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Suppose $z = a+bi$ and $w = u+iv$. Let $\displaystyle a = \left(\frac{|w|+u}{2} \right)^{1/2}$ and $\displaystyle b = \left(\frac{|w|-u}{2} \right)^{1/2}$. Show that $z^2 = w$ if $v \geq 0$ and $(\bar{z})^{2} = w$ if $v \leq 0$.

So $|w| = \sqrt{u^2+v^2}$. So this is just a matter of computing $(a+bi)^2$ and $(a-bi)^2$ and substituting in the values?

Source: Chapter 1, Problem 10 from Principles of Mathematical Analysis by Rudin

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    Math enthusiasm is always good in my book! :)2010-12-21

2 Answers 2

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This is related to what Gunnar Magnusson said above. First suppose $|w| = 1$. Then $u = \cos(\theta)$ and $v = \sin(\theta)$ for some $\theta \in (-\pi,\pi]$, so that $w = e^{i\theta}$. Then $a = ({1 + \cos(\theta) \over 2})^{1 \over 2}$ and $b = ({1 - \cos(\theta) \over 2})^{1 \over 2}$. By the half-angle formulas, $a = \cos{\theta \over 2}$. Also, $ b= \sin{\theta \over 2}$ when $v \geq 0$, while $b = -\sin{\theta \over 2}$ when $v < 0$.

So if $v \geq 0$, $z = a + ib = \cos{\theta \over 2}+ i\sin{\theta \over 2} = e^{i{\theta \over 2}}$, while if $v < 0$, $\bar{z} = a - ib = \cos{\theta \over 2}+ i\sin{\theta \over 2} = e^{i{\theta \over 2}}$. Thus in the first case, $z^2 = w$, while in the second case $(\bar{z})^2 = w$. This is what you want.

If $|w| \neq 1$, the above proves it for ${w \over |w|}$ in place of $w$. Multiplying everything through by $|w|$ then gives your result.

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HINT:

The key is: if $v \geq 0 \Longrightarrow (v^{2})^\frac{1}{2}=v$ and if $v \leq 0 \Longrightarrow (v^{2})^\frac{1}{2}=-v$

For the first part Note that :

$z^{2}=a^{2}-b^{2}+2ab \imath = \frac{|w|+u}{2}-\frac{|w|-u}{2}+2\imath \left(\frac{|w|^{2}-u^{2}}{4} \right)^\frac{1}{2} = $ $ u + 2 \imath \left(\frac{\left(\sqrt{u^{2}+v^{2}}\right)^{2}-u^{2}}{4}\right)^\frac{1}{2}= u+2 \imath \left(\frac{v^{2}}{4}\right)^\frac{1}{2} = u + \imath (v^{2})^\frac{1}{2} = u+v\imath = w$

The other is same, using $-v$ (in the key) and expanding $\bar{z}$.

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    Hmm... if I was doing just rough maths then I'd agree that saying root of 1 is +1 is acceptable. However, while doign a rigorous mathematical proof its just not accurate from what I can see and so there must be something somewhere to say that the square root must be greater than 0 if that is indeed what must be happening. It just feels like an assumption has been made somewhere that I am missing.2010-12-23