请求求解满足方程z⁴+4z³+7z²+6z+3=0的所有复数解

Hey there! Let's break down how to find all complex solutions to this quartic equation. The key here is to factor the quartic into quadratics first, since solving quadratics is straightforward with the quadratic formula.

Step 1: Factor the Quartic into Quadratics #

We start by assuming the quartic can be written as a product of two quadratics:
[
(z^2 + az + b)(z^2 + cz + d) = 0
]
Expanding this product and equating coefficients with the original equation gives us a system of equations:

1. (a + c = 4) (coefficient of (z^3))
2. (ac + b + d =7) (coefficient of (z^2))
3. (ad + bc =6) (coefficient of (z))
4. (bd=3) (constant term)

Testing integer pairs for (b) and (d) (since (bd=3)), we find (b=1) and (d=3) works perfectly. Substituting these into the system:

• From equation 3: (3a + c =6). Using (c=4-a) (from equation1), we get (a=1) and (c=3).
• Checking equation2: (1*3 +1+3=7), which matches.

So the quartic factors to:
[
(z^2 + z +1)(z^2 +3z +3)=0
]

Step2: Solve Each Quadratic #

Now we solve each quadratic separately using the quadratic formula (z=\frac{-B\pm\sqrt{B^2-4AC}}{2A}):

First Quadratic: (z^2 + z +1=0)

[
z = \frac{-1\pm\sqrt{1-4}}{2} = \frac{-1\pm i\sqrt{3}}{2}
]

Second Quadratic: (z^2 +3z +3=0)

[
z = \frac{-3\pm\sqrt{9-12}}{2} = \frac{-3\pm i\sqrt{3}}{2}
]

All Complex Solutions #

Putting it all together, the four complex roots are:

• (z_1 = \frac{-1 + i\sqrt{3}}{2})
• (z_2 = \frac{-1 - i\sqrt{3}}{2})
• (z_3 = \frac{-3 + i\sqrt{3}}{2})
• (z_4 = \frac{-3 - i\sqrt{3}}{2})

You can verify each root by plugging it back into the original equation—they all satisfy the equality!

内容的提问来源于stack exchange，提问作者Nashenas