# Roots of complex polynomial

I have a complex polynomial $z^2 + 2z + 2$.

From trying out roots repeatedly I can see the correct roots are $z = -1 + i$ and $z = -1 - i$

But what is the standard method for getting roots of a complex polynomial like this?

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Depending on what you mean by "like this," I'd say the quadratic formula! – Cam McLeman Feb 11 '12 at 22:08

1. The quadratic formula works! You just need to interpret the square root as the complex square root. So the two roots are given by the two values of $$\frac{-2 +\sqrt{4-8}}{2} = -1 +\frac{\sqrt{-4}}{2} = -1+\sqrt{-1}.$$ Since $\sqrt{-1}$ takes two values, $i$ and $-i$, the two roots are $-1-i$ and $-1+i$.

2. Complex roots to polynomials with real coefficients always come in "conjugate pairs"; that is, if $a+bi$ is a root, so is $a-bi$.

3. The same "tricks" as for real numbers also work for complex numbers: if $r_1$ and $r_2$ are the roots of $az^2+bz+c$, then we will have $az^2+bz+c = a(z-r_1)(z-r_2)$. So you are looking for numbers $r_1$ and $r_2$ such that $r_1r_2 = \frac{c}{a}$ and $r_1+r_2 = -\frac{b}{a}$. Here, you are looking for two complex numbers whose product is $2$ and whose sum is $-2$. If you can find them "by inspection" (staring at the polynomr ealizing what the right answer is), then that's it.Since the polynomial has real coefficients, either you have two real roots which it is easy to verify don't exist), or they are two complex numbers of the form $a+bi$ and $a-bi$, with $a,b$ reals. So we need $(a+bi)+(a-bi) = -2$ and $(a+bi)(a-bi) = 2$. Or, $2a=-2$, $a^2+b^2 = 2$. Hence $a=-1$, $b^2=1$, giving the two solutions.

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If you're asking, it's probably worthwhile looking at Wikipedia's page on the quadratic equation, specifically, the definition at the top, the sections on the quadratic formula (how to solve), as pointed out by Cam McLeman, and the derivations (why it works, and the best way to remember it if you can reconstruct the derivation). Here's the synopsis:

The equation $$ax^2+bx+c=0$$ has solution(s) $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\qquad(a\neq0)$$ because $$ax^2+bx+c=0$$ $$\iff x^2+\frac{b}ax+\frac{c}a=0$$ $$\iff x^2+\frac{b}{a}x+\frac{c}a+\frac{b^2}{4a^2}=\frac{b^2}{4a^2}$$ $$\iff x^2+\frac{b}{a}x+\frac{b^2}{4a^2}=\frac{b^2}{4a^2}-\frac{c}a$$ $$\iff \left(x+\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a^2}$$ $$\iff \left(x+\frac{b}{2a}\right)=\pm\frac{\sqrt{b^2-4ac}}{2a}$$ $$\iff x=-\frac{b}{2a}\pm\frac{\sqrt{b^2-4ac}}{2a}.$$

This is widely touted as the "standard" method for quadratic equations, with real or complex coefficients. Arturo Magidin's comments are also very helpful.

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Generally speaking 1) roots are that of an equation, so you shouldn't really say roots of a polynomial; although the phrase is used and it implies that the polynomial = 0;

For quadratic equations; the root is found using the quadratic formula. Which can be simply derived by completing the square method:

$$ax^2 + bx + c = 0\qquad \mbox{(1)}$$

Now, since

$$(g+h)^2 = g^2 + 2gh + h^2$$

we can divide $\mbox{(1)}$ by $a$ (since $a\ne0$)

$${x^2} + {{b}\over{a}}x + {c\over{a}} = 0$$ $$\implies {x^2} + {{b}\over{a}}x = -{c \over a}$$

Now complete the square by adding $\left({1 \over 2}{b \over a}\right)^2$ to both sides

$$\therefore {x^2} + {{b}\over{a}}x + \left({1 \over 2}{b \over a}\right)^2 = -{c \over a} + \left({1 \over 2}{b \over a}\right)^2$$

$$\left(x + {b \over {2a} }\right)^2 = - {c \over a} + {{b^2} \over {4a^2}}$$

takings square root of both sides:

$$x + {b \over {2a} } = \pm {{\sqrt{b^2 - 4ac}}\over {2a}}$$

Or,

$$x = -{b \over {2a} } \pm {{\sqrt{b^2 - 4ac}}\over {2a}}$$

Simplifying:

$${ x = {-b \pm {{\sqrt{b^2 - 4ac}}} \over {2a}} }$$

There are other ways of deriving the formula; i encourage you to see if you can figure them out.

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