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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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    It seems so. If you want a slightly more geometric way of thinking about the same problem, then invent polar coordinates, start drawing unit circles, and think about what multiplication does to complex numbers of modulus 1. Once you understand that, reduce to this case by considering $z/|z|$ instead of $z$.2010-12-21
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    Yes, it's proven by computing $(a+bi)^2$ and $(a-bi)^2$.2010-12-21
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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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    I follow all the maths here but its your hint I am not getting... if v is 1 then v squared is 1 and the square root of that can be positive or negative. Why are we saying it has to be the positive form? likewise for the v<= form... In fact, now I look at the question I wonder if my confusion might be coming from the fact that z seems to be ambiguous since the real and imaginary parts can both be positive or negative. Maybe its just too long since I last played with complex numbers so I'm forgetting something...2010-12-21
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    This reminds me of Cauchy Schwartz.2010-12-21
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    @Chris: Consider $v=-2$, note that $(v^2)^\frac{1}{2}= -(-2)=-v$ . It is the key for the problem2010-12-21
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    @Trevor: can solve the problem with this?2010-12-21
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    @Chris: The square root of 1 is by convention +1, not -1. That's why we write the solutions of an equation like $x^2=3$ as $x=\sqrt{3}$ and $x=-\sqrt{3}$; the second solution is the negative solution, because the minus sign makes it so, but the actual symbol $\sqrt{3}$ denotes the *positive* solution.2010-12-21
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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