复数 $z$ 满足方程 $\mathrm{f}(z) = 0$，其中
$\mathrm{f}(z) = (1 - a\mathrm{i})z^2 - 2\mathrm{i}z - 10 - 20\mathrm{i}$
且 $a$ 是一个实数。已知其中一个根的形式为 $1 + b\mathrm{i}$，其中 $b$ 是一个整数。

1. Substitute $z = 1 + b\mathrm{i}$ into the given equation and expand the squared term.

$\mathrm{f}(1+b\mathrm{i}) = (1 - a\mathrm{i})(1 + b\mathrm{i})^2 - 2\mathrm{i}(1 + b\mathrm{i}) - 10 - 20\mathrm{i} = 0$
$(1 - a\mathrm{i})(1 - b^2 + 2b\mathrm{i}) - 2\mathrm{i} - 2b\mathrm{i}^2 - 10 - 20\mathrm{i} = 0$
$(1 - a\mathrm{i})(1 - b^2 + 2b\mathrm{i}) - 2\mathrm{i} + 2b - 10 - 20\mathrm{i} = 0$

2. Expand the product and group the real and imaginary parts.

$(1 - b^2) + 2b\mathrm{i} - a(1 - b^2)\mathrm{i} - 2ab\mathrm{i}^2 + 2b - 10 - 22\mathrm{i} = 0$
$(1 - b^2 + 2ab + 2b - 10) + \mathrm{i}(2b - a(1 - b^2) - 22) = 0$
$(-b^2 + 2ab + 2b - 9) + \mathrm{i}(2b - a(1 - b^2) - 22) = 0$

3. Equate the real and imaginary parts to zero to form two equations.

For a complex number to be zero, both its real and imaginary parts must be zero.
Real part:
$-b^2 + 2ab + 2b - 9 = 0 \implies 2ab = b^2 - 2b + 9 \quad \text{--- (1)}$
Imaginary part:
$2b - a(1 - b^2) - 22 = 0 \implies a(1 - b^2) = 2b - 22 \quad \text{--- (2)}$

4. Eliminate $a$ to show the required equation for $b$.

Multiply equation (1) by $(1 - b^2)$:
$2ab(1 - b^2) = (b^2 - 2b + 9)(1 - b^2)$
Multiply equation (2) by $2b$:
$2ab(1 - b^2) = 2b(2b - 22) = 4b^2 - 44b$
Equating the two expressions for $2ab(1 - b^2)$:
$4b^2 - 44b = (b^2 - 2b + 9)(1 - b^2)$
$4b^2 - 44b = b^2 - b^4 - 2b + 2b^3 + 9 - 9b^2$
$4b^2 - 44b = -b^4 + 2b^3 - 8b^2 - 2b + 9$
Rearranging all terms to one side:
$b^4 - 2b^3 + 12b^2 - 42b - 9 = 0 \quad \text{(shown)}$