\(i^\frac12 = p + \iota q, \;p,q\, \in R\)

Then \((p + \iota q)^2 = i\)

⇒ \(p^2 - q^2 + 2\iota\, pq = i\)

\(\therefore 2 pq = 1\) & \(p^2 - q^2 = 0\)

⇒ \(p^2 = (\frac1{2p})^2 = 0\)

⇒ \(p^2 - \frac1{4p^2} = 0\)

⇒ \(p^4 - \frac14 = 0\)

⇒ \(p^2 = \sqrt{\frac14} = \pm \frac12\)

\(\therefore p = \pm \frac1{\sqrt2}\)       \((\because p \in R,\, \therefore p^2 \ge 0)\)

⇒ \(q = \frac1{2p} = \pm \frac1{2\sqrt 2}\)

\(\therefore i^\frac12 = \frac1{\sqrt 2}(1 + \frac\iota2) \; or\; i^\frac12= \frac{-1}{\sqrt2}(1+ \frac\iota2)\)

Now, 

\(i^\frac12 + (-i)^\frac12 = i^\frac12 + i^\frac12(-1)^\frac12\)    \((\because (ab)^m = a^m b^m)\)

\(= i^\frac12(1 + i) \)    \((\because \sqrt{-1} = i)\)

\(=\frac1{\sqrt2} (1 + \frac\iota 2) (1 + \iota)\)

\(= \frac1{\sqrt 2}(1 + \frac{3\iota}2 - \frac12)\)

\(= \frac1{2\sqrt 2} + \frac{3\iota}{2\sqrt2}\)

or

\(i^\frac12 + (-i)^\frac12 = i^\frac12 (1 + i)\)

\(= -\frac1{\sqrt2} (1 + \frac i2) (1 + i)\)

\(= -\frac1{\sqrt2} (\frac12 + \frac{3i}2)\)

\(= \frac{-1}{2\sqrt 2} - \frac{3i}{2\sqrt2}\)

Hence, \(i^\frac12 + (-i)^\frac12 = \frac1{2\sqrt2} + \frac{3i}{2\sqrt2}\) or \( \frac{-1}{2\sqrt2} - \frac{3i}{2\sqrt 2}\).