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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.