解答本题时请勿使用图形计算器。 复数 $w$ 是方程 $z^4 + a z^3 + 46 z^2 + b z + 125 = 0 ,$ 的一个根，其中 $a,b \in \mathbb{R}$。

1. Expand the given product of the two linear factors.

$(z - w)(z - w^*) = z^2 - w^*z - wz + ww^*$
$= z^2 - (w + w^*)z + ww^*$

2. Use the Cartesian form of a complex number to show that the coefficients are real.

Let $w = x + iy$, where $x, y \in \mathbb{R}$. Then $w^* = x - iy$.

Sum of roots:
$w + w^* = (x + iy) + (x - iy) = 2x$
Since $x \in \mathbb{R}$, $2x \in \mathbb{R}$.

Product of roots:
$ww^* = (x + iy)(x - iy) = x^2 - (iy)^2 = x^2 + y^2$
Since $x, y \in \mathbb{R}$, $x^2 + y^2 \in \mathbb{R}$.

Therefore, $z^2 - (2x)z + (x^2 + y^2)$ is a quadratic polynomial with real coefficients.

1. Use the given argument to establish the relationship between the real and imaginary parts of $w$.

Let $w = x + iy$, where $x, y \in \mathbb{R}$.
Given $\arg(w) = -\tan^{-1}(2)$ where $-\frac{\pi}{2} < -\tan^{-1}(2) < 0$, $w$ lies in the fourth quadrant, which means $x > 0$ and $y < 0$.
$\tan(\arg w) = \frac{y}{x}$
$\tan(-\tan^{-1}(2)) = -2$
$\frac{y}{x} = -2 \implies y = -2x$

2. Use the given modulus to solve for the exact values of $x$ and $y$, and hence show the value of $w$.

Given $|w| = \sqrt{5}$:
$\sqrt{x^2 + y^2} = \sqrt{5} \implies x^2 + y^2 = 5$
Substitute $y = -2x$ into the equation:
$x^2 + (-2x)^2 = 5$
$x^2 + 4x^2 = 5$
$5x^2 = 5 \implies x^2 = 1$
Since $x > 0$, we take $x = 1$.
$y = -2(1) = -2$
Therefore, $w = 1 - 2i$ (shown).

1. Identify the conjugate root and form the first quadratic factor.

Since the polynomial equation has real coefficients, the complex conjugate $w^* = 1 + 2i$ must also be a root.
The quadratic factor corresponding to these two roots is:
$(z - (1-2i))(z - (1+2i)) = ((z - 1) + 2i)((z - 1) - 2i)$
$= (z - 1)^2 - (2i)^2$
$= z^2 - 2z + 1 - 4(-1)$
$= z^2 - 2z + 5$

2. Set up an identity with a second quadratic factor and compare coefficients to find it.

Let the other quadratic factor be $z^2 + pz + q$. Then,
$z^4 + a z^3 + 46 z^2 + b z + 125 = (z^2 - 2z + 5)(z^2 + pz + q)$
Comparing the constant terms:
$5q = 125 \implies q = 25$
Comparing the coefficients of $z^2$:
$q - 2p + 5 = 46$
Substituting $q = 25$:
$25 - 2p + 5 = 46$
$30 - 2p = 46 \implies -2p = 16 \implies p = -8$

3. Solve the second quadratic factor to find the remaining roots.

The second quadratic factor is $z^2 - 8z + 25 = 0$.
Using the quadratic formula:
$z = \frac{8 \pm \sqrt{(-8)^2 - 4(1)(25)}}{2}$
$z = \frac{8 \pm \sqrt{64 - 100}}{2}$
$z = \frac{8 \pm \sqrt{-36}}{2}$
$z = \frac{8 \pm 6i}{2} = 4 \pm 3i$
Therefore, all the roots of the equation are $1 - 2i$, $1 + 2i$, $4 + 3i$, and $4 - 3i$.

4. Compare the remaining coefficients to find $a$ and $b$.

Comparing the coefficients of $z^3$:
$p - 2 = a$
$a = -8 - 2 = -10$
Comparing the coefficients of $z$:
$5p - 2q = b$
$b = 5(-8) - 2(25)$
$b = -40 - 50 = -90$