Q. polynomial/complex (1 Viewer)

[marsenal]

This question is from Patel's book:

Given that cot(theta)=y + 1 and a and b are the roots of
z^2 - 2z + 2=0, prove that



(y + a)^n - (y + b)^n
-------------------------- =
a - b


sin n(theta)
------------------
[sin(theta)]^n



Thanks

 

[spice girl]

Originally posted by marsenal
This question is from Patel's book:

Given that cot(theta)=y + 1 and a and b are the roots of
z^2 - 2z + 2=0, prove that



(y + a)^n - (y + b)^n
-------------------------- =
a - b


sin n(theta)
------------------
[sin(theta)]^n



Thanks

Now we know b is the conjugate of a.

~a is a-bar, or the conjugate of a, so ~a = b, ~b = a

so since y is real
(y+b)^n = (y+~a)^n = {~(y + a)}^n = ~{(y+a)^n}
thus (y + a)^n and (y + b)^n are conjugates.

Now solving the quadratic, we know a, b = 1 + i, 1 - i
We know z - ~z = 2i*Im(z)
So
{(y + a)^n - (y + b)^n}/(a-b) = 2i*Im{(y+a)^n}/2i*Im(a) or = 2i*Im{(y+b)^n}/2i*Im(b)

=-Im{(y+b)^n)} = Im{(y+a)^n)} = Im{(cot@ + i)^n} (if a = 1+i, b=1-i)

cot@ = cos@/sin@
so (cot@ + i)*sin@ = cos@+isin@ = cis@
so (sin@)^n(cot@ + i)^n = (cis@)^n = cisn@ (De-moivre's)
(cot@ + i)^n = cisn@/(sin@)^n

Take imaginary parts of both sides:
Im{(cot@ + i)^n} = Im{cisn@/(sin@)^n} = sin(n@)/(sin@)^n

I think it's pretty much proven...

 

[marsenal]

Thanks spice girl