I am trying to solve the following set of equations for $L'$ as described in this publication excerpt:

enter image description here

So basically, to re-write that, I have the following series of equations:


$b=-(75X^2 + 147Y^2)/80$

$c = -9X^3/16$

$P = 2a^3−9ab+ 27c$

$Q = P^2 -4(a^2-3b)^3$

$w_1=−0.5 +\frac{i\sqrt{3}}{2}$

$w_2=−0.5 -\frac{i\sqrt{3}}{2}$

Which lead to the following three possible solutions:

$L'_1 = -\frac{1}{3} * [a+\sqrt[3]{0.5*(P+\sqrt{Q})} +\sqrt[3]{0.5*(P-\sqrt{Q})}]$

$L'_1 = -\frac{1}{3} * [a+w_2\sqrt[3]{0.5*(P+\sqrt{Q})} +w_1\sqrt[3]{0.5*(P-\sqrt{Q})}]$

$L'_1 = -\frac{1}{3} * [a+w_1\sqrt[3]{0.5*(P+\sqrt{Q})} +w_2\sqrt[3]{0.5*(P-\sqrt{Q})}]$

I am now trying to solve $L'$, but I'm not actually sure how to do it.

My problem is for any input of $X$ or $Y$, $P$ I believe can end up positive or negative, but $Q$ is always a negative.

So let's say hypothetically $P = 2$ and $Q = -3$, then how do you solve:

$\sqrt[3]{0.5*(P-\sqrt{Q})} = \sqrt[3]{1-0.5*i\sqrt{3}}$

What would be the next step?

Basically, I don't know how to solve these equations if $Q$ is negative or if $0.5*(P-\sqrt{Q})$ is negative. I don't know much about complex math. I've used Euler's formula in other situations but I'm not sure here.

Any help? Thanks.


2 Answers 2


For this equation, I should strongly recommend to use the trigonometric method for three real roots since $$\Delta=6914880 \left(175 X^4 Y^2+230 X^2 Y^4+147 Y^6\right) >0 \quad \forall X,Y$$ $$p=-\frac{49}{240} \left(5 X^2+9 Y^2\right)\qquad\qquad q=\frac{49}{4320} X \left(27 Y^2-35 X^2\right)$$ which gives for the roots $$L'_{k}=-\frac X 6+\frac{7 \sqrt{5 X^2+9 Y^2}}{6 \sqrt{5}}\times $$ $$\cos \left(\frac{2 \pi k}{3}-\frac{1}{3} \cos ^{-1}\left(-\frac{\sqrt{5} X \left(27 Y^2-35 X^2\right)}{7 \left(5 X^2+9 Y^2\right)^{3/2}}\right)\right)$$ with $k=0,1,2$.

  • $\begingroup$ Oh WOW Claude! If that works, that is far easier. I have just been working through all the Euler equations with complex numbers and endless atan, sin, cos, and i functions based on the first answer I received, which was starting to seem unsolvable and insanely complex. This is much simpler and mathematically less costly. I need to calculate this a lot so I can't afford so many esoteric functions. Now just to be clear, the cos and inverse cos functions here are radian based functions, right? I mean in C++ cos and inverse cos are radian based, so this should work smoothly, right? Thanks so much. $\endgroup$
    – mike
    Nov 16, 2020 at 4:37
  • $\begingroup$ Can you also explain if you have a moment what the first line means or where that came from? What's Δ or ∀X,Y represent? Where did 6914880 come from? Sorry if this is too stupid. The only math course I took in undergrad was "intro to stats". :) If it's too much bother don't worry. Thanks either way. $\endgroup$
    – mike
    Nov 16, 2020 at 4:45
  • $\begingroup$ @mike. I only know radians. For the remaining go to en.wikipedia.org/wiki/Cubic_equation and just follow the steps. $\endgroup$ Nov 16, 2020 at 4:54
  • $\begingroup$ Thanks again Claude. Very appreciated. $\endgroup$
    – mike
    Nov 16, 2020 at 5:02

In the next step, you want the cube root of the complex number $z = 1-i\sqrt{3}/2$. First express this number in the form $z = re^{i\theta}$ using Euler's identity. For a number $z = a + i b$ we have $ r = \sqrt{a^2 + b^2}$ and $\theta = \tan^{-1}(b/a)$:

$$ z = \sqrt{1^2 + (\sqrt{3}/2)^2} e^{i\tan^{-1}(-\sqrt{3}/2) + 2\pi i n} = \frac{1}{2}\sqrt{7}e^{i \tan^{-1}(\sqrt{3}/2) + i 2\pi n} .$$ Here $n$ is an arbitrary integer which accounts for the periodicity of $e^{i x}$.

Then you can take the cube root:

$$ z^{1/3} = \frac{7^{1/6}}{2^{1/3}} e^{i \tan^{-1}(-\sqrt{3}/2)/3 + i 2 \pi n/3} = \frac{7^{1/6}}{2^{1/3}} \big[ \cos( \tan^{-1}(-\sqrt{3}/2)/3) + i \sin (\tan^{-1}(-\sqrt{3}/2)/3)\Big] e^{i 2 \pi n/3} $$

which you can simplify as desired. Notice it's multi-valued ($n = 0, \pm 1, \dots$). These different values are branches.

  • $\begingroup$ Wow thanks okay. I didn't know about that relationship or those equations. I understand the principle roughly of what you were doing with $n$ but I will just be setting $n=0$ as I'm just looking for the simplest solution I believe. $\endgroup$
    – mike
    Nov 16, 2020 at 2:24

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