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Suppose $z = a + bi$, $w = u + iv$, 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 that $(\bar{z})^2 = w$ if $v \leq 0$. Conclude that every complex number (with one exception!) has two complex square roots.

I have two problems:

1) I completed a proof but it assumed $|w|+u > 0$ and $|w| - u > 0$ so that $a,b \in \mathbb{R}$. Should I have been assuming that $a,b,u,v \notin \mathbb{R}$ was possible? Or is their some kind of proof that $a,b \in \mathbb{R}$ always? Does it matter?

2) What exactly does proving the first part have to do with concluding that every complex number has two roots? It seems like the real conclusion is "there are two expressions for the square root depending on whether the imaginary part is non-negative or non-positive". I've seen a couple of solution guides (yes I peeked sue me) do the same proof I did (not sure if it's a proof because of objection (1)) and then either 1) immediately conclude the bold statement without justification or 2) conclude that $z$ and $\bar{z}$ are the roots of $w$. But answer (2) looks wrong, since if $v>0$, $z$ is a square root of $w$ but $\bar{z}$ isn't and vice versa for $v < 0$. Although the statement is still true since if $z$ is a square root then so is $-z$, but then this just makes the preceding theorem seem irrelevant. Am I missing something?

Thanks.

sqrtoiler
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  • Edited fixing problem statement. It's correct now. – sqrtoiler Jan 06 '20 at 09:17
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    What does "There are two expressions for THE square root" mean? If $0\ne z\in \Bbb C$, there are exactly two complex numbers whose squares are equal to $ z.$ You are to take $ a $ and $b$ as non-negative; note that $|w|\pm u=\sqrt {u^2+v^2}\pm u\ge \sqrt {u^2}\pm u=|u|-u\ge 0$ when $u,v\in \Bbb R,$ where $\sqrt {.}$ denotes non-negative square root of a non-negative real and $|w|,|u|$ denote absolute values of the real numbers $w,u$. – DanielWainfleet Jan 06 '20 at 09:54

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1) This is a good point, but you can be sure that $|w| + u \ge 0$ and $|w|-u \ge 0$ because

$(|w|+u)(|w|-u)=|w|^2-u^2=v^2 \ge 0$

If $v^2=0$ then $v=0$ and $|w|=u$. If $v^2>0$ then $|w|+u$ and $|w|-u$ are both either strictly positive or strictly negative. But they cannot both be strictly negative because $(|w|+u)+(|w|-u)=2|w|\ge0$.

2) I think there is an implicit assumption that positive square roots are used in the definitions of $a$ and $b$, so $a,b \ge 0$. Then we have the following cases:

  • If $v > 0$ the square roots of $w$ are $z$ and $-z$ in the first and third quadrants
  • If $v < 0$ the square roots of $w$ are $\bar{z}$ and $-\bar{z}$ in the second and fourth quadrants
  • If $v=0$ then $z=\bar{z}=a=\sqrt{u}$ and the square roots of $w$ are $\pm \sqrt{u}$
  • The exception occurs when $w=0$, when the two square roots coincide
gandalf61
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  • I have selected this as the answer as it seems to give the most satisfactory reply that will be found. Your reply to my first objection is great, and your reply to the second is pretty much what I was thinking. Also, after examining some errata and professor commentaries on Rudin, it seems like little ambiguities like this might be something I just have to get used to. – sqrtoiler Jan 07 '20 at 02:47