# z=(1+i)^n+(1-i)^nshow that z=2(square root 2)^n*cosnpi/4

sciencesolve | Teacher | (Level 3) Educator Emeritus

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Put `z = z_1^n + z_2^n`

You need to transform the actual rectangular form of the complex numbers `z_1`  and `z_2`  numbers in the polar form. The polar form of a complex number does permit you to use De Moivre's theorem to solve the problem.

The rectangular form of complex number is `z = x + i*y`

The polar form of complex number is`z = v*(cos alpha + isin alpha)`

```v = sqrt(x^2 + y^2)`

`` `tan alpha = y/x` Compare the rectangular form of `z_1` to the standard rectangular form.

`z_1= 1+i =gt x = 1 and y = 1 =gt v_1 = sqrt 2`

`tan alpha = 1 =gt alpha = pi/4`

`z_1 = sqrt 2*(cos pi/4 + i*sin pi/4)`

Raise to n:

`z_1^n = (sqrt 2)^n*(cos pi/4 + i*sin pi/4)^n`

Use Moivre's theorem:

`z_1^n = (sqrt 2)^n*(cos n*pi/4 + i*sin n* pi/4)`

Compare the rectangular form of `z_2`  to the standard rectangular form.

`z_2= 1-i =gt x = 1 and y = -1 =gt v_1 = sqrt 2`

`tan alpha = -1 =gt alpha = -pi/4`

`cos (-pi/4) = cos pi/4`  (even function)

`sin (-pi/4) = -sin pi/4`  (odd function)

`z_1 = sqrt 2*(cos pi/4- i*sin pi/4)`

Raise to n:

`z_2^n = (sqrt 2)^n*(cos pi/4- i*sin pi/4)^n`

Use Moivre's theorem:

`z_2^n = (sqrt 2)^n*(cos n*pi/4- i*sin n* pi/4)`

`z = (sqrt 2)^n*(cos n*pi/4 + i*sin n* pi/4) + (sqrt 2)^n*(cos n*pi/4- i*sin n* pi/4)` `z = (sqrt 2)^n*(cos n*pi/4 + i*sin n* pi/4 + cos n*pi/4- i*sin n* pi/4)`

Eliminate opposite members:

`z = 2*(sqrt 2)^n*(cos n*pi/4)`

The last line proves that `(1+i)^n + (1-i)^n =2*(sqrt 2)^n*(cos n*pi/4).`