The distinct roots of complex number

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I am trying to find the z0 to z6 roots of this equation but I am stuck here. Anyone care to show the step by step on how to procced?

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The Attempt at a Solution

 

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  • #2
Simon Bridge
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Welcome to PF;

is it that z=x+iy: z6-1=c6 (c=a+ib): a,b,x,y are real, a,b are constants?
so you need to find:

[itex](x+iy)^6-(a+ib)^6-1=0[/itex]

You could multiply out the polynomials and then solve for the real and imaginary parts ... or express each of the complex numbers as exponentials. Which have you tried?
 
  • #3
SammyS
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I am trying to find the z0 to z6 roots of this equation but I am stuck here. Anyone care to show the step by step on how to proceed?

Homework Statement


Homework Equations


The Attempt at a Solution

Here's your image, below.

attachment.php?attachmentid=50796&d=1347591428.gif


It can be easier to take the power of a complex number when it's expressed in "polar" form.

Assuming your last line is correct:

[itex]\displaystyle \frac{192}{5}-j\frac{64}{5}=\frac{64\sqrt{10}}{5}e^{-j\,\tan^{-1}(1/3)}[/itex]
 
  • #4
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Hi SammyS,

How do you derive to that equation?
 
  • #5
SammyS
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Hi SammyS,

How do you derive to that equation?
That's polar form for a complex number. (By the Way, I'm not too sure it will be all that helpful here, what with the -1 part of the z6 - 1, and with the arctan in the exponent.)

In general, you can express the complex number, a + b j as
[itex]
a+bj=\sqrt{a^2+b^2\ }\left(\cos(\theta)+j\sin(\theta)\, \right)\,,[/itex]
where [itex]\displaystyle \tan(\theta)=\frac{b}{a}\ .[/itex]

Using Taylor series, it can be shown that [itex]\displaystyle e^{j\,\theta}=\cos(\theta)+j\sin(\theta)\ .[/itex]

[itex]\displaystyle \cos(\theta)+j\sin(\theta)[/itex] is also called the cis() function, but the exponential form makes it very handy for taking a complex number to a power.
[itex]\displaystyle \frac{192}{5}-j\frac{64}{5}=\frac{64}{5}\left(3-1j\right)[/itex]


[itex]\displaystyle 3-j=\sqrt{3^2+1^2}\left(\cos(\theta)+j\sin(\theta) \right)\,,\ \text{ where }\theta=\arctan\left(\frac{-1}{\ 3}\right)\ [/itex]
[itex]\displaystyle =\sqrt{10}\,e^{\,j\,\arctan(-1/3)} \ .[/itex]​
 
  • #6
Simon Bridge
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caveat: you have to be real careful with the signs inside the arctangent. ##\arctan(\frac{-1}{2}) = \pi + \arctan(\frac{1}{-2})##
(arctan is not a good description of the calculation - what you want is the "argument" of the complex number)

If you have a complex number z, then it can be expressed as
$$z = a+jb=|z|e^{j\arg(z)}$$ ... and you know how to find the modulus and argument of a complex number?

So if z is a ratio of powers of complex numbers, eg. $$z=\frac{x^m}{y^n}$$ then you can write: $$z=\frac{(|x|e^{j\arg(x)})^m}{(|y|e^{j\arg(y)})^n }=\frac{|x|^m e^{jm\arg(x)}}{|y|^n e^{jn\arg(y)} } = \frac{|x|^m}{|y|^n} e^{\left [ j(m\arg(x)-n\arg(y)) \right ]} = Ae^{j\theta} = A\cos(\theta) + jA\sin(\theta)$$
 
  • #7
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Thanks Simon for the warning.

Hey Sammys. I have work out and this is what I came out with.. it looks messy but how do I proceed next?
 

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  • #8
SammyS
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Thanks Simon for the warning.

Hey Sammys. I have work out and this is what I came out with.. it looks messy but how do I proceed next?
I'll try to get to it soon.

Anyone else, feel free to help.
 
