Find the solutions to z^2 = i

Here z is a complex number, and therefore, z = a + bi. Thus, our equation becomes (a + bi)^2 = i. Expanding the brackets we get a^2 + 2abi + (b^2)(i^2) = i. Since i is the square root of -1, i^2 = the whole number -1. Then (b^2)(i^2) is equal to (b^2)(-1) or -b^2. Substituting and rearranging above we have a^2 - b^2 + 2abi = i. Now we compare real and imaginary parts on both sides. The real part on RHS is a^2 - b^2, whereas the real part on LHS is 0 (it's missing). From here follows that a^2 -b^2 = 0 or a^2 = b^2 or a = +/- b. The imaginary part on RHS is 2ab, whereas the imaginary part on LHS is 1 (notice that the imaginary part doesn't include i, only its multiple). We see that 2ab = 1 or ab = 1/2 or substituting for b we get a^2 = 1/2.
Simplifying we are left with a = square root of 1/2. Since a and b have the same values we can conclude that b = square root of 1/2 as well. Keep in mind that the equation will hold true if both a and b have negative values. a = b = +/- square root of 1/2. Therefore, our solutions for z in the form of a + b*i are: z = sqr (1/2) + sqr (1/2) * i and z = -sqr (1/2) - sqr (1/2) * i.