# Every complex number has 2 square roots - Rudin

I am using Baby Rudin for self study and as many of you are aware, it is one thing to follow Rudin's calculations to verify that his proof is correct. It is quite another to work backwards to see how he came up with his proof in the first place. I find that unless I do the latter, the former does absolutely nothing for me. Chapter 1 exercise 10 is as follows: Suppose $z = a + bi$, $w = u + vi$ and $a = \left(\frac{|w| + u}{2}\right)^{1/2}$, $b = \left(\frac{|w| - u}{2}\right)^{1/2}$. Prove that $z^2 = w$ if $v \geq 0$ and $\overline{z}^2 = w$ if $v \leq 0$. Conclude that every complex number(with one exception) has two complex square roots. It is pretty easy to verify this calculation: If $v \geq 0$ then $(a+bi)^2 = (-a - bi)^2= a^2 - b^2 + 2abi = \frac{|w| + u}{2} + \frac{|w| - u}{2} + 2i\left(\frac{|w| + u}{2}\right)^{1/2}\left(\frac{|w| - u}{2}\right)^{1/2} = u + 2i \sqrt{\frac{|w|^2 - u^2}{4}} = u + 2i\sqrt{\frac{u^2 + v^2 - u^2}{4}} = u + 2i \sqrt{\frac{v^2}{4}} = u + 2i\frac{|v|}{2} = u + |v|i = u + vi$ On the other hand if $v \leq 0$ then it is similar to show that $(a - bi)^2 = (-a + bi)^2 = u - |v|i$

This completes the proof. Now for the backwards derivation attempt:

We want $(a + bi)^2 = u + vi \implies a^2 - b^2 + 2abi = u + vi \implies a^2 - b^2 = u$ and $2ab = v$. So essentially we have a system of equations to solve for $a$ and $b$. I have been stuck on this for a couple hours. Maybe I am missing something obvious, but any help would be greatly appreciated.

## 2 Answers

I figured it out:

$a^2 - b^2 = u$ and $2ab = v$

$b = \frac{v}{2a}$

$a^2 - \frac{v^2}{4a^2} - u = 0$

$4a^4 -4ua^2 - v^2 = 0$

By the quadratic formula:

$a^2 = \frac{4u \pm \sqrt{16^2 - 4(4)(-v^2)}}{8} = \frac{4u \pm \sqrt{16u^2 + 16v^2}}{8} = \frac{4u+4\sqrt{u^2 + v^2}}{8} = \frac{u \pm |w|}{2}$

So that $a = \sqrt{\frac{u + |w|}{2}}$ since we are only interested in real values of $a$. Then

$a^2 - b^2 = u \implies b^2 = a^2 - u = \frac{|w| + u}{2} - u = \frac{|w| - u}{2}$

so that $b = \sqrt{\frac{|w| - u}{2}}$

• +1 for figuring this out. For self study figuring out things for yourself is a must. – Paramanand Singh Apr 24 '17 at 9:38

Hint: Use this $$(x^2+y^2)^2=(x^2-y^2)^2+(2xy)^2$$

• I'm going to keep looking at this, but totally lost so far... – Joe Apr 23 '17 at 17:09