algebra-precalculuscomplex numbers
For something like:
$$
z^4 + 8z^2 + 3
$$
how can I find all the complex roots using the quadratic equation, or is there a better method?
I tried a u substitution using $u = z^2$, but then when I applied the quadratic formula, I was getting real roots and there was no where to sub u back in.
The quadratic functions are generally in the form $f(x)=Ax^2+Bx+C$.
The zeroes of this function is then the solution to the equation $Ax^2+Bx+C=0$.$($where $A\ne 0)$
Hence we try to solve this equation.
If you try to solve the equation (just like you'd solve a linear equation), you'd run into trouble when you realize that the $x$s can't be as easily isolated like in the case with linear equations. So we use a different approach, what is called the "completion of squares".
If you are familiar with the method of "completing the squares" (and it isn't very difficult,) you'd know that our goal is to somehow "arrange" the equation in the form $(\text{linear binomial})^2+\text{constant term}$.
When you arrange the equation in this way, isolating the $x$s is very easy.
I'll not show you the entire process of completing the square for our given function but I presume that you are somewhat familiar with it (if not, I suggest you look it up on the internet).
The end result of our method will yield a very familiar expression, what we are used to calling as the "quadratic formula". You should end up with something like this:
$$x=\frac{-B}{2\cdot A} \pm \frac{\sqrt{B^2-4\cdot A\cdot C}}{2\cdot A}$$
where $A$, $B$ and $C$ are the coeffients of our function $f(x)$.
The relation above should give you the zeroes of any quadratic function with coefficients $A$, $B$ and $C$ ( Also note that $A$, $B$, $C \in \mathbb R$). Note that this particular relation has many take-aways. First it hints you of the existence of $2$ zeroes (notice the $\pm$ sign in our relation). Second it kind of explains why $ A \ne 0$ (if $A=0$ then the whole relation goes meaningless, for when you divide any thing by $0 $; in our case $2 \times 0 $, the result is undefined.)
The third, and probably the most important one, is that it tells you when $x$ is a Real number or when it is not (i.e when it is a Complex number). Remember that while $A$, $B$ and $C$ must always be Real numbers, it isn't necessary for $x$ to be Real. It is a well known fact square roots of negative numbers are not Real number, instead, we call them the imaginary numbers $($ and the complex of Real and Imaginary numbers are called the Complex Numbers, usually in the form $a+b\mathbb i $ and $\mathbb i =\sqrt{-1})$.
So,
The geometric interpretation of these three cases is also pretty interesting and gives a deeper insight at what is going on (Although your question has already been answered, i suggest you keep reading, I'm almost done :)).
You should know that the zeroes of a function are those values of $x$ which results in $f(x)$ being equal to zero. Or, if you graph the function, It is that point on the graph(or that value for $x$ ), where the function crosses the x-axis. So, what do these three cases imply on the graph then?
As far as your question is concerned, If you'd like to know if the zeroes are Real or not, check if $B^2 -4AC$ is positive, negative or zero, the nature of the roots (zeroes) should be apparent.
I bothered to write down such long an answer because I really wanted you to know what's going on behind the scenes. You might already know this stuff but still, I Hope it helps
If a polynomial is given numerically, (coefficients $a_0..a_n$ are given), the resultant method can be used to get at a numerical value of the discriminant. The coefficients of the polynomial and its derivative are put in a (n+2)- squared Sylvester matrix. Then the determinant is the discriminant wanted. Where writing out the discriminant of a matrix containing symbols is prohibitive, the discriminant can be calculated swiftly numerically using existing matrix packages in $O(N^3)$ time.
It is very nicely explained in
http://www2.math.uu.se/~svante/papers/sjN5.pdf
This is example 4.7
If $f(x) = ax^4 + bx^3 + cx^2 + dx + e$, then Theorem 3.3 yields
\begin{align} \begin{array}{| ccccccc |} & a & b & c & d & e & 0 & 0 &\\ & 0 & a & b & c & d & e & 0 \\ & 0 & 0 & a & b & c & d & e \\ & 4a& 3b& 2c& d & 0 & 0 & 0 \\ & 0 & 4a& 3b& 2c& d& 0 & 0 \\ & 0 & 0 & 4a& 3b& 2c& d & 0 \\ & 0 & 0 & 0 & 4a& 3b& 2c& d \\ \end{array} \end{align}
= $b^2c^2d^2 - 4b^2c^3 e - 4b^3d^3 + 18b^3cde - 27b^4e^2 - 4ac^3d^2 + 16ac^4e + 18abcd^3 - 80abc^2de - 6ab^2d^2e + 144ab^2ce^2 - 27a^2d^4 + 144a^2cd^2e - 128a^2c^2e^2 - 192a^2bde^2 + 256a^3e^3$
Best Answer
Presumably you got $u=z^2=-4\pm\sqrt{13}$. Now you need to solve for $z$. Both of our values of $z^2$ are negative. The solutions of $z^2=-4+\sqrt{13}$ are $z=\pm i \sqrt{4-\sqrt{13}}$, and the solutions of $z^2=-4-\sqrt{13}$ are $z=\pm i\sqrt{4+\sqrt{13}}$.
Remark: Things get somewhat more complicated when you want to find the square roots of a general complex number. For example, suppose that we want to find the square roots of $3+4i$. One usual way is to first rewrite $z$ as $$\sqrt{3^2+4^2} \left( \frac{3}{\sqrt{3^2+4^2}}+ \frac{4}{\sqrt{3^2+4^2}}i\right).$$ Note that $\sqrt{3^2+4^2}=5$. Let $\theta$ be an angle whose cosine is $\frac{3}{5}$ and whose sine is $\frac{4}{5}$. Then the square roots of $z$ are $$\pm \sqrt{5}\left(\cos\frac{\theta}{2}+\sin\frac{\theta}{2}i\right).$$