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

1 Answer
Jun 11, 2016

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)# 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#