# Roots of polynomials

A square root of $$c=a+ib$$ is a number $$z=x+iy$$ so that $$$z^2= c.$$$

Nonzero **complex numbers** have two square roots. This differs from real numbers, where only positive numbers have (real) square roots.

It is possible to compute the root of a complex number in **Cartesian form**, although it is simpler in **polar form**, which will be covered later.

In Cartesian form, equating real and imaginary parts, we find $$$ x^2-y^2 = a,\quad 2xy = b.$$$ We can solve this system of equations.

Let's illustrate this with the square roots of $$-3+4i.$$ The equations are $$$x^2 - y^2 = -3,\quad 2xy = 4.$$$ From the second equation we get $$x=2/y$$. From the first, we obtain $$(2/y)^2 - y^2 = -3$$ or $$y^4 - 3y^2 - 4 = 0$$. The positive solution is $$y^2 = 4$$. This gives $$y = \pm 2$$, and the second equation gives $$x=\pm1$$. So the roots are $$\pm (1 + 2i)$$.

The equation $$$ a z^2 + b z +c = 0 $$$ with complex coefficients $$a\ne 0$$, $$b$$ and $$c$$ is solved the same way as real quadratic equations. The solutions are, just as for real quadratics, $$$ z = \frac{-b\pm \delta}{2a},\quad \delta^2 = \Delta = b^2 - 4ac. $$$

If the solutions are equal then $$z$$ is said to be a repeated root. Otherwise they are said to be distinct.

The discriminant $$\Delta$$ is a complex number. The equation always has complex solutions.

As an example, we solve the equation $$$2iz^2 - 3z - 5 = 0$$$

We have $$a=2i$$, $$b=-3$$ and $$c=-5$$ so roots are $$$ z = \frac{3 \pm \delta}{4i},\qquad \delta ^2 = 9+40i. $$$ We compute $$\delta = 5+4i$$ and obtain $$z = \big(3 \pm (5+4i)\big)/(4i)$$. This simplifies to $$$ z = \frac{8+4i}{4i}= 1-2i,\quad z = \frac{-2-4i}{4i} = -1 + \frac{i}{2}. $$$

**Polynomial equations always have complex solutions**. For instance, $$z^2+1 = 0$$ has two complex solutions, $$i$$ and $$-i$$, but no real solutions.

This is the fundamental theorem of algebra. It is equivalent to saying that every non-constant polynomial $$P_n$$ of degree $$n$$ can be factorised into a product of $$n$$ linear factors: $$$P_ n(z)=a(z-z_1)(z-z_2)\cdots (z-z_n).$$$ The $$z_i$$ are the **roots** of $$P_n$$.

The number of repetitions of a root among the $$z_i$$ is called the multiplicity of that root. For example, in $$(x-i)^5$$ the root $$i$$ has multiplicity $$5$$. A root of multiplicity $$1$$ is called simple.

When all the coefficients of the polynomial are **real**, the roots are either **real** or come in pairs as **complex conjugates** with the same multiplicities.

To see this, if $$z_0$$ is a non-real root of polynomial with real coefficient, we have $$$P(z_0)=a_nz_0^n+ \cdots+a_1 z_0+a_0 = 0.$$$ $$$\begin{align*} P(\overline{z}_0) &= a_n\overline{z}_0^n+ \cdots+a_1 \overline{z}_0+a_0\\ &= \overline{a_nz}_0^n+ \cdots+\overline{a_1z}_0+\overline{a}_0\\ &= \overline{ a_nz_0^n+ \cdots+a_1 z_0+a_0}\\ &= \overline{P(z_0)} = \overline{0} = 0\\ \end{align*}$$$ We conclude that $$\overline{z}_0$$ is a root of the polynomial.