On the meaning of multiple-valued functions

Question

I have a very weak understanding of what does it mean for a function to be multiple-valued. We have: $$z^{\frac{1}{2}}=|z|^{\frac{1}{2}}[\cos(\theta/2+k\pi)+i\sin(\theta/2+k\pi)]$$ for $k=0,1$. We have two outputs for every value of $z$ so I have no doubt this is a multiple-valued function. Now, the two branches are: $$f_1(z)=|z|^{\frac{1}{2}}e^{\frac{i\theta}{2}} ,\space f_2(z)=-|z|^{\frac{1}{2}}e^{\frac{i\theta}{2}}$$But we have: $$f_1(|z|,\theta)\not=f_1(|z|,\theta+2k\pi), \space f_2(|z|,\theta)\not=f_2(|z|,\theta+2k\pi)$$ My question is: are $f_1$ and $f_2$ multiple-valued functions themselves? If so, is branch cutting the means by which we can make them single-valued ?

Answer 1

I think your problem is essentially a problem of definition. If you want

$$f_1(z)=|z|^{1/2}e^{i\theta/2} ,\\ f_2(z)=-|z|^{1/2}e^{i\theta/2}$$

to be two distinct values of the square root, then you need an appropriate definition of $\theta$.

If you say that $\theta$ in each expression must take on all values in a set of the form $\{\theta+2k\pi \mid k\in\mathbb R\}$, then as you (almost?) discovered, $f_1$ will produce both values of the square root (one square root for even $k$ and the other square root for odd $k$), and so will $f_2$. So you are back where you started, but now with three names for the same function.

On the other hand, if you set $\theta=\arg(z)$ in your definitions of $f_1$ and $f_2$, then there is only one value of $\theta$ and therefore the functions $f_1$ and $f_2$ are single-valued.

Answer 2

In fact, $\theta$ is nothing more than the complex argument ($\arg$) of $z$ ($\theta = \arg\left( z \right)$). To say that $f$ from $\left| z \right|$ and $\arg\left( z \right)$ depends is the same as saying it depends on $z$ because those two properties were instrumental in defining the complex number uniquely, so I'd say it is still kinda single-valued. However, we consider them to be multivalued. But they are multi-valued.

Branch cuts do not arise for the single-valued trigonometric, hyperbolic, integer power, and exponential functions. However, their multivalued inverses do require branch cuts. - Weisstein, Eric W. "Branch Cut." from Mathworld

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A branch cut simply occurs when a multi-valued function is discontinuous. We call the points from which these branches rise branch points. More specifically, such a point exists when an n-valued function analytic in a neighborhood has more than $n$ values ​​in the immediate vicinity of that point.

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The branches differ in the $k$ as in $e^{x \cdot i} = e^{x \cdot i + 2 \cdot k \cdot \pi \cdot i}$ ('cause $e^{x}$ is periodic to $2 \cdot k \cdot \pi \cdot i$). If $k$ is even then one branch is present, if $k$ is odd then the second branch is present. This becomes obvious when we use the identity just mentioned for the derivation: $$ \begin{align*} z^{\frac{1}{2}} &\overset{\text{Polar Form}}{\equiv} \left( \left| z \right| \cdot e^{\arg\left( z \right) \cdot i} \right)^{\frac{1}{2}} \quad\mid\quad e^{\mathbb{R} \cdot i} = e^{\mathbb{R} \cdot i + 2 \cdot k \cdot \pi \cdot i}\\ z^{\frac{1}{2}} &= \left( \left| z \right| \cdot e^{\left( \arg\left( z \right) + 2 \cdot k \cdot \pi \right) \cdot i} \right)^{\frac{1}{2}}\\ z^{\frac{1}{2}} &= \left| z \right|^{\frac{1}{2}} \cdot \left( e^{\left( \arg\left( z \right) + 2 \cdot k \cdot \pi \right) \cdot i} \right)^{\frac{1}{2}}\\ z^{\frac{1}{2}} &= \left| z \right|^{\frac{1}{2}} \cdot e^{\frac{\arg\left( z \right) + 2 \cdot k \cdot \pi}{2} \cdot i}\\ z^{\frac{1}{2}} &\overset{\text{Euler's Formula}}{=} \left| z \right|^{\frac{1}{2}} \cdot \left( \cos\left( \frac{\arg\left( z \right) + 2 \cdot k \cdot \pi}{2} \right) + \sin\left( \frac{\arg\left( z \right) + 2 \cdot k \cdot \pi}{2} \right) \cdot i \right)\\ \end{align*} $$

$$ \begin{align*} z^{\frac{1}{2}} &= \begin{cases} \left| z \right|^{\frac{1}{2}} \cdot \left( \cos\left( \frac{\arg\left( z \right)}{2} \right) + \sin\left( \frac{\arg\left( z \right)}{2} \right) \cdot i \right),\, &\text{if}\, k\, \text{is}\, \text{even}\\ -\left| z \right|^{\frac{1}{2}} \cdot \left( \cos\left( \frac{\arg\left( z \right)}{2} \right) + \sin\left( \frac{\arg\left( z \right)}{2} \right) \cdot i \right),\, &\text{if}\, k\, \text{is}\, \text{odd}\\ \end{cases}\\ z^{\frac{1}{2}} &= \begin{cases} \left| z \right|^{\frac{1}{2}} \cdot e^{\frac{\arg\left( z \right)}{2} \cdot i},\, &\text{if}\, k\, \text{is}\, \text{even}\\ -\left| z \right|^{\frac{1}{2}} \cdot e^{\frac{\arg\left( z \right)}{2} \cdot i},\, &\text{if}\, k\, \text{is}\, \text{odd}\\ \end{cases}\\ \end{align*} $$