Question

What is `(sqrt(5)+i)^6` in the form `a+bi`?

Answer 1

`(sqrt(5)+i)^6 = -176+56sqrt(5)i`

Explanation:

`(sqrt(5)+i)^2 = (sqrt(5))^2+2sqrt(5)i+i^2 = 4+2sqrt(5)i`

`(4+2sqrt(5)i)^2 = 4^2+2(4)(2sqrt(5)i)+(2sqrt(5)i)^2`

`color(white)((4+2sqrt(5)i)^2) = 16+16sqrt(5)i-20`

`color(white)((4+2sqrt(5)i)^2) = -4+16sqrt(5)i`

So:

`(sqrt(5)+i)^6 = (4+2sqrt(5)i)(-4+16sqrt(5)i)`

`color(white)((sqrt(5)+i)^6) = -16+64sqrt(5)i-8sqrt(5)i-160`

`color(white)((sqrt(5)+i)^6) = -176+56sqrt(5)i`

Check

We can make a rudimentary check of the answer by making sure that:

`abs(-176+56sqrt(5)i) = abs(sqrt(5)+i)^6`

We find:

`abs(sqrt(5)+i) = sqrt((sqrt(5))^2+1^2) = sqrt(5+1) = sqrt(6)`

`abs(-176+56sqrt(5)i) = sqrt((-176)^2+(56sqrt(5))^2) = sqrt(30976+15680)`

`= sqrt(46656) = sqrt(6^6) = (sqrt(6))^6`

Answer 2

`-176 + 56sqrt5i`

Explanation:

`6`th row of Pascal's triangle:

`1` `6` `15` `20` `15` `6` `1`

`(sqrt(5)+i)^6 = (sqrt5)^6 +6((sqrt5)^5 * i) + 15((sqrt5)^4 * i^2) + 20((sqrt5)^3 * i^3) + 15((sqrt5)^2 * i^4) + 6((sqrt5) * i^5 )+ i^6`

`(sqrt5)^6 = 5^3 = 125`

`15(sqrt5)^4 * i^2 = 15 ((sqrt5)^4 * -1) = 15(5^2 * -1) = 15 * -25 = -375`

`15(sqrt5)^2 * i^4 = 15 * 5 * 1 = 75`

`i^6 = i^2 = -1`

`125-375+75-1 = -176`

`(sqrt(5)+i)^6 = -176 + 6((sqrt5)^5 * i) + 20((sqrt5)^3 * i^3) + 6((sqrt5) * i^5 )`

`6(sqrt5)^5 * i = 6 *25sqrt5 i = 150sqrt5i`

`20(sqrt5)^3 * i^3 = 20*5sqrt5 * -i = -100sqrt5i`

`6(sqrt5) * i^5 = 6sqrt5 * i = 6sqrt5i`

`150sqrt5 i -100sqrt5i + 6sqrt5i = 56sqrt5 * i = 56sqrt5i`

`(sqrt(5)+i)^6 = -176 + 56sqrt5i`, in `a + bi` form.