Well Ricky, that was a cryptic answer.
You got it right!

Well done !!!

Yes, we are, Ricky.
You gave Euler's identity as the solution, I was referring that.

CN # 2

Find the fifth roots of unity and their sum.

You can express this so:


<==+1

Hi namealreadychosen,

In polar form, 1 = cos(2πk) + i sin(2πk) for any integer k.
==> The 5 fifth roots of unity are given by
cos(2πk/5) + i sin(2πk/5) for k = 0,1,2,3,4.
----------------------
If you want the answer not in trigonometric form, we need to be more crafty.

Since x^5 - 1 = 0 is the equation for the fifth roots of unity:
(x - 1)(x^4 + x^3 + x^2 + x + 1) = 0.

The first factor yields x = 1.

As for the second factor, rewrite it as
x^2 + x + 1 + 1/x + 1/x^2 = 0 (divide both sides by x^2)
==> (x^2 + 1/x^2) + (x + 1/x) + 1 = 0
==> [(x + 1/x)^2 - 2] + (x + 1/x) + 1 = 0.

Letting z = x + 1/x yields
z^2 + z - 1 = 0.

Now, we have a quadratic in z!
==> z = (-1 ± √5) / 2.

Now, we solve for x.
Since z = x + 1/x = (-1 ± √5) / 2,

2x^2 - x[-1 ± √5] + 2 = 0.

The plus sign yields
x = [(1 - √5) ± sqrt((6 - 2√5) - 16)] / 4
= [(1 - √5) ± i * sqrt(10 + 2√5)] / 4.

The negative sign yields
x = [(1 + √5) ± sqrt((6 + 2√5) - 16)] / 4
= [(1 + √5) ± i * sqrt(10 - 2√5)] / 4.

In summary, the five fifth roots of unity (in radical form) are
x = 1, [(1 - √5) ± i * sqrt(10 + 2√5)] / 4, [(1 + √5) ± i * sqrt(10 - 2√5)] / 4.

Hi anonimnystefy,

OK. I shall post problems here too.

CN#3.  If

hi ganesh

thanks

Hi ganesh;

Yes, you are right, I had that and forgot about the root of 3! A really stupid mistake!

Hi ganesh,

CN #5. Find the real and imaginary parts of the complex number

hi ganesh

Hi ganesh;