Solve the polynomial equation in the complex numbers

The Fundamental Theorem of Algebra assures us that any polynomial with real number coefficients can be factored completely over the field of complex numbers .

In the case of quadratic polynomials , the roots are complex when the discriminant is negative.

Example 1:

Factor completely, using complex numbers.

x 3 + 10 x 2 + 169 x

First, factor out an x .

x 3 + 10 x 2 + 169 x = x ( x 2 + 10 x + 169 )

Now use the quadratic formula for the expression in parentheses, to find the values of x for which x 2 + 10 x + 169 = 0 .

Here a = 1 , b = 10 and c = 169 .

x = − b ± b 2 − 4 a c 2 a

x = − 10 ± 10 2 − 4 ( 1 ) ( 169 ) 2 ( 1 ) = − 10 ± 100 − 676 2 = − 10 ± − 576 2

Write the square root using imaginary numbers.

x = − 10 ± 24 i 2 = − 5 ± 12 i

We now know that the values of x for which the expression

x 2 + 10 x + 169

equals 0 are x = − 5 + 12 i   and   x = − 5 − 12 i .

So, the original polynomial can be factored as

x 3 + 10 x 2 + 169 x = x ( x − [ − 5 + 12 i ] ) ( x − [ − 5 − 12 i ] )

You can verify this using FOIL .

Sometimes, you can factor a polynomial using complex numbers without using the quadratic formula. For instance, the difference of squares rule:

x 2 − a 2 = ( x + a ) ( x − a )

This can also be used with complex numbers when a 2 is negative, as follows:

x 2 + 25 = ( x + 5 i ) ( x − 5 i )

Example 2:

Factor completely, using complex numbers.

9 x 2 y + 64 y

First, factor out y .

9 x 2 y + 64 y = y ( 9 x 2 + 64 )

Now, use the difference of square rule to factor 9 x 2 + 64 .

9 x 2 + 64 = 9 x 2 − ( − 64 ) = ( 3 x ) 2 − ( 8 i ) 2 = ( 3 x + 8 i ) ( 3 x − 8 i )

Therefore, 9 x 2 y + 64 y = y ( 3 x + 8 i ) ( 3 x − 8 i ) .

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1. Solve: $$ z^3 - 3z^2 + 6z - 4 = 0$$

How do I solve this?

Can I do it by basically letting $ z = x + iy$ such that $ i = \sqrt{-1}$ and $ x, y \in \mathbf R $ and then substitute that into the equation and get a crazy long equation? If I did that I suspect I wouldn't be able to decipher the imaginary part of the equation.

Or should I change it to one of the forms below:

$$ z^n = r^n \mathbf{cis} n \theta $$ $$ z^n = r^n e^{n\theta i} $$

And then plug that into the equation? I did that. But it looked unsolvable. I'm so confused.

Adam

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asked Sep 5, 2013 at 14:17

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Since complex number field $\mathbb{C}$ is algebraically closed, every polynomials with complex coefficients have linear polynomial decomposition. In this case, it's $$ z^3 - 3z^2 + 6z - 4 = (z - 1)(z - 1 + \sqrt{3}i)(z - 1 - \sqrt{3}i). $$ So you can see the solution of the equation easily from this representation.

One way to find out such decomposition is simply put $$ z^3 - 3z^2 + 6z - 4 = c(z - \alpha)(z - \beta)(z - \gamma) $$ to find out the coefficients by equating coefficients of power of $z$.

answered Sep 5, 2013 at 14:31

OratOrat

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If you substitute in $z=x+iy$ the real and imaginary parts are each cubics-you just sort out the terms that have a factor $i$ for the imaginary part and you have a real equation. But as Daniel Fischer says, there is an easier approach here. To see what happens, we have $$(x+iy)^3-3(x+iy)^2+6(x+iy)-4=0\\x^3+3ix^2y-3xy^2-iy^3-3x^2-6ixy+3y^2+6x+6iy-4=0\\x^3-3xy^2-3x^2+3y^2+6x-4=0\\3x^2y-y^3-6xy+6y=0$$ where the second came from the imaginary part. It is no worse than the real part, which is the next to last line.

answered Sep 5, 2013 at 14:31

Ross MillikanRoss Millikan

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$z=x+1$ $\Longrightarrow$ $x^{3}+3x=0$

$\therefore$ $x=0$ , $\pm\sqrt{3}i$

$\therefore$ $z=1$ , $1\pm\sqrt{3}i$

answered Sep 5, 2013 at 14:57

chloe_shichloe_shi

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The easiest thing is just try to guest a root of the polynomial first. In this case, for

$$p(z) = z^3 - 3z^2 + 6z - 4,$$ we have that $p(1) = 0$.

Therefore, you can factorize it further and get $$z^3 - 3z^2 + 6z - 4 = (z-1)(z^2 - 2z + 4)$$

$$= (z-1)((z-1)^2 + 3).$$

Their roots are just $$z_{1} = 1, \hspace{10pt}z_{2} = 1 + i\sqrt{3}, \hspace{10pt}z_{3} = 1 - i\sqrt{3}.$$

answered Sep 5, 2013 at 14:54

Petch PuttichaiPetch Puttichai

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@Petch Puttichai , after we've guessed the root say $p(1)$, how do we find the other factors , is it that we divide the polynomial with first factor , are there any other methods ?

For example if I have a polynomial of $z^4+z^3+... $ after we find the first root by guessing, how to get the other polynomial which I think should be of the form $z^3+z^2+...$.

Thanks for all your help.

Arif

answered Feb 11, 2016 at 5:54

ArifArif

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