  • #9
Simon Bridge
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I'm wondering if it isn't just easier to multiply it out ... we just start with $$z^6-1=\left [ \frac{64j(1-j)}{1+2j} \right ]^6 = 64^6\left [ \frac{j+1}{2j+1} \right ]^6$$

After all:
##(j+1)^2=(j+1)(j+1)=-1+j+j+1=2j##
##(j+1)^4=(2j)(2j)=-4##
##(j+1)^6=-4(2j)=-8j##

... that doesn't look so bad.

Same for the denominator; then rationalize the denominator to get the RHS in form ##\alpha+j\beta## and procede polar of whatever from there ... ##z^6## is likely to be a pain in terms of ##a+jb## but it's doable in polar form - 2 equations, solve for a and b. Or solve for the amplitude and phase that give zeros and convert afterwards?
 
  • #10
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I'm wondering if it isn't just easier to multiply it out ... we just start with $$z^6-1=\left [ \frac{64j(1-j)}{1+2j} \right ]^6 = 64^6\left [ \frac{j+1}{2j+1} \right ]^6$$

After all:
##(j+1)^2=(j+1)(j+1)=-1+j+j+1=2j##
##(j+1)^4=(2j)(2j)=-4##
##(j+1)^6=-4(2j)=-8j##

... that doesn't look so bad.

Same for the denominator; then rationalize the denominator to get the RHS in form ##\alpha+j\beta## and procede polar of whatever from there ... ##z^6## is likely to be a pain in terms of ##a+jb## but it's doable in polar form - 2 equations, solve for a and b. Or solve for the amplitude and phase that give zeros and convert afterwards?


Hi Simon,

Is it possible to show me your method starting from $$z^6-1=\left [ \frac{64j(1-j)}{1+2j} \right ]^6$$ till the end equation $$z^6$$= ##a+jb##
please?
 
  • #11
Simon Bridge
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That last bit is notation.

If I just wanted to find the roots of ##z^6-1## I'd put ##z^6-1=0##, then expand z in terms of a and b as in ##(a+jb)^6-1=0##.

This is a bit like that, only you have another complex number, which is a constant, in there:

##z^6+1=C##

So I want to be able to write ##C=\alpha+j\beta## then I can just equate the real and imaginary parts for two simultaneous equations with two unknowns. That's the idea - the rest is strategy to make it look like that.

I was just wondering if the difficult-looking complex number is actually much simpler than it looks. We've been scared of multiplying out the power of six. After all, it should be easier to handle ##z^6-1=\alpha+j\beta## shouldn't it?

Anyway - to illustrate the point, I multiplied out the numerator for you ... it turns out to be quite simple. You'll have to do the rest I'm afraid.

Where does this come from - it seems unusually obnoxious for homework?
 
  • #12
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That last bit is notation.

If I just wanted to find the roots of ##z^6-1## I'd put ##z^6-1=0##, then expand z in terms of a and b as in ##(a+jb)^6-1=0##.

This is a bit like that, only you have another complex number, which is a constant, in there:

##z^6+1=C##

So I want to be able to write ##C=\alpha+j\beta## then I can just equate the real and imaginary parts for two simultaneous equations with two unknowns. That's the idea - the rest is strategy to make it look like that.

I was just wondering if the difficult-looking complex number is actually much simpler than it looks. We've been scared of multiplying out the power of six. After all, it should be easier to handle ##z^6-1=\alpha+j\beta## shouldn't it?

Anyway - to illustrate the point, I multiplied out the numerator for you ... it turns out to be quite simple. You'll have to do the rest I'm afraid.

Where does this come from - it seems unusually obnoxious for homework?
Will try it out. Actually, this is not homework. I am just trying out past year exam questions. Sad to say, solutions is not provided. So I will never know whether I am doing it right or wrong.
 
  • #13
Simon Bridge
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Exam question??? ... then, either look for a trick in the course-notes for the year, or that year covered similar examples in assignments so they'd remember some special approach.
 
  • #14
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SOLVED!! Thanks a lot guys!! Appreciate it.
 

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