The easy way
$|z|^2 = \frac {1}{2}^2 + (\frac {\sqrt 2}{2})^2 = \frac {3}{4}\\
|z| = \frac {\sqrt 3}{2}\\
z^{\frac 12} = \frac {3^\frac 14}{2^{\frac12}} e^{\frac {i}{2}\arctan \sqrt 2}$
You really could stop here. Or, you can keep going:
$z^{\frac 12} = \frac {3^\frac 14}{2^{\frac12}}(\cos(\frac {1}{2}\arctan \sqrt 2) + i\sin(\frac {1}{2}\arctan \sqrt 2))\\
z^{\frac 12} = \frac {3^\frac 14}{2^{\frac12}}\left(\sqrt {\frac {\sqrt 3+ 1}{2\sqrt 3}} + i\sqrt {\frac {\sqrt 3- 1}{2\sqrt 3}}\right)\\
z^{\frac 12} = \frac{\sqrt {\sqrt 3+ 1}}{2} + i\frac{\sqrt {\sqrt 3- 1}}{2}$
And we can compare this to our other approach
$z^{\frac 12} = x+iy\\
(x+iy)^2 = \frac {1}+i\frac {\sqrt 2}{2}\\
x^2 - y^2 = \frac 12\\
2xy = \frac {\sqrt 2}{2}\\
y = \frac {\sqrt 2}{4x}\\
x^2 - \frac {1}{8 x^2} = \frac 12\\
8x^4 - 4x^2 - 1 = 0\\
x^2 = \frac {4 + \sqrt {48}}{16}\\
x^2 = \frac {1 + \sqrt {3}}{4}\\
x = \frac {\sqrt {1 + \sqrt {3}}}{2}\\
y = \frac {\sqrt {-1 + \sqrt {3}}}{2}\\
z^{\frac 12}= \frac {\sqrt {1 + \sqrt {3}}}{2} + i\frac {\sqrt {-1 + \sqrt {3}}}{2}$