Question

Complex numbers question!
One of the roots of z is a=cos2-isin2.
Find the other roots and trace them on the graph.

Answer 1

idk, sowy

Answer 2

z has fourth degree, so the solutions are four

Answer 3

clearly z lies on the unit circle. Also, as we are finding fourth roots, the arguments of two successive roots will differ by Pi/2 on the circle. You are already given that one root is e^(-2i), so the other three roots are,
\[e ^{i (-2+\pi/2)},e ^{i (-2+\pi)},e ^{i (-2+3\pi/2)}\]

Answer 4

Can you explain to me why the arguments differ by pi/2?

Answer 5

And why is it e^(-2i) and not e^(2i)?

Answer 6

you must be aware of the fact that for any complex number on the unit circle (in the complex plane of course), the nth roots form a regular n-gon inscribed in the unit circle. This also means that the angle subtended at the centre of the circle by each side of the n-gon will be equal.
So, as the there are n sides, the measure of each angle will be 2Pi/n. Here, n=4. So, the roots form a square. So, angle subtended at the centre by each side of the square will be 2Pi/4 which is Pi/2. So, the difference of arguments between two successive roots will be Pi/2.
You have cos2-i sin2. e^(ix)=cosx+i sinx. So, e^(-ix)=cosx+i sin(-x)=cosx-i sin(x)

Answer 7

Ok i get it. In addition is it okay if I reason by adding as a solution the conjugate number of the given root?

Answer 8

Thanks you by the way.

Answer 9

On my book the results are:
sin2-icos2
-cos2+isin2
sin2+icos2

Answer 10

@naylah no you can't add a conjugate. That's because you're finding the root of a non real number. It'd have been fine had it been a real number.
and you're welcome :)

Answer 11

But why aren't the results similar to the ones of my book?

Answer 12

My professor is very strict about the method so he really wants stuff done his way...

Answer 13

your first answer is not correct. are you sure that's not a typo? it should be -sin2-icos2 instead of sin2-icos2

Answer 14

yeah I forgot a sign, but how do I manage to come this far?