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

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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

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

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

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

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

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

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

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