6.4 Square Roots

Let $u$ be a direction in $C(0,1)$, with $u\neq -1$. Then we know that $1,0,u,1+u$ are the vertices of a parallelogram.

Since $\vert u\vert=\vert 1\vert=1$, 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 $0$ to $1+u$ bisects angle $(1$-$0$-$u)$. Hence I expect that the direction of $1+u$ (i.e., $${ {{1+u}\over {\vert 1+u\vert}})}$$ is a square root of $u$. I can prove that this is the case without using any geometry.

6.23   Theorem. Let $u$ be a direction in $\mbox{{\bf C}}$ with $u\neq -1$. Then $${ {{1+u}\over {\vert 1+u\vert}}}$$ is a square root of $u$.

Proof: I just need to square $${ {{1+u}\over {\vert 1+u\vert}}}$$. Well,

$$\left( {{1+u}\over {\vert 1+u\vert}}\right)^2={{(1+u)^2}\over... ...\vert^2}}={{(1+u)^2}\over {(1+u)(1+u)^*}}={{1+u}\over {1+u^*}}.$$

Now since $u$ is a direction, we know that $uu^*=1$, and hence

$${{1+u}\over {1+u^*}}={{uu^*+u}\over {1+u^*}}={{u(u^*+1)}\over {(1+u^*)}}=u.\mbox{ $\mid\!\mid\!\mid$}$$

6.24   Corollary. Every complex number has a square root.

Proof: Let $\alpha\in\mbox{{\bf C}}$. If $\alpha=0$, then clearly $\alpha$ has a square root. If $\alpha\neq 0$, let $ru$ be the polar decomposition for $\alpha$. If $u\neq -1$, then $${\pm r^{1\over 2}\left( {{1+u}\over {\vert 1+u\vert}}\right)}$$ are square roots of $\alpha$. If $u=-1$, then $${\pm r^{1\over 2}i}$$ are square roots of $\alpha$. $\mid\!\mid\!\mid$

6.25   Example. We will find the square roots of $21 - 20i$. Let $\alpha = 21 - 20i$. Then

$$\vert\alpha\vert = \sqrt{21^2 + 20^2} = \sqrt{441+400} = \sqrt{841} = 29.$$

Hence the polar decomposition for $\alpha$ is

$$\alpha = 29\left({21 - 20i \over 29}\right) = ru \mbox{ where } r = 29 \mbox{ and }u = {21 - 20i \over 29}.$$

The square roots of $\alpha$ are

$$\begin{eqnarray*} \pm r^{1\over 2}\left({1+u\over \vert 1+u\vert}\right) &=& \pm... ...right) = \pm \sqrt{29}\left({5-2i\over \vert 5-2i\vert}\right). \end{eqnarray*}$$

Now $\vert 5 - 2i\vert = \sqrt{25+4} = \sqrt{29}$, so the square roots of $\alpha$ are $\pm(5-2i)$.

6.26   Exercise. Find the square roots of $12+5i$. Write your answers in the form $a+bi$, where $a$ and $b$ are real. A

Let $a,b\in\mbox{{\bf R}}$. There is a formula for the square root of $a+bi$ that allows you to say

$$\mbox{ the square roots of } 2+4i \mbox{ are } \pm\left(\sqrt{\sqrt 5+1}+i\sqrt{\sqrt 5-1}\right)$$ (6.27)

and

$$\mbox{ the square roots of } 6-2i \mbox{ are } \pm\left(\sqrt{\sqrt{10}+3}-i\sqrt{\sqrt{10}-3}\right).$$ (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 $c+di$, and equate the real and imaginary parts of $(c+di)^2$ and $a+bi$. Then solve for $c$ and $d$ and show that your solution works.

b)

Let $ru$ be the polar decomposition of $a+bi$. You know how to find a square root $v$ for $u$, and $r^{1\over 2}v$ will be a square root of $a+bi$. Write this in the form $c+di$.

c)

On the basis of (6.27) and (6.28), guess the formula, and show that it works.