential form.

$\frac{z_1}{z_2} = \frac{2e^{i\frac{\pi}{5}}}{2e^{i\frac{3\pi}{5}}}$
$= e^{i\left(\frac{\pi}{5} - \frac{3\pi}{5}\right)}$
$= e^{-i\frac{2\pi}{5}}$

3. Apply the condition for $\left(\frac{z_1}{z_2}\right)^n$ to be real and positive to find the smallest positive integer $n$.

$\left(\frac{z_1}{z_2}\right)^n = \left(e^{-i\frac{2\pi}{5}}\right)^n = e^{-i\frac{2n\pi}{5}}$
For a complex number to be real and positive, its argument must be an integer multiple of $2\pi$:
$-\frac{2n\pi}{5} = 2m\pi \quad \text{for some integer } m$
$n = -5m$
To find the smallest positive integer $n$, we substitute $m = -1$:
$n = 5$

1. Identify the complex numbers $z_1$ and $z_3$ from the roots of $z^5+32=0$.

$z^5 = -32 = 32e^{i\pi}$
The roots are $z = 2e^{i\left(\frac{\pi+2k\pi}{5}\right)}$.
For roots with $0 < \arg z \le \pi$:
$k=0 \implies z = 2e^{i\frac{\pi}{5}}$
$k=1 \implies z = 2e^{i\frac{3\pi}{5}}$
$k=2 \implies z = 2e^{i\pi} = -2$
Given $0 < \arg z_1 < \arg z_2 < \arg z_3 \le \pi$, we have $z_1 = 2e^{i\frac{\pi}{5}}$ and $z_3 = -2$.

2. Formulate the relationship for a clockwise rotation by $\frac{\pi}{2}$ about point $B$.

A clockwise rotation of $\frac{\pi}{2}$ corresponds to multiplication by $e^{-i\frac{\pi}{2}} = -i$.
The vector $\vec{BA}$ is represented by $z_1 - z_3$, and $\vec{BA'}$ is represented by $z_{A'} - z_3$.
Therefore,
$z_{A'} - z_3 = -i(z_1 - z_3)$

3. Substitute $z_1$ and $z_3$ and convert to Cartesian form to solve for $z_{A'}$.

$z_{A'} - (-2) = -i\left( 2e^{i\frac{\pi}{5}} - (-2) \right)$
$z_{A'} + 2 = -i\left( 2\cos\frac{\pi}{5} + 2i\sin\frac{\pi}{5} + 2 \right)$
$z_{A'} + 2 = -2i\cos\frac{\pi}{5} - 2i^2\sin\frac{\pi}{5} - 2i$
Using $i^2 = -1$:
$z_{A'} + 2 = 2\sin\frac{\pi}{5} - i\left(2\cos\frac{\pi}{5} + 2\right)$
$z_{A'} = -2 + 2\sin\frac{\pi}{5} - i\left(2\cos\frac{\pi}{5} + 2\right)$

4. Extract the real part of $z_{A'}$.

The real part of $z_{A'}$ is the component without the $i$:
$\operatorname{Re}(z_{A'}) = -2 + 2\sin\frac{\pi}{5}$