Put the complex solutions of an equation under polar form

B

bluemath

New member
Joined
Sep 20, 2015
Messages
43
Hello,
I solved x8 + x6 + x2 + 1 = 0

I found +- i , + - root [(1 + i root(3))/2] , + - root [(1 - i root(3))/2]

I would like to put now these solutions under polar form (I know how to make it in the standard cases)

If we take the solution + root [(1 + i root(3))/2] , I have a problem with the great root and I'm not able to find the polar form.

How to find the remarkable value of the angle please ?

Thanks
 
In general, a complex number of the form a + bi can also be written as (a, bi), and then the standard polar transformations can be used. With that in mind, let's look at the root you specify. First, I'd note the root:

\(\displaystyle \displaystyle \sqrt{\frac{1}{2}\left(1+i\sqrt{3}\right)}=\sqrt[3]{i}=\frac{\sqrt{3}}{2}+\frac{1}{2}\cdot i\)

Then we have: \(\displaystyle \displaystyle a=\frac{\sqrt{3}}{2}\) and \(\displaystyle \displaystyle b=\frac{1}{2}\)

So the polar form of the root is:

\(\displaystyle z=r\left(cos\left(\theta \right)+i\cdot sin\left(\theta \right)\right)\)

Now you can calculate r based on a and b, and \(\displaystyle \theta\) based on the known tangent, and continue from there...
 
bluemath said:
Hello,
I solved x8 + x6 + x2 + 1 = 0

I found +- i , + - root [(1 + i root(3))/2] , + - root [(1 - i root(3))/2]

I would like to put now these solutions under polar form (I know how to make it in the standard cases)

If we take the solution + root [(1 + i root(3))/2] , I have a problem with the great root and I'm not able to find the polar form.

How to find the remarkable value of the angle please ?

Thanks
Click to expand...
Given a complex number c
c = a + i b
and the same complex number C
C = M \(\displaystyle e^{i\, \theta}\, =\, M [cos(\theta)\, +\, i\, sin(\theta)]\)

Now, since C=c,
\(\displaystyle a\, =\, M cos(\theta)\)
and
\(\displaystyle b\, =\, M sin(\theta)\)
We have
\(\displaystyle M\, =\, \sqrt{a^2 + b^2}\)
with the non-negative root taken as a matter of convention. Also the angle theta is generally taken as the Principle Angle given by
\(\displaystyle \theta\, =\, atan2(b, a )\)
where atan2 takes into account the signs of a and b, see
https://en.wikipedia.org/wiki/Atan2
for example. This makes sure the cosine and sine have the proper sign when using a positive M.
 
ksdhart, Denis and Ishuda : many thanks :D

I'm sorry but how to make real root (or mathematical symbols) on the forum please ?
 
Last edited:
bluemath said:
ksdhart, Denis and Ishuda : many thanks :D

I'm sorry but how to make real root (or mathematical symbols) on the forum please ?
Click to expand...
Are you asking how we get the equations, etc. such as
\(\displaystyle C\, =\, e^{i\, \theta}\, =\, M [cos(\theta)\, +\, i\, sin(\theta)]\)
If so, we tend to use LaTex, see
ftp://ftp.ams.org/pub/tex/doc/amsmath/short-math-guide.pdf
for example [it also has other references]. You need to enclose the code in the tag pair [ tex ] [/ tex ] (without the spaces)

You can hit click Reply With Quote to see the actual code. As an example the above equation is coded as
Code: 
[tex]C\, =\, e^{i\, \theta}\, =\, M [cos(\theta)\, +\, i\, sin(\theta)][/tex]
 
ksdhart said:
In general, a complex number of the form a + bi can also be written as (a, bi), and then the standard polar transformations can be used. With that in mind, let's look at the root you specify. First, I'd note the root:

\(\displaystyle \displaystyle \sqrt{\frac{1}{2}\left(1+i\sqrt{3}\right)}=\sqrt[3]{i}\).
Click to expand...

I'm sorry but how did you find this equality please ?
 
bluemath said:
Hello,
I solved x8 + x6 + x2 + 1 = 0

I found +- i , + - root [(1 + i root(3))/2] , + - root [(1 - i root(3))/2]

I would like to put now these solutions under polar form (I know how to make it in the standard cases)

If we take the solution + root [(1 + i root(3))/2] , I have a problem with the great root and I'm not able to find the polar form.

How to find the remarkable value of the angle please ?

Thanks
Click to expand...

First thing I would do is note that if we write \(\displaystyle \displaystyle \begin{align*} X = x^2 \end{align*}\) then the equation is

\(\displaystyle \displaystyle \begin{align*} x^8 + x^6 + x^2 + 1 &= 0 \\ X^4 + X^3 + X + 1 &= 0 \\ X^3 \, \left( X + 1 \right) + 1 \, \left( X + 1 \right) &= 0 \\ \left( X + 1 \right) \left( X^3 + 1 \right) &= 0 \\ \left( X + 1 \right) ^2 \, \left( X^2 - X + 1 \right) &= 0 \\ X + 1 = 0 \textrm{ or } X^2 - X + 1 &= 0 \\ X = -1 \textrm{ or } X &= \frac{1 \pm \sqrt{ \left( -1 \right) ^2 - 4 \cdot 1 \cdot 1 }}{2 \cdot 1} \\ X = -1 \textrm{ or } X &= \frac{1 \pm \sqrt{ -3 }}{2} \\ X = -1 \textrm{ or } X &= \frac{1 \pm \mathrm{i}\,\sqrt{3}}{2} \end{align*}\)

So case 1:

\(\displaystyle \displaystyle \begin{align*} X &= -1 \\ x^2 &= -1 \\ x &= \pm \, \mathrm{i} \end{align*}\)

Case 2:

