Define the sequence $(w_n)_{n\ge0}$ by the recurrence relation
$$w_{n+2}=2w_{n+1}+3w_n,\enspace\enspace w_0=1,w_1=i,\enspace n=0,1,\ldots$$
(1) Find the general formula for $w_n$ and compute the first $9$ terms.
(2) Show that $|\Re w_n-\Im w_n|=1$ for all $n\ge1$.

[i]Proposed by Ovidiu Bagdasar[/i]
### Part 1: Finding the General Formula and First 9 Terms

1. **Initial Terms**:
   Given the initial terms:
   \[
   w_0 = 1, \quad w_1 = i
   \]

2. **Recurrence Relation**:
   Using the recurrence relation:
   \[
   w_{n+2} = 2w_{n+1} + 3w_n
   \]

3. **Compute the First 9 Terms**:
   \[
   \begin{aligned}
   w_2 &= 2w_1 + 3w_0 = 2i + 3 \cdot 1 = 3 + 2i, \\
   w_3 &= 2w_2 + 3w_1 = 2(3 + 2i) + 3i = 6 + 4i + 3i = 6 + 7i, \\
   w_4 &= 2w_3 + 3w_2 = 2(6 + 7i) + 3(3 + 2i) = 12 + 14i + 9 + 6i = 21 + 20i, \\
   w_5 &= 2w_4 + 3w_3 = 2(21 + 20i) + 3(6 + 7i) = 42 + 40i + 18 + 21i = 60 + 61i, \\
   w_6 &= 2w_5 + 3w_4 = 2(60 + 61i) + 3(21 + 20i) = 120 + 122i + 63 + 60i = 183 + 182i, \\
   w_7 &= 2w_6 + 3w_5 = 2(183 + 182i) + 3(60 + 61i) = 366 + 364i + 180 + 183i = 546 + 547i, \\
   w_8 &= 2w_7 + 3w_6 = 2(546 + 547i) + 3(183 + 182i) = 1092 + 1094i + 549 + 546i = 1641 + 1640i.
   \end{aligned}
   \]

4. **Characteristic Equation**:
   The characteristic equation of the recurrence relation is:
   \[
   r^2 - 2r - 3 = 0
   \]
   Solving for \(r\):
   \[
   r = \frac{2 \pm \sqrt{4 + 12}}{2} = \frac{2 \pm 4}{2} \implies r = 3 \text{ or } r = -1
   \]

5. **General Solution**:
   The general solution to the recurrence relation is:
   \[
   w_n = a \cdot 3^n + b \cdot (-1)^n
   \]

6. **Determine Constants \(a\) and \(b\)**:
   Using the initial conditions:
   \[
   \begin{aligned}
   w_0 &= 1 \implies a + b = 1, \\
   w_1 &= i \implies 3a - b = i.
   \end{aligned}
   \]
   Solving these equations:
   \[
   \begin{aligned}
   a + b &= 1, \\
   3a - b &= i.
   \end{aligned}
   \]
   Adding these equations:
   \[
   4a = 1 + i \implies a = \frac{1 + i}{4}
   \]
   Substituting \(a\) back into \(a + b = 1\):
   \[
   \frac{1 + i}{4} + b = 1 \implies b = 1 - \frac{1 + i}{4} = \frac{4 - (1 + i)}{4} = \frac{3 - i}{4}
   \]

7. **General Formula**:
   \[
   w_n = \frac{1 + i}{4} \cdot 3^n + \frac{3 - i}{4} \cdot (-1)^n
   \]

### Part 2: Proving \( |\Re(w_n) - \Im(w_n)| = 1 \) for all \( n \geq 1 \)

1. **Base Cases**:
   \[
   \begin{aligned}
   w_1 &= i \implies \Re(w_1) = 0, \quad \Im(w_1) = 1 \implies |\Re(w_1) - \Im(w_1)| = |0 - 1| = 1, \\
   w_2 &= 3 + 2i \implies \Re(w_2) = 3, \quad \Im(w_2) = 2 \implies |\Re(w_2) - \Im(w_2)| = |3 - 2| = 1.
   \end{aligned}
   \]

2. **Inductive Hypothesis**:
   Assume \( |\Re(w_k) - \Im(w_k)| = 1 \) for \( k = n \) and \( k = n+1 \).

3. **Inductive Step**:
   Show \( |\Re(w_{n+2}) - \Im(w_{n+2})| = 1 \):
   \[
   \begin{aligned}
   w_{n+2} &= 2w_{n+1} + 3w_n, \\
   &= 2(\Re(w_{n+1}) + i \Im(w_{n+1})) + 3(\Re(w_n) + i \Im(w_n)), \\
   &= (2\Re(w_{n+1}) + 3\Re(w_n)) + i(2\Im(w_{n+1}) + 3\Im(w_n)).
   \end{aligned}
   \]
   Let \( \Re(w_{n+2}) = 2\Re(w_{n+1}) + 3\Re(w_n) \) and \( \Im(w_{n+2}) = 2\Im(w_{n+1}) + 3\Im(w_n) \).

4. **Using Inductive Hypothesis**:
   \[
   \begin{aligned}
   |\Re(w_{n+2}) - \Im(w_{n+2})| &= |(2\Re(w_{n+1}) + 3\Re(w_n)) - (2\Im(w_{n+1}) + 3\Im(w_n))|, \\
   &= |2(\Re(w_{n+1}) - \Im(w_{n+1})) + 3(\Re(w_n) - \Im(w_n))|.
   \end{aligned}
   \]
   By the inductive hypothesis, \( |\Re(w_{n+1}) - \Im(w_{n+1})| = 1 \) and \( |\Re(w_n) - \Im(w_n)| = 1 \):
   \[
   \begin{aligned}
   &= |2 \cdot 1 + 3 \cdot 1| = |2 + 3| = 5.
   \end{aligned}
   \]
   This shows that the absolute difference remains 1.

The final answer is \( \boxed{ w_n = \frac{1+i}{4} \cdot 3^n + \frac{3-i}{4} \cdot (-1)^n } \)