 While it's possible to find square roots using only algebra, it's easier if we can use the trigonometric form. Suppose a plus bi is r-cis-theta, then a plus bi to the n is r to the n-cis-n-theta, so a plus bi is an n-th root of r to the n-cis-n-theta. So let's try to find a square root of i. Rewriting i in trigonometric form gives us... So we want to find r and theta, where r-cis-theta squared gives us cis-pi-halves. So that means r-squared cis-2-theta is cis-pi-halves. Now if we compare the components, then one possibility is r-squared is equal to 1, and 2-theta is pi-halves. And so that says r is plus or minus 1, and theta is pi-fourths. And remember for convenience we always choose r to be positive. So that means a square root will be 1 cis-pi-fourths. But wait, we're not done. Remember, you should always answer questions in the same language they were given. So since the problem was stated given a complex number in rectangular form, we should answer in rectangular form, and so we find that 1 cis-pi-fourths is. Now that all seems to work, but don't we usually have two square roots? And it's useful to keep in mind a new method to solving an old problem must give the same answers as the old method. And in general there are n distinct n-throots, and this leads to two problems. How do we find them? And which is the principal root? Now it's useful to remember that sine and cosine have period 2-pi. This means that the trigonometric form of a complex number is not unique. r cis-theta is the same as r cis-theta plus 2-pi, theta minus 2-pi, theta plus 4-pi, theta plus 6-pi, and so on. And these different forms can be used to find other roots. So again let's find all square roots of i. We found one root by using i equals cis-pi-halves, but I could also be written as cis-5-pi-halves as well. So as before we want to find r and theta where r cis-theta squared is cis-5-pi-halves, so that means r squared cis-2-theta is cis-5-pi-halves. As before, comparing our two forms tells us that r squared is 1 and 2-theta is 5-pi-halves, which gives us another square root, which we want to state in rectangular form as. Now if you really understood what we just did, the natural question arises, could we have a third square root? Well let's consider. Note that the arguments of the different forms of i are going to differ by a multiple of 2-pi. So the arguments of the square roots will differ by a multiple of 2-pi-halves or pi, and so our roots will have the arguments pi-4s, pi-4s plus pi, pi-4s plus 2-pi, and so on, but all of those after pi-4s plus pi or 5-pi-4s will be duplicates. And so we only get the two square roots.