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Here's your image, below.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
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 z^{6} - 1, and with the arctan in the exponent.)Hi SammyS,
How do you derive to that equation?
I'll try to get to it soon.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'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?
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.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?