The two square roots of a complex $w$ with module $\left\vert w\right\vert $ and $\theta =\arg (w)$, i.e $w=\left\vert w\right\vert e^{i\theta }$ are $ \sqrt{w}=\sqrt{\left\vert w\right\vert }e^{i\frac{\theta +2k\pi }{2}},k\in \{0,1\}.$
For $w=i$, we have $\left\vert w\right\vert =\left\vert i\right\vert =1$, $% \theta =\arg (i)=\frac{\pi }{2}$. Thus $\sqrt{e^{i\pi /2}}=e^{i\frac{\pi /2+2k\pi }{2}},k\in \{0,1\}.$ One of the roots is
$z_{1}=e^{i\frac{\pi /2}{2}}=e^{i\pi /4}=\cos \left( \frac{\pi }{4}\right) +i\sin \left( \frac{\pi }{4}\right) =\frac{1}{2}\sqrt{2}+\frac{1}{2}i\sqrt{2} $ and the other $z_{2}=e^{i\frac{\pi /2+2\pi }{2}}=e^{i5\pi /4}=\cos \left( \frac{5}{4% }\pi \right) +i\sin \left( \frac{5}{4}\pi \right) =-\frac{1}{2}\sqrt{2}-% \frac{1}{2}i\sqrt{2}.$
Only $z_{2}$ has negative imaginary part. Observing that $\frac{1}{2}\sqrt{2}% =\frac{1}{\sqrt{2}}$, we get $z_{2}=\frac{1}{\sqrt{2}}\left( -1-i\right) .$
Added 2: $w=i,z_1,z_2$

Added: The $n^{th}$ roots of the complex $w=\left\vert w\right\vert e^{i\theta }$ are
$\sqrt[n]{w}=\sqrt[n]{\left\vert w\right\vert }e^{i\frac{\theta +2k\pi }{n}},k\in \left\{ 0,1,2,\ldots ,n-1\right\} ,$
because
$\begin{eqnarray*} \left( \sqrt[n]{w}\right) ^{n} &=&\left( \sqrt[n]{\left\vert w\right\vert }% e^{i\frac{\theta +2k\pi }{n}}\right) ^{n}=\left( \sqrt[n]{\left\vert w\right\vert }\right) ^{n}\left( e^{i\frac{\theta +2k\pi }{n}}\right) ^{n} \\ &=&\left\vert w\right\vert e^{i\left( \theta +2k\pi \right) }=\left\vert w\right\vert e^{i\theta }e^{i2k\pi }=\left\vert w\right\vert e^{i\theta }\cdot 1 \\ &=&\left\vert w\right\vert e^{i\theta }=w. \end{eqnarray*}$