# Complex Numbers - Finding Roots

Hi there I was wondering if someone could help me? I am struggling to find the roots of the polynomial

$z^4+2z+3=0$

It is not a quadratic so can't use the quadratic formula so am not quite sure what to do. What is normally a good way to tackle complex polynomials of this form?

Thanks for your help.

Edit: I needed its roots to answer the question 67(f), but looks like it would be very difficult to find the reisudes from those singularities. I am guessing there might have been a mistake in the question and that they meant $x^2$:

So wanted a way to do it without using wolfram alpha

• Why do you need its sick roots? Aug 10, 2014 at 12:54
• You can try mediacru.sh/xdemgsRCo3xK Aug 10, 2014 at 12:55
• @flawr Really cool. I would up vote you posted it as an answer. Aug 10, 2014 at 12:57
• I too suspect $x^2$ was meant. Do you mind saying what book is this from? Aug 10, 2014 at 13:25
• It was from a Hmk sheet at university but annoyingly that question didn't have a solution! Aug 10, 2014 at 13:54

Here is a neat little trick which I stole from some chap named Ferrari. It works for all equations of the form $x^4 + ax^2 + bx + c = 0$, with the only difference being a completing-of-the-square at the first step. See here for a reference (I chose this link because the story is almost as interesting as the maths!).

\begin{align*} z^4+2z+3&=0\\ \Rightarrow (z^2)^2&=-2z-3\\ \Rightarrow (z^2+y)^2&=-2z-3+2yz^2 + y^2&\forall\: y\\ &=(2y)z^2 - 2z + (- 3 + y^2) \end{align*}

Now the right hand side is a quadratic in $z$ and we can choose $y$ so that the right hand side has the form $(\alpha z+\beta)^2$. So, take the almighty formula and stipulate that $b^2-4ac=0$: \begin{align*} (-2)^2 -4(2y)(- 3 + y^2) &= 0\\ \Rightarrow4 + 24y - 8y^3 &= 0 \\ \Rightarrow1 + 6y - 2y^3 &= 0 \end{align*} This is a cubic in $y$, which we can solve$^{\dagger}$. We get $y=-1.6418$, $y=-0.16825$ and $y=1.8100$. These values then mean you have the following formula, where $\alpha$ and $\beta$ are numbers dependent on our value of $y$. \begin{align*} (z^2+y)^2&=(\alpha z+\beta)^2\\ \Rightarrow z^2+y&=\alpha z+\beta \end{align*} Solve as usual.

$^{\dagger}$ Okay, I admit, I did this bit using Wolfram-Alpha...but you can do it by hand, and it isn't too difficult. I am just running late for my tea...

Here's a crazy attempt:

If you write down the equation as

$(z^2)^2 + 2z + 3 = 0$,

and you behave as if it's a quadratic polynomial $p(z)$, i.e. use the formula for quadratic equations, you would get that

$$z = \dfrac{-1 + \sqrt{1-3z^2}}{z^2}$$ $$z = \dfrac{-1 - \sqrt{1-3z^2}}{z^2}$$

Now, if you solve for $z$ in these equations you will get the solutions for the original equation.

• No, it does not. If you checked the solutions of the equations I wrote, you would see that the roots are the same. Aug 10, 2014 at 13:34
• I figured out what you meant, the notation with $z1$ and $z2$ both used for meaning $z$ was really confusing. I've made a suggested edit which should make it more clear what you mean. Aug 10, 2014 at 13:43
• I don’t understand. You’re taking an equation of form $T^2+2z+3=0$ and saying it’s quadratic in ? Aug 10, 2014 at 13:53
• Darksonn: Understood, thanks for the edit; @Lubin, I'm taking an equation in the form $az^2+bz+c$, only with the difference that $a$ is not a constant, but $a = z^2$. I don't really know if that is proper, but it seems to be working. I was hoping someone more knowledgeable than me would discuss the idea. Aug 10, 2014 at 13:57
• @Eutherpy Your idea works. It's not exactly helpful, but it is correct as per the following. For all $z\in \mathbb C$ with $z\neq 0$ and with a proper interpretation of $\sqrt \cdot$ \begin{align} z^4+2z+3=0&\iff z^2+\dfrac 2 z+\dfrac 3{z^2}=0\\ &\iff z^2+\dfrac 2 z=-\dfrac 3{z^2}\\ &\iff z^2+\dfrac 2 z+\dfrac 1{z^4}=\dfrac 1{z^4}-\dfrac 3{z^2}\\ &\iff \left(z+\dfrac 1{z^2}\right)^2=\dfrac{1-3z^2}{z^4}\\ &\iff z=-\dfrac 1{z^2}\pm \sqrt{\dfrac{1-3z^2}{z^4}}. \end{align} A discussion by cases yields what you wrote. Aug 10, 2014 at 16:37