How do you simplify #(1−(sqrt3)*i)^(1/2)#?

Question

3794 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-12-11T03:39:42

Apply the identity
#e^(i theta) = cos(theta)+isin(theta)#

to find
#(1-sqrt(3)i)^(1/2)=sqrt(6)/2 - sqrt(2)/2i#

We will be using the identity

#e^(i theta) = cos(theta)+isin(theta)#

First, we can make some slight modifications to our original value to make it easier to find #theta#.

#(1-sqrt(3)i)^(1/2) = (2(1/2-sqrt(3)/2i))^(1/2)#

and now, using

#cos(-pi/3) = 1/2# and #sin(-pi/3) = -sqrt(3)/2#

together with the above identity, and that #(x^a)^b = x^(ab)#, we get

#(2(1/2-sqrt(3)/2i))^(1/2) = (2cos(-pi/3)+isin(-pi/3))^(1/2)#

#= (2e^(i(-pi/3)))^(1/2)#

# = sqrt(2)e^(i(-pi/6))#

#= sqrt(2)(cos(-pi/6)+isin(-pi/6))#

#= sqrt(2)(sqrt(3)/2 - 1/2i)#

#= sqrt(6)/2 - sqrt(2)/2i#