已知 z=2−iz = 2 - \text{i} 是方程 4z4−12z3+17z2+pz+q=04z^4 - 12z^3 + 17z^2 + pz + q = 0 的一个根,其中 pp 和 qq 是实数,求 pp 和 qq 的值。 利用所求得的 pp 和 qq 的值,求方程 4z4−12z3+17z2+pz+q=04z^4 - 12z^3 + 17z^2 + pz + q = 0 的其它根的精确形式。

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完整解法

10(a)

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

Since the equation has real coefficients (p,q∈Rp, q \in \mathbb{R}), the complex conjugate z=2+iz = 2 + \text{i} is also a root.
The quadratic factor containing these two roots is:
(z−(2−i))(z−(2+i))=((z−2)+i)((z−2)−i)(z - (2 - \text{i}))(z - (2 + \text{i})) = ((z - 2) + \text{i})((z - 2) - \text{i})
=(z−2)2−i2=(z2−4z+4)−(−1)=z2−4z+5= (z - 2)^2 - \text{i}^2 = (z^2 - 4z + 4) - (-1) = z^2 - 4z + 5

2. Set up a polynomial identity with an unknown quadratic factor.

Let the quartic polynomial be identically equal to the product of this quadratic factor and another quadratic factor. By observing the leading coefficient, the second factor must start with 4z24z^2:
4z4−12z3+17z2+pz+q≡(z2−4z+5)(4z2+cz+d)4z^4 - 12z^3 + 17z^2 + pz + q \equiv (z^2 - 4z + 5)(4z^2 + cz + d)

3. Expand the right-hand side and compare the coefficients of $z^3$ and $z^2$ to find $c$ and $d$.

Expanding the right-hand side:
=4z4+cz3+dz2−16z3−4cz2−4dz+20z2+5cz+5d= 4z^4 + cz^3 + dz^2 - 16z^3 - 4cz^2 - 4dz + 20z^2 + 5cz + 5d
=4z4+(c−16)z3+(d−4c+20)z2+(5c−4d)z+5d= 4z^4 + (c - 16)z^3 + (d - 4c + 20)z^2 + (5c - 4d)z + 5d

Comparing the coefficients of z3z^3:
c−16=−12  ⟹  c=4c - 16 = -12 \implies c = 4

Comparing the coefficients of z2z^2:
d−4c+20=17d - 4c + 20 = 17
Substitute c=4c = 4:
d−4(4)+20=17  ⟹  d+4=17  ⟹  d=13d - 4(4) + 20 = 17 \implies d + 4 = 17 \implies d = 13

4. Compare the remaining coefficients to solve for $p$ and $q$.

Comparing the coefficients of zz:
p=5c−4dp = 5c - 4d
Substitute c=4c = 4 and d=13d = 13:
p=5(4)−4(13)=20−52=−32p = 5(4) - 4(13) = 20 - 52 = -32

Comparing the constant terms:
q=5dq = 5d
Substitute d=13d = 13:
q=5(13)=65q = 5(13) = 65

答案

p=−32p = -32, q=65q = 65

10(b)

1. Concisely derive the values of $p$ and $q$ required from part (a) by substituting the given root.

Substitute z=2−iz = 2 - \text{i} into 4z4−12z3+17z2+pz+q=04z^4 - 12z^3 + 17z^2 + pz + q = 0:
4(2−i)4−12(2−i)3+17(2−i)2+p(2−i)+q=04(2-\text{i})^4 - 12(2-\text{i})^3 + 17(2-\text{i})^2 + p(2-\text{i}) + q = 0
4(−7−24i)−12(2−11i)+17(3−4i)+2p−pi+q=04(-7-24\text{i}) - 12(2-11\text{i}) + 17(3-4\text{i}) + 2p - p\text{i} + q = 0
Equating the real and imaginary parts to zero:
−32−p=0  ⟹  p=−32-32 - p = 0 \implies p = -32
−1+2p+q=0  ⟹  q=65-1 + 2p + q = 0 \implies q = 65

2. Identify the complex conjugate root and form the corresponding quadratic factor.

Since the polynomial has real coefficients, the complex conjugate z=2+iz = 2 + \text{i} is also a root.
The quadratic factor corresponding to these two roots is:
(z−(2−i))(z−(2+i))=(z−2)2−i2=z2−4z+5(z - (2-\text{i}))(z - (2+\text{i})) = (z-2)^2 - \text{i}^2 = z^2 - 4z + 5

3. Compare coefficients to find the remaining quadratic factor.

Using p=−32p = -32 and q=65q = 65, the polynomial is 4z4−12z3+17z2−32z+654z^4 - 12z^3 + 17z^2 - 32z + 65.
Let the remaining quadratic factor be 4z2+Bz+C4z^2 + Bz + C:
4z4−12z3+17z2−32z+65=(z2−4z+5)(4z2+Bz+C)4z^4 - 12z^3 + 17z^2 - 32z + 65 = (z^2 - 4z + 5)(4z^2 + Bz + C)
Comparing the constant terms:
5C=65  ⟹  C=135C = 65 \implies C = 13
Comparing the z3z^3 coefficients:
B−16=−12  ⟹  B=4B - 16 = -12 \implies B = 4
Thus, the other factor is 4z2+4z+134z^2 + 4z + 13.

4. Solve the remaining quadratic equation to find the remaining roots.

4z2+4z+13=04z^2 + 4z + 13 = 0
z=−4±16−4(4)(13)8z = \frac{-4 \pm \sqrt{16 - 4(4)(13)}}{8}
z=−4±−1928=−4±8i38=−12±i3z = \frac{-4 \pm \sqrt{-192}}{8} = \frac{-4 \pm 8\text{i}\sqrt{3}}{8} = -\frac{1}{2} \pm \text{i}\sqrt{3}
The other roots are 2+i2+\text{i}, −12+i3-\frac{1}{2} + \text{i}\sqrt{3}, and −12−i3-\frac{1}{2} - \text{i}\sqrt{3}.

答案

2+i,−12+i3,−12−i32+\text{i}, \quad -\frac{1}{2} + \text{i}\sqrt{3}, \quad -\frac{1}{2} - \text{i}\sqrt{3}

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