Let $G$ be the set of polynomials of the form
\[P(z)=z^n+c_{n-1}z^{n-1}+\cdots+c_2z^2+c_1z+50,\]
where $c_1,c_2,\cdots, c_{n-1}$ are integers and $P(z)$ has $n$ distinct roots of the form $a+ib$ with $a$ and $b$ integers. How many polynomials are in $G$?

${ \textbf{(A)}\ 288\qquad\textbf{(B)}\ 528\qquad\textbf{(C)}\ 576\qquad\textbf{(D}}\ 992\qquad\textbf{(E)}\ 1056 $
1. **Understanding the problem**: We need to find the number of polynomials of the form 
   \[
   P(z) = z^n + c_{n-1}z^{n-1} + \cdots + c_2z^2 + c_1z + 50
   \]
   where \(c_1, c_2, \ldots, c_{n-1}\) are integers, and \(P(z)\) has \(n\) distinct roots of the form \(a + ib\) with \(a\) and \(b\) being integers.

2. **Factorization into irreducible polynomials**: Each polynomial \(P(z)\) can be factored into irreducible polynomials over \(\mathbb{Q}[z]\). Since the roots are of the form \(a + ib\) with \(a, b \in \mathbb{Z}\), the irreducible factors will either be linear (if the root is real) or quadratic (if the roots are complex conjugates).

3. **Constant term condition**: The constant term of \(P(z)\) is 50. Therefore, the product of the constant terms of the irreducible factors must be 50.

4. **Counting irreducible factors**: We need to count the number of ways to form the product 50 using the constant terms of the irreducible factors. The possible constant terms (factors of 50) are \(\pm 1, \pm 2, \pm 5, \pm 10, \pm 25, \pm 50\).

5. **Distinct roots condition**: Since \(P(z)\) has \(n\) distinct roots, the irreducible factors must have distinct roots. This means that the constant terms can only be repeated if they are squares of factors of 50 and have more than one possible factorization.

6. **Counting valid factorizations**:
   - For \(d = 1\), there are 2 possible factorizations: \(1\) and \(-1\).
   - For \(d = 2\), there are 3 possible factorizations: \(2\), \(-2\), and \((1+i)(1-i)\).
   - For \(d = 5\), there are 5 possible factorizations: \(5\), \(-5\), \((2+i)(2-i)\), \((1+2i)(1-2i)\), and \((1+i\sqrt{5})(1-i\sqrt{5})\).
   - For \(d = 10\), there are 5 possible factorizations: \(10\), \(-10\), \((3+i)(3-i)\), \((1+3i)(1-3i)\), and \((1+i\sqrt{10})(1-i\sqrt{10})\).
   - For \(d = 25\), there are 6 possible factorizations: \(25\), \(-25\), \((4+i)(4-i)\), \((2+3i)(2-3i)\), \((1+4i)(1-4i)\), and \((1+i\sqrt{25})(1-i\sqrt{25})\).
   - For \(d = 50\), there are 7 possible factorizations: \(50\), \(-50\), \((5+i)(5-i)\), \((3+4i)(3-4i)\), \((2+5i)(2-5i)\), \((1+5i)(1-5i)\), and \((1+i\sqrt{50})(1-i\sqrt{50})\).

7. **Combining factorizations**: We need to combine these factorizations to form the product 50. We do casework on the unique even constant term \(d \in \{\pm 2, \pm 10, \pm 50\}\) in our product. For convenience, let \(F(d)\) be the number of ways to get a product of \(50/d\) without using \(\pm 1\) (so only using \(\pm 5, \pm 25\)).

8. **Calculating \(F(d)\)**:
   - \(F(-50) = 0\)
   - \(F(50) = 1\)
   - \(F(-10) = f(5) = 5\)
   - \(F(10) = f(-5) = 1\)
   - \(F(-2) = f(-25) + f(-5)f(5) = 6\)
   - \(F(2) = f(25) + \binom{f(5)}{2} = 16\)

9. **Final calculation**:
   \[
   2^{f(1)} \sum_{d \in \{2, 10, 50\}} (f(-d) + f(d))(F(-d) + F(d)) = 4 \left[ (1+3)(6+16) + (1+5)(1+5) + (1+7)(0+1) \right] = 4 \times 132 = 528
   \]

The final answer is \(\boxed{528}\)