Since , all four sides of the parallelogram are equal, and thus the parallelogram is a rhombus. Since the diagonals of a rhombus bisect its angles, the segment from to bisects angle --. Hence I expect that the direction of (i.e., is a square root of . I can prove that this is the case without using any geometry.

Proof: I just need to square
. Well,

Now since is a direction, we know that , and hence

Proof: Let . If , then clearly has a square root. If , let be the polar decomposition for . If , then are square roots of . If , then are square roots of .

Hence the polar decomposition for is

The square roots of are

Now , so the square roots of are .

Let
. There is a formula for the square root of that allows you to
say

- a)
- Suppose the square root is , and equate the real and imaginary parts of and . Then solve for and and show that your solution works.
- b)
- Let be the polar decomposition of . You know how to find a square root for , and will be a square root of . Write this in the form .
- c)
- On the basis of (6.27) and (6.28), guess the formula, and show that it works.