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# 6.4 Square Roots

Let be a direction in , with . Then we know that are the vertices of a parallelogram.

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.

6.23   Theorem. Let be a direction in with . Then is a square root of .

Proof: I just need to square . Well,

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

6.24   Corollary. Every complex number has a square root.

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 .

6.25   Example. We will find the square roots of . Let . Then

Hence the polar decomposition for is

The square roots of are

Now , so the square roots of are .

6.26   Exercise. Find the square roots of . Write your answers in the form , where and are real. A

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

 (6.27)

and
 (6.28)

6.29   Exercise. Verify that assertions (6.27) and (6.28) are correct.

6.30   Entertainment. Find the square root formula, and prove that it is correct. (There are at least three ways to do this. Method c) is probably the easiest.)
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.

Next: 6.5 Complex Functions Up: 6. The Complex Numbers Previous: 6.3 Roots of Complex   Index