[剑桥/9709/2009] 复数 $-2 + i$ 记为 $u$。

1. Substitute the root $u = -2 + i$ into the polynomial equation.

Since $u = -2 + i$ is a root of $x^3 - 11x - k = 0$, it must satisfy the equation:
$(-2 + i)^3 - 11(-2 + i) - k = 0$

2. Expand the cubic term $(-2 + i)^3$ and simplify the equation.

Expanding using the binomial theorem (or by repeated multiplication):
$[(-2)^3 + 3(-2)^2(i) + 3(-2)(i)^2 + i^3] - 11(-2 + i) - k = 0$
$[-8 + 12i - 6(-1) - i] + 22 - 11i - k = 0$
$(-8 + 12i + 6 - i) + 22 - 11i - k = 0$
$(-2 + 11i) + 22 - 11i - k = 0$

3. Group the real and imaginary parts to solve for $k$.

Combining the real and imaginary terms:
$(-2 + 22 - k) + (11i - 11i) = 0$
$(20 - k) + 0i = 0$

Equating the real parts to zero:
$20 - k = 0$
$k = 20$

1. Apply the conjugate root theorem to find the other complex root

Since the equation $x^3 - 11x - k = 0$ has real coefficients ($k$ is real), its complex roots must occur in conjugate pairs.

The given root is $u = -2 + i$.

The other complex root is its conjugate: $u^* = -2 - i$.

1. Calculate the modulus of the complex number $u$.

$|u| = \sqrt{(-2)^2 + 1^2}$
$|u| = \sqrt{4 + 1} = \sqrt{5}$

2. Calculate the argument of the complex number $u$.

Since the real part is negative and the imaginary part is positive, $u$ lies in the second quadrant.
The basic angle $\alpha = \tan^{-1}\left(\frac{1}{2}\right) \approx 0.4636$ rad.
$\arg(u) = \pi - 0.4636 \approx 2.68$ rad (or $153.4^\circ$)