\(\displaystyle \displaystyle \begin{align*} X &= \frac{1 - \mathrm{i}\,\sqrt{3}}{2} \\ X &= \frac{1}{2} - \mathrm{i}\,\frac{\sqrt{3}}{2} \end{align*}\)

X is obviously in the fourth quadrant. Its magnitude is \(\displaystyle \displaystyle \begin{align*} \sqrt{ \left( \frac{1}{2} \right) ^2 + \left( -\frac{\sqrt{3}}{2} \right) ^2 } = \sqrt{ \frac{1}{4} + \frac{3}{4} } = \sqrt{ 1 } = 1 \end{align*}\) and its angle is \(\displaystyle \displaystyle \begin{align*} -\arctan{ \left( \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} \right) } = -\arctan{ \left( \sqrt{3} \right) } = -\frac{\pi}{3} \end{align*}\)

also any added or subtracted integer multiple of \(\displaystyle \displaystyle \begin{align*} 2\,\pi \end{align*}\) will work. So that means

\(\displaystyle \displaystyle \begin{align*} X &= \frac{1}{2} - \mathrm{i}\,\frac{\sqrt{3}}{2} \\ X &= \mathrm{e}^{ \mathrm{i}\,\left( 2\,\pi\, n - \frac{\pi}{3} \right) } \textrm{ where } n \in \mathbf{Z} \\ x^2 &= \mathrm{e}^{ \mathrm{i}\,\left( 2\,\pi\,n - \frac{\pi}{3} \right) } \\ x &= \left[ \mathrm{e}^{ \mathrm{i}\,\left( 2\,\pi\,n - \frac{\pi}{3} \right) } \right] ^{\frac{1}{2}} \\ x &= \mathrm{e}^{ \mathrm{i}\,\left( \pi\,n - \frac{\pi}{6} \right) } \end{align*}\)

The two principal solutions are

\(\displaystyle \displaystyle \begin{align*} x &= \mathrm{e}^{ \mathrm{i}\,\left( -\frac{\pi}{6} \right) } \\ &= \cos{ \left( -\frac{\pi}{6} \right) } + \mathrm{i}\sin{ \left( -\frac{\pi}{6} \right) } \\ &= \frac{\sqrt{3}}{2} - \frac{1}{2}\,\mathrm{i} \end{align*}\)

and

\(\displaystyle \displaystyle \begin{align*} x &= \mathrm{e}^{ \mathrm{i}\,\left( \frac{5\,\pi}{6} \right) } \\ &= \cos{ \left( \frac{5\,\pi}{6} \right) } + \mathrm{i}\sin{ \left( \frac{5\, \pi}{6} \right) } \\ &= -\frac{\sqrt{3}}{2} + \frac{1}{2}\,\mathrm{i} \end{align*}\)

Can you try and find the solutions for Case 3, where \(\displaystyle \displaystyle \begin{align*} X = x^2 = \frac{1}{2} + \mathrm{i}\,\frac{\sqrt{3}}{2} \end{align*}\)?
 
bluemath said:
I'm sorry but how did you find this equality please ?
Click to expand...

You guys had me worried there for a second that I made a mistake and that what I posted wasn't true. Whew! :) But, to the answer the question, you can prove this equality by rewriting the given form:

\(\displaystyle \sqrt{\frac{1}{2}\left(1+i\sqrt{3}\right)}=\sqrt{ \frac{1}{2}}\cdot \sqrt{1+\sqrt{3}i}\)

If we apply De Moivre's Theorem to the second term in the above expression, we get:

\(\displaystyle \displaystyle \sqrt{1+\sqrt{3}i}=\sqrt{2}\cdot \left(cos\left(\frac{\frac{5\pi }{3}+2k\pi }{2}\right)+i\cdot sin\left(\frac{\frac{5\pi \:}{3}+2k\pi}{2}\right)\right)=\sqrt{2}\cdot \left(cos\left(\frac{5\pi}{6}+k\pi\right)+i\cdot \:sin\left(\frac{5\pi}{6}+k\pi\right)\right)\)

Valid values of k are 0 and 1. The result we need comes when k=1, so plug that in:

\(\displaystyle \sqrt{2}\cdot \left(cos\left(\frac{5\pi }{6}+\pi \right)+i\cdot sin\left(\frac{5\pi }{6}+\pi \right)\right)=\sqrt{2}\cdot \left(cos\left(\frac{11\pi }{6}\right)+i\cdot sin\left(\frac{11\pi }{6}\right)\right)=\sqrt{2}\cdot \left(\frac{\sqrt{3}}{2}+\frac{1}{2}i\right)\)

Returning to the original expression, we now have:

\(\displaystyle \sqrt{\frac{1}{2}}\cdot \sqrt{2}\cdot \left(\frac{\sqrt{3}}{2}+\frac{1}{2}i\right)=\frac{\sqrt{2}}{2}\cdot \sqrt{2}\cdot \left(\frac{\sqrt{3}}{2}+\frac{1}{2}i\right)=\frac{\sqrt{3}}{2}+\frac{1}{2}i\)

And I have this value memorized, through repeated encounters in my own course work, as one of the cubic roots of i. This also can be proven using several methods, one of which is shown in this post on Math Stack Exchange
 
Something to always remember:

The easiest way to exponentiate a complex number is when it is in its EXPONENTIAL form, because then you can follow your index laws!
 
@ prove it : thanks for your solution, really nice.

@ ksdhart : thanks for the complement.

Still many thanks to all for your great help :)
 
bluemath said:
@ prove it : thanks for your solution, really nice.

@ ksdhart : thanks for the complement.

Still many thanks to all for your great help :)
Click to expand...

My solution is not complete though, you still have some work to do...
 
You must log in or register to reply here.
Top