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Here's your image, below.timeforchg said: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 z6 - 1, and with the arctan in the exponent.)timeforchg said:Hi SammyS,
How do you derive to that equation?
I'll try to get to it soon.timeforchg said: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?
Simon Bridge said: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.Simon Bridge said: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?
Complex numbers are numbers that have both a real and an imaginary component. They are written in the form a + bi, where a and b are real numbers and i is the imaginary unit, which is equal to the square root of -1.
Unlike real numbers, which can be plotted on a number line, complex numbers cannot be represented on a standard number line. They require a two-dimensional plane, known as the complex plane, to plot them.
The distinct roots of a complex number refer to the different solutions when trying to find the square root of a complex number. This is because complex numbers have two square roots, known as the principal root and the negative root.
To find the distinct roots of a complex number, you can use the formula z = √r(cosθ + isinθ), where r is the absolute value of the complex number and θ is the angle formed between the positive real axis and the complex number on the complex plane. The principal root is found when θ is between 0 and 180 degrees, while the negative root is found when θ is between 180 and 360 degrees.
Complex numbers are used in many scientific fields, including physics, engineering, and mathematics. They are essential in solving equations with imaginary solutions, such as in quantum mechanics and electrical engineering. They also have applications in signal processing, fluid dynamics, and many other areas of science.