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September 11th, 2016, 02:47 PM   #1
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Hey everybody, I'm taking an introduction to complex variables course, and I've run into a roadblock in my understanding of the material. It may seem trivial, but I'm having trouble understanding, the function w = z^2. I'm basically confused about this whole page of notes that I wrote down (along with what the "slit plane" is; I have no idea what this is). The way that my professor dealt with it in notes made it seem like w was a function of z, but it should be the other way around, I guess. If anyone can help explain (or surmise as much as they can from my notes), I would appreciate it. I guess to be more specific, I'm confused with the mapping: why the negative real axis isn't included, what the slit plane is, and why one of the 2 solutions to the z^2 = w function has pi being added to the angle. The images for what I'm talking about are attached. Thanks.
Attached Images WIN_20160911_18_40_08_Pro.jpg (94.5 KB, 5 views) WIN_20160911_18_41_27_Pro.jpg (93.1 KB, 6 views) September 11th, 2016, 03:24 PM #2 Senior Member   Joined: Sep 2016 From: USA Posts: 609 Thanks: 378 Math Focus: Dynamical systems, analytic function theory, numerics A slit plane is the usual complex plane with a (one-sided) infinite arc removed. In the example in your notes, the slit plane is $\mathbb{C} \setminus (-\infty,0]$ which means the entire complex plane but you remove the real interval $(-\infty,0]$. As for the function $w = z^2$, you are correct that $w$ is a complex variable which depends on another complex variable, $z$. $w = z^2$ is the formula which tells you what $w$ is for any given $z$. For example, $w(2) = 4$ and $w(.5) = .25$ so $w$ acts on reals exactly as the usual squaring function. However, $w(1+i) = (1+i)^2 = 2i$ and $w(i) = -1$, etc. The reason the two solutions to $z^2 = w$ differ by $\pi$ is because multiplication of complex numbers can be regarded as a rotation and a scaling. If we consider $a,b \in \mathbb{C}$ as polar vectors of the form $z = |z|\exp(i\theta)$ where $\theta$ is the polar angle and $|z|$ the length of the vector, then the product of 2 complex numbers is given by $ab = |a||b|\exp(i(Arg(a) + Arg(b))$. In the case $a = b$ this amounts to the formula $z \mapsto z^2$ by $z^2 = |z|^2\exp(2*Arg(z))$. September 11th, 2016, 03:36 PM #3 Newbie   Joined: Sep 2016 From: United States Posts: 2 Thanks: 0 Thanks for answering but I still have one question. What was the rationale for why that segment of the complex plane was removed (basically the negative real part removed), though? I don't see where that's coming from. September 11th, 2016, 03:46 PM   #4
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 Originally Posted by pirateprogrammer Thanks for answering but I still have one question. What was the rationale for why that segment of the complex plane was removed (basically the negative real part removed), though? I don't see where that's coming from.
This is called choosing a "branch". Its actually something you've seen before without realizing it. Back in high school we claimed that $\sqrt{16} = 4$ which is correct but why not $\sqrt{16} = -4$? Its clear that $(-4)^2 = 16$ as well and the fact is that it would serve equally well as the defined square root.

What is important is that if we want to define an inverse function for squaring, we have to make a choice. If we further want this inverse function to be holomorphic, then making that choice amounts to "deleting" the complex values at which the inverse would fail to be continuous. For this function, this is the negative real line.

For other functions, different choices will be made. Tags complex, elementary, functions, questions Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post feelthemath Complex Analysis 3 April 30th, 2015 12:47 AM wolfblood Algebra 4 November 12th, 2012 11:16 AM mathbalarka Real Analysis 2 August 30th, 2012 04:37 AM swtdelicaterose Linear Algebra 1 November 26th, 2009 04:06 AM rinomato1 Algebra 2 October 19th, 2009 12:43 PM

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