# Prove that the number (cospi/4+i*sinpi/4)^2008 is real.

hala718 | Certified Educator

(cospi/4 + i*sinpi/4)^ 2008

Let us find the values for the numbers inside the brackets.

==> cospi/4 = sqrt2/2

==> sinpi/4 = sqrt2/2

==> ( sqrt2/2 + (sqrt2/2)*i ) ^ 2008.

Let us rewrite the power.

We know that a^b*c = a^b)^c

==> [(sqrt2/2 + (sqrt2/2)*i ]^2*1004

==> (sqrt2/2 + sqrt2/2 * i)^2 ]^1004.

==> (sqrt2/2)^2 + 2sqrt2/2*sqrt2/2*i + sqrt2/2)^2*i^2 ]^ 1004.

==> (2/4 + 2*2/4*i - 2/4) ^1004

==> (1/2 + i - 1/2) ^1004

==> (i)^1004

Now we will rewrite the power.

==> (i)^2]^502

==> (-1)^502 = 1

Then the values of the number ( cospi/4 + sinpi/4*i)2008 is 1.

Then the number is real.

justaguide | Certified Educator

We have to prove that (cos pi/4 + i*sin pi/4)^2008 is real.

Now (cos pi/4 + i*sin pi/4)^2008

=> (cos pi/4 + i*sin pi/4)^2^1004

=> [ (cos pi/4)^2 + i^2*(sin pi/4)^2 + 2*i*(cos pi/4)(sin pi/4)]^1004

Now i^2 = -1

=> [ (cos pi/4)^2 - (sin pi/4)^2 + 2*i*(cos pi/4)(sin pi/4)]^1004

(cos x)^2 - (sin x)^2 = cos 2x

=> [(cos pi/2) + 2*i*(cos pi/4)(sin pi/4)]^1004

cos (pi/2) = 0

=> [2*i*(cos pi/4)(sin pi/4)]^2^502

=> [4*i^2 *(cos pi/4)^2*(sin pi/4)^2]^502

=> [-16*(cos pi/4)^2*(sin pi/4)^2]^502

This is real as i has been eliminated.

Therefore (cos pi/4 + i*sin pi/4)^2008 is real.

neela | Student

To prove that (cospi/4 +isipi/4)^2008) is real.

We know that (cosa+isina)^n= (cosna+isinb) by De Moivre's theorem.

Therefore (cospi/4+isinpi/4) = (cos2008*pi/4+isin2008*pi/4).

(cospi/4+isinpi/4) = cos502pi) + isin(502)pi

(cospi/4+isinpi/4) = 1 + 0. as cos2npi = cos0 = 1  for  any integer n and sin2npi =   sin0 = 0  for any  integer n.

Therefore (cospi/4 +isipi/4)^2008) = 1 which is real.

giorgiana1976 | Student

Since the complex number inside the brackets is written in the polar form, we can use Moivre's rule and we'll get:

(cospi/4+i*sinpi/4)^2008 = (cos 2008pi/4+i*sin 2008pi/4)

We'll simplify and we'll get:

(cos 2008pi/4+i*sin 2008pi/4) = (cos 502pi+i*sin 502pi)

Since 502 is multiple of 2, we'll re-write 502pi as:

502pi = 251*2pi

(cos 2pi+i*sin 2pi) = 1 + i*0

(cos 502pi+i*sin 502pi) = 1

Since the imaginary part is 0, the given number is a real number:

(cospi/4+i*sinpi/4)^2008 = 1