Be careful with signs

$i$ may be written as $cos(\frac{\pi}{2})+isin(\frac{\pi}{2})$ . Using de Moivre's formula $\sqrt{i} = cos(\frac{\pi}{2\times2})+isin(\frac{\pi}{2\times2}) = cos(\frac{\pi}{4})+isin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i$ Yet, when you find square root, only the positive value will come up (in terms of complex value, it will be the root whose $arg(z)$ is close to zero). We will then multiply with $-1$ to produce another root, which is $-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}i$ . Therefore, $\sqrt{i} = \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i, -\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}i$

We find a complex number which satisfies the equation

$(a+ bi)^{2} = i$

or

$a^{2} - b^{2} + 2abi = i$

separating the real and imaginary parts, we get a system of real numbers $a$ and $b$ such that

$a^{2} - b^{2} = 0$

$2ab = 1$

which means the two possible solutions are

$a = b$ and $a = -b$

putting that in mind, and substituting that in the second equation, we see that only the first relationship ( $a=b$ ) holds so that both $a$ and $b$ become real numbers.

Thus,

$a = b = \pm \frac {\sqrt{2}}{2}$

So, it will be clear that the solution for this equation is

$\boxed{ \large \pm( \frac {\sqrt{2}}{2} + \frac {\sqrt{2}}{2}i) }$