2

In my recent post I have a problem with the following function: $x^4-4x^2+16$, and what I need is to find the complex roots.

Here is my answer:

First step, I make the substitution $x^2=y$ which involving $y^2-4y^2+16$, with $x^2=2\pm i\sqrt{12}$. Therfore:

$x^4-4x^2+16=(x^2-2-i\sqrt{12})(x^2-2+i\sqrt{12}) =(x+\sqrt{2+i\sqrt{12}})(x+\sqrt{2-i\sqrt{12}})(x-i\sqrt{2+i\sqrt{12}})(x-\sqrt{2-i\sqrt{12}})$

factorized irreducible over $\mathbb{R}\subset\mathbb{C}$.

Here is author's answer:

$x{_1,_2,_3,_4}=\pm(\sqrt{3}\pm i)$.

I realize the problem is at factorization: $x^4-4x^2+16=(x^2+bx+c)(x^2+dx+e)$. If I pair the complex roots from author's answer I'll obtain:

$x^4-4x^2+16=(x+\sqrt{3}+i)(x+\sqrt{3}-i)(x-\sqrt{3}-i)(x-\sqrt{3}+i)=(x^2+\sqrt{12}x+4)(x^2-\sqrt{12}x+4)$

factorized irreducible over $\mathbb{Q}\subset\mathbb{R}$, but reducible over $\mathbb{C}$.

My final question: Why $x_1{,_2,_3,_4}=\pm\sqrt{2\pm i\sqrt{12}}$ isn't enough to make the assessment that it is the final answer ? How can I get author's answer $x{_1,_2,_3,_4}=\pm(\sqrt{3}\pm i)$ begining from my complex roots found ? Why we obtain 2 different form of complex roots, who's belong in the same field $\mathbb{C}$ when we change factorization ?

Lucas
  • 1,865

4 Answers4

2

Hint:

with your substitution, the solutions that you have are: $$ x^2=2(1\pm i\sqrt{3})=4 e^{\pm i \frac{\pi}{3}} $$ and this gives immediately the autor's answer.

added:

The complex number $a+ib=2+2i\sqrt{3} \rightarrow $ can be write in polar form $\rho e^{i\theta}$ with: $$ \rho=\sqrt{a^2+b^2}=\sqrt{4+12}=4 $$ $$ \theta= \arctan\left(\dfrac{b}{a}\right)=\arctan\left(\dfrac{2\sqrt{3}}{2}\right)= \dfrac{\pi}{3} $$

And if $x^2=4 e^{\pm i \frac{\pi}{3}}$ we have $x=\pm 2 e^{\pm i \frac{\pi}{6}}$, that, in binomial form, are the solutions.

Emilio Novati
  • 62,675
  • which is the phase with this method? is my first time when I see... can you give me more details? – Lucas May 02 '15 at 15:20
  • how you obtain this equality $2\pm i\sqrt{12})=4 e^{\pm i \frac{\pi}{3}}$ ? about what method is it? – Lucas May 02 '15 at 15:22
  • I add to my answer. Just a moment – Emilio Novati May 02 '15 at 15:24
  • can you tell me how can I put in binominal form $\pm 2e^{\pm i \frac{\pi }{6}}$ ? – Lucas May 02 '15 at 15:43
  • http://en.wikipedia.org/wiki/Complex_number#Polar_form. In you case $a+ib=2 \cos (\pi/6)\pm i \sin (\pi/6)=\sqrt{3}\pm i$. – Emilio Novati May 02 '15 at 15:53
  • if you look I obtained 4 roots: $\pm\sqrt{2\pm i\sqrt{12}}$, but wasn't enough to finish, why we need to put $\pm\sqrt{2\pm i\sqrt{12}}$ in this form $\pm\sqrt{3}\pm i$? – Lucas May 02 '15 at 23:28
1

Note that your answer gives that same result as author's answer, but the more important thing to pay attention with is when you use $\sqrt z$ with $z$ a complex numbers this is not defind so in I would say that your answer is incomplete because you did not find the complex square roots of $2\mp i\sqrt {12}$, and when you did this you will find the same result as author's.

Another way to see things is that any complex number is of the form $x+iy$ where $x,y\in \Bbb R$ which is the canonical form. and your answer does not give the roots of this form.

Maybe we can use that: $$\begin{align}x^4-4x^2+16&=x^2-2\cdot4x^2+16+4x^2\\ &=(x^2-4)^2+4x^2\\ &=(x^2-2ix-4)(x^2+2ix-4)\\ &=((x-i)^2-3)((x+i)^2-3)\\ &=(x-i-\sqrt3)(x-i+\sqrt3)(x+i-\sqrt3)(x+i+\sqrt3)\end{align}$$

this is how the author did it, but you can simplify your expression, for instance: $$\pm\sqrt{2\pm i\sqrt{12}}=\mp \sqrt{2+\mp 2i\sqrt 3}=\mp \sqrt{3^2+\mp 2i\sqrt 3 +i^2}=\mp \sqrt{(i\mp\sqrt 3)^2}=\pm(\sqrt{3}\pm i)$$

Elaqqad
  • 13,725
  • Can you give me a complete answer, they are more unclear , can you answer at all ? – Lucas May 02 '15 at 15:17
  • I added some explication even if I don't really understand your problem – Elaqqad May 02 '15 at 15:25
  • if you say that I didn't find the complex square roots, than what I found: $\pm\sqrt{2+\pm i\sqrt{12}}$ ? – Lucas May 02 '15 at 15:32
  • You did not solve the equation $x^2=2+\pm i\sqrt{12}$ the sqyare root has no sense for complex numbers you have to take $x=a+ib $ et find $a$ and $b$ such that $(a+ib)^2=2+\pm i\sqrt{12}$ and $\cdots\cdots$ – Elaqqad May 02 '15 at 15:41
1

To denest the square roots you can employ the simple, memorable algorithm here.

Then $\:2\pm i\sqrt{12}\:$ has norm $= 16.\:$ $\rm\ \color{blue}{subtracting\ out}\ \sqrt{norm}\ = 4\ $ yields $\ {-2}\pm i\sqrt{12}\:$

with $\ {\rm\ \sqrt{trace}}\: =\: \sqrt{-4}\ =\ 2i.\quad\ \ \ \rm \color{brown}{Dividing\ it\ out}\ $ of the above yields $\quad\ \ \ i\pm\sqrt 3$

Community
  • 1
Bill Dubuque
  • 272,048
0

all you have to do is show that $\sqrt{2\pm i\sqrt 12} = \sqrt 3 \pm i$ squaring both side should give you that.

abel
  • 29,170