Hi guys.
"Find all complex numbers satisfying "
The hint is to factorise as a difference of squares.
This seems incorrect.
The bracketed terms are not quadratics of standard form and two z terms are identical, leading only to five solutions, from what I can see.
I've seen a solution involving De Moivre's theorem, in which a full rotation is split into six pieces, taking the where .
I'm very interested in understanding the solution with De Moivre's theorem. It was explained in a course I took last semester but I simply cannot remember a thing about it.
I've asked several people about this problem, but they can't seem to distinguish copying from understanding.
I would love to really understand the solution to this problem.
Can you help?
math1051 assignment 1 @ UQ?
I'm stuck, too. I think i'm way out of my element here, but I just plugged in some values into de moivre's theorem.
I don't fully understand complex numbers, and I too am keen to know HOW to solve a question like this, rather than have someone say "here's the answer, off you go". Will be watching this thread.
First, we express the RHS in the form | , where is of course the modulus, and the argument. Since , and , we have:
De Moivre's theorem states that:
Or more relevantly:
So
We just plug to get the six values and then convert them into the Cartesian form, i.e. .
Note that all the subsequent values after n = 5 are repetitions, i.e n = 0 has the same value as n = 6.
DeMoivre's Theorem isn't too bad as long as you remember that the roots are all evenly spaced around a circle.
So, in this case, remember that .
Therefore:
.
Since they are evenly spaced around the circle and there are 6 roots, that means they are all separated by an angle of .
So the other roots are:
and
.
If you convert each of these to cartesians, you should get the same answers as those found by factorising and using Null Factor Law.
I've now done it both ways (ie; diff of squars and DM theorum).
Both worked.
to anyone else doing the 1051 assignment, what in the hell is he asking for in question 10?
"Suppose y = cos(x), where 0<= x <= pi/2, obtain sin(x) as a function of y for the same interval.
Is he just asking, what sin function will give us the same "plot" as the cos function? something like 1-sin(x), i've done it, i just can't find it. But I don;t know if that's what we have to do?
Thankyou so much Prove It. I'm kind of a rope-learner, for things I have never fully encountered before I need to be shown how to do it. In fact, all I had to do was look at your post, remember than sin^2x + cos^2x = 1 and I had a go. compared my answer to yours and it came out the same!
Thanks again!