How Do You Form a Degree 4 Polynomial with Given Complex Zeros?

[caters]

Homework Statement

Form a polynomial whose zeros and degree are given below. You don't need to expand it completely but you shouldn't have radical or complex terms.

Degree 4: No real zeros, complex zeros of 1+i and 2-3i

Homework Equations

(-b±√b^2-4ac)/2a

The Attempt at a Solution

I want something along the lines of 1 + sqrt(-1) and 1 - sqrt(-1)

I tried x^2 + x + 1 and here is what I got

(-1+sqrt(3)i)/2 and (-1-sqrt(3)i)/2

There has to be a better way than plugging coefficients into ax^2 + bx + c to find 2 factors, 1 for each complex zero(I only need 2 since complex numbers come in pairs)

I know for real zeros + y intercept I can just solve for a at x = 0

Is there a shortcut to trial and error with complex zeros that aren't a simple multiple of i(multiples of i would mean a sum of squares and that's easy)?

 

[fresh_42]

You said that complex numbers come in pairs. What does this mean?
And do you know how a polynomial can be written, if you know all its roots?

 

[member 587159]

Homework Statement

Form a polynomial whose zeros and degree are given below. You don't need to expand it completely but you shouldn't have radical or complex terms.

Degree 4: No real zeros, complex zeros of 1+i and 2-3i

Homework Equations

(-b±√b^2-4ac)/2a

The Attempt at a Solution

I want something along the lines of 1 + sqrt(-1) and 1 - sqrt(-1)

I tried x^2 + x + 1 and here is what I got

(-1+sqrt(3)i)/2 and (-1-sqrt(3)i)/2

There has to be a better way than plugging coefficients into ax^2 + bx + c to find 2 factors, 1 for each complex zero(I only need 2 since complex numbers come in pairs)

I know for real zeros + y intercept I can just solve for a at x = 0

Is there a shortcut to trial and error with complex zeros that aren't a simple multiple of i(multiples of i would mean a sum of squares and that's easy)?

You can choose 2 zeros yourself? What do you know about:

$z \overline{z}$ where $z \in \mathbb{C} \backslash {R}$

 

[caters]

You said that complex numbers come in pairs. What does this mean?
And do you know how a polynomial can be written, if you know all its roots?

Complex numbers coming in pairs means that I only need 2 factors for 4 zeros instead of 4 factors and that each factor is quadratic.

And I know that the polynomial with complex terms would be (x-(1+i))(x-(1-i))(x-(2+3i))(x-(2-3i))

 

[Ray Vickson]

Homework Statement

Form a polynomial whose zeros and degree are given below. You don't need to expand it completely but you shouldn't have radical or complex terms.

Degree 4: No real zeros, complex zeros of 1+i and 2-3i

Homework Equations

(-b±√b^2-4ac)/2a

The Attempt at a Solution

I want something along the lines of 1 + sqrt(-1) and 1 - sqrt(-1)

I tried x^2 + x + 1 and here is what I got

(-1+sqrt(3)i)/2 and (-1-sqrt(3)i)/2

There has to be a better way than plugging coefficients into ax^2 + bx + c to find 2 factors, 1 for each complex zero(I only need 2 since complex numbers come in pairs)

I know for real zeros + y intercept I can just solve for a at x = 0

Is there a shortcut to trial and error with complex zeros that aren't a simple multiple of i(multiples of i would mean a sum of squares and that's easy)?

Complex roots come in (conjugate) pairs IF the coefficients of the polynomial are real numbers. If the polynomial has complex number coefficients, the roots do not need to come in pairs. For example, the polynomial $p = x^3 - 4 i x^2 -(6-5i)x +5-i$ has roots $1$, $i+1$ and $-2+3i$.

 

[fresh_42]

Complex numbers coming in pairs means that I only need 2 factors for 4 zeros instead of 4 factors and that each factor is quadratic.

Almost. It's not exactly a square, i.e. not exactly the same two zeros. See your own answer below.

And I know that the polynomial with complex terms would be (x-(1+i))(x-(1-i))(x-(2+3i))(x-(2-3i))

So why don't you simply multiply this out?

 

[haruspex]

So why don't you simply multiply this out

... which you will find easier if you first regroup the terms in the factors to separate out the complex parts, e.g. ((x-1)-i) instead of (x-(1+i)).

 

[SammyS]

Complex numbers coming in pairs means that I only need 2 factors for 4 zeros instead of 4 factors and that each factor is quadratic.

What you have above is not quite right, as others have stated.

However, what you have below is correct.

And I know that the polynomial with complex terms would be (x-(1+i))(x-(1-i))(x-(2+3i))(x-(2-3i))

To expand the first two factors and the last two factors, regroup them as follows, then use difference of squares (difference times sum) or product of complex conjugates.

(x-(1+i))(x-(1-i)) = ((x-1) - i)((x-1) + i)

(x-(2+3i))(x-(2-3i)) = ((x-2) - 3i)((x-2) + 3i)