Question

How would one show that `(\frac { 1+ i } { \sqrt { 2} } ) ^ { 8} + ( \frac { 1- i } { \sqrt { 2} } ) ^ { 8} = 2`?

Original syntax:

How do you solve `(\frac { 1+ i } { \sqrt { 2} } ) ^ { 8} + ( \frac { 1- i } { \sqrt { 2} } ) ^ { 5} = 2`?

Answer 1

See below.

Explanation:

Rewriting:

`(1+i)^8/(sqrt(2))^8+(1-i)^8/(sqrt(2))^8=2`

`(1+i)^8/(2^(1/2))^8+(1-i)^8/(2^(1/2))^8=2`

`(1+i)^8/(2^4)+(1-i)^8/(2^4)=2`

`(1+i)^8+(1-i)^8=2*2^4=2^5`

Expanding brackets:

`(1+i)^8=1+8i+28i^2+56i^3+70i^4+56i^5+28i^6+8i^7+i^8`

`(1+i)^8=1+8i+28(-1)+56(-i)+70(1)+56(i)+28(-1)+8(-i)+1`

`16`

`(1-i)^8=1-8i+28i^2-56i^3+70i^4-56i^5+28i^6-8i^7+i^8`

`(1-i)^8=1-8i+28(-1)-56(-i)+70(1)-56(i)+28(-1)-8(-i)+1`

`16`

We then have:

`16+16= 2^5=> 32=32`

There is nothing to solve here, unless `i` wasn't the imaginary unit.

Answer 2

I presume the intent is to "show" rather than "solve" as there are no variables in the expression.

Thus we wish to show that:

`( (1+i)/sqrt(2) )^8 + ( (1-i)/sqrt(2) )^8 = 2`

We could use the binomial theorem to expand the brackets but due to the higher powers and number of terms DeMoivre's Law provides a much faster solution.

We can represent the LHS by:

`L = ( (1+i)/sqrt(2) )^8 + ( (1-i)/sqrt(2) )^8`

`\ \ = (1+i)^8/(sqrt(2) )^8 + (1-i)^8/(sqrt(2) )^8`

And now if we plot the two points `1+i` and `1-i` on the argand diagram we can very quickly put them into polar form:

So we can write:

`L = (sqrt(2)(cos(pi/4)+isin(pi/4)))^8/(sqrt(2) )^8 + (sqrt(2)(cos(-pi/4)+isin(-pi/4)))^8/(sqrt(2) )^8`

`\ \ = (sqrt(2))^8(cos(pi/4)+isin(pi/4))^8/(sqrt(2) )^8 + (sqrt(2))^8(cos(-pi/4)+isin(-pi/4))^8/(sqrt(2) )^8`

`\ \ = (cos(pi/4)+isin(pi/4))^8 + (cos(-pi/4)+isin(-pi/4))^8`

And now we apply DeMoivre's Law to get:

`L = cos((8pi)/4)+isin((8pi)/4) + cos(-(8pi)/4)+isin(-(8pi)/4)`

`\ \ = 1+0i + 1+0i`

`\ \ = 2` QED

Answer 3

See below.

Explanation:

Using `x+iy = sqrt(x^2+y^2)e^(i phi)` with `phi = arctan(y/x)` we have

`((1pmi)/sqrt2)^8 = (sqrt2/sqrt2 e^(pm pi/4))^8 = e^(pm i 2pi)` and then

`((1+i)/sqrt2)^8+((1-i)/sqrt2)^8 = e^(i 2pi)+e^(-i 2pi)`

and now using de Moivre's identity

`e^(ix) = cosx+i sinx` we have

`cosx = (e^(ix)+e^(-ix))/2` and then

`((1+i)/sqrt2)^8+((1-i)/sqrt2)^8 = 2cos(2pi) = 2`