How do you simplify #w = [ ( 1 + i ) / ( 2 - i ) ] ^ 0.25#?
1 Answer
#((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) = ((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
#sqrt(a+bi) = (sqrt((sqrt(a^2+b^2)+a)/2)) + (sqrt((sqrt(a^2+b^2)-a)/2))i#
So, with
#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^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#