Question

How do you simplify `w = [ ( 1 + i ) / ( 2 - i ) ] ^ 0.25`?

Answer 1

`((1+i)/(2-i))^0.25`

`=root(4)(1/5+3/5i)`

`=1/2sqrt(((2+sqrt(sqrt(10)+1))sqrt(10))/5) + 1/2sqrt(((2-sqrt(sqrt(10)+1))sqrt(10))/5)i`

Explanation:

First simplify `(1+i)/(2-i)` by multiplying both numerator and denominator by the Complex conjugate `2+i` of the denominator:

`(1+i)/(2-i) = ((1+i)(2+i))/((2-i)(2+i)) = (2+3i+i^2)/(2^2-i^2) = (1+3i)/5 = 1/5+3/5i`

Then note that:

`((1+i)/(2-i))^0.25 = root(4)(1/5+3/5i) = sqrt(sqrt(1/5+3/5i))`

Then (see https://socratic.org/s/avemtNYy) since `a, b > 0` we have:

`sqrt(a+bi) = (sqrt((sqrt(a^2+b^2)+a)/2)) + (sqrt((sqrt(a^2+b^2)-a)/2))i`

So, with `a=1/5` and `b=3/5`, we find:

`sqrt(1/5+3/5i)`

`= (sqrt((sqrt((1/5)^2+(3/5)^2)+1/5)/2)) + (sqrt((sqrt((1/5)^2+(3/5)^2)-1/5)/2))i`

`= (sqrt((sqrt(10/25)+1/5)/2)) + (sqrt((sqrt(10/25)-1/5)/2))i`

`= (sqrt((sqrt(10)+1)/10)) + (sqrt((sqrt(10)-1)/10))i`

Then with `a=(sqrt((sqrt(10)+1)/10))` and `b=(sqrt((sqrt(10)-1)/10))` we have:

`a^2+b^2`

`= (sqrt((sqrt(10)+1)/10))^2+(sqrt((sqrt(10)-1)/10))^2`

`=((sqrt(10)+1)/10)+((sqrt(10)-1)/10)=(2sqrt(10))/10 = sqrt(10)/5`

`(sqrt(a^2+b^2)+a)/2 = (sqrt(10)/5 + (sqrt((sqrt(10)+1)/10)))/2`

`= (sqrt(10)/5 + (sqrt((10(sqrt(10)+1))/100)))/2`

`= (2sqrt(10)+sqrt(10)sqrt(sqrt(10)+1))/20`

`=((2+sqrt(sqrt(10)+1))sqrt(10))/20`

Similarly:

`(sqrt(a^2+b^2)-a)/2 = ((2-sqrt(sqrt(10)+1))sqrt(10))/20`

So:

`((1+i)/(2-i))^0.25`

`=sqrt(((2+sqrt(sqrt(10)+1))sqrt(10))/20) + sqrt(((2-sqrt(sqrt(10)+1))sqrt(10))/20)i`

`=1/2sqrt(((2+sqrt(sqrt(10)+1))sqrt(10))/5) + 1/2sqrt(((2-sqrt(sqrt(10)+1))sqrt(10))/5)i`