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It's my first encounter with multivalued function .

In book i am reading it's given that $$ \log z = \ln r + i (\theta + 2n\pi)$$

So first tell me if this thinking right

If $x = \sqrt 4$ then $x = 2$

And

if $x^2 = 4 $ then $ x= \pm2$

So square root of 4 is 2 as square root of positive number is always positive . So even if we replace x by complex z then same thing should apply in above equation that $z = \sqrt 4$ then $z = 2$ and $z^2 = 4$ then $z = \pm2$. Then

why it's that $\sqrt z$ is multi valued .

Now back to logarithmic function

I want to calculate log 1 so putting z = 1 we get $$ \log 1 = (2nπ )i $$

So isn't $ \log 1 $ a fix complex number how is so that it occupies different infinite places on complex plane . Doesn't this imply that all those points are same which obviously isn't true ?

So should not be it like that $\log 1 = 0 $ is fix point on complex plane but equation $\log z = 0$ has many solutions ?

So atleast when z is real then should not $\log z $ single valued ?

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    $\begingroup$ If $x$ is real and positive, then there are two numbers $y$ such that $y^2=x$, but one of those two numbers stands out as it is positive, so when we only have real numbers to worry about, we define $\sqrt x$ for positive $x$ to be the positive number whose square is $x$. When $x$ is not real, or real but negative, there are still two numbers $y$ with $y^2=x$, but now there is no good way to point to one of those two numbers and say, that's the one we want. So when we leave the reals behind, we're stuck with $\sqrt x$ being multi-valued. Similar remarks apply to the logarithm. $\endgroup$ May 14 at 4:17
  • $\begingroup$ Sorry sir but i am still confused . If i compare it with two variable function then , according to me $ y = \sqrt {1-x^2}$ is single valued whereas $ y^2 = 1-x^2 $ is multi valued ? Is it right ? Isn't whole problem rises when we take roots . But if already square root is taken in equation then there shouldn't be multi values . So isn't f(z) in $ f^2(z)=z$ is multivalued but single valued in $ f(z) = \sqrt z$ ? Are both same ? $\endgroup$
    – Rishi
    May 14 at 4:28
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    $\begingroup$ As a function of a complex variable, $\sqrt z$ has a "branch point" at the origin. If you delete from the domain a curve going from the origin to infinity – for example, the negative real axis – then you can define a single-valued, analytic function $\sqrt z$ on the new domain. In fact, you can define two of them, so you have to "choose a branch" if you really want a well-defined analytic $\sqrt z$ on the domain. If you insist on the domain being the entire complex plane, then you have to live with $\sqrt z$ being multi-valued, or else accept it being discontinuous. (continued) $\endgroup$ May 14 at 6:21
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    $\begingroup$ So, your choices are, for both $\sqrt z$ and $\log z$, to have it single-valued and analytic on the plane with a curve deleted, or to have it continuous on the entire plane but multivalued, or to have it single-valued on the entire plane but discontinuous. $\endgroup$ May 14 at 6:24
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    $\begingroup$ The picture here may be helpful in visualizing. $\endgroup$ May 14 at 13:17

1 Answer 1

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In Real Analysis, $\sqrt{4}$ refers unambiguously to $(+2)$, rather than the set $\{-2,2\}$.

In Complex Analysis, $\sqrt{4}$ refers to the set of all complex numbers of the form $(x + iy)$ where $(x + iy)^2 = 4$. Therefore, in Complex Analysis, $\sqrt{4}$ refers to the set of values $\{-2, +2\}$.


In Complex Analysis, for $z = (x + iy) : ~x,y \in \Bbb{R}$, if $z = [0 + i(0)]$, then $\sqrt{z}$ refers unambiguously to $(0)$.

Without loss of generality, assume that $z \in \Bbb{C}$, such that $z \neq 0$.

Then, there exists a unique value $r \in \Bbb{R^+}$, and a unique value $\theta \in (-\pi, \pi]$, such that

$z = r[\cos(\theta) + i\sin(\theta)] = re^{i\theta}.$

For a given non-zero complex $z_0 = re^{i\theta}$, $\sqrt{z_0}$ refers to the set of values

$\displaystyle \left\{\sqrt{r}e^{i(\theta/2)}, \sqrt{r}e^{i(\pi + \theta/2)}\right\}.$

The point is that both of the elements in the above set satisfy the equation $z^2 = z_0$.


The remainder of your posting, which asks about $\log(1)$ and (in general), $\log(z)$ is best discussed in the general case.

First of all, for $z\in \Bbb{C}, \log(z)$ only has meaning if $z \neq 0.$

For a given non-zero complex $z_0, \log(z_0)$ refers to the set of all complex numbers $z$ such that $e^z = z_0$.

Before discussing this for the general case, temporarily assume that $z_0$ is a positive real number. Then, consider the distinction between

  • Real Analysis : $\log(z_0)$
  • Complex Analysis : $\log(z_0)$.

In a sense, both of the references above refer to the complete set of values $z$ such that $e^z = z_0$. However, in Real Analysis, complex numbers with a non-zero imaginary component are outlawed. Therefore, in Real Analysis, for each positive real number $z_0$, there is exactly $(1)$ real number $z$ such that $e^z = z_0$.

In Real Analysis, this value is designated as $\log(z_0).$

In Complex Analysis, things change. This is because for each of the elements $w$ in the following set, $e^w = 1$:

$\text{Set} ~A = \{\cdots, -6i\pi, -4i\pi, -2i\pi, 0, 2i\pi, 4i\pi, 6i\pi, \cdots\}.$

This is because (for example), by definition, $e^{2i\pi} = \cos(2\pi) + i\sin(2\pi) = [1 + i(0)] = 1.$

Therefore, in the example above, when trying to determine the complete set of all values $z$ such that $e^z = z_0$, you can take any element $w$ from the set $A$, and add it to the Real Analysis specification of $\log(z_0)$.

Then, you have that $e^{w + \log(z_0)} = e^w \times e^{\log(z_0)} = 1 \times z_0 = z_0$.

This explains why, for any positive real number $z_0$, in Complex Analysis, $\log(z_0)$ is multi-valued. It is because any element $w$ could be chosen from the set $A$.


For any non-zero Complex $z_0$, you have virtually the same analysis. That is $z_0$ can be uniquely expressed as $re^{i\theta} ~: ~\theta \in (-\pi, \pi]$.

Then, one of the complex numbers $z$ that satisfies the constraint $e^z = z_0$ will be

$z = \log(r) + i\theta$
where $\log(r)$ refers to the Real Analysis specification for $\log(r)$.

Then, you have that

$\displaystyle e^z = e^{\log(r) + i\theta} = e^{\log(r)} \times e^{i\theta} = re^{i\theta} = z_0$.

Then, once $z$ has been identified, you can (again) couple it with any element $w$ from the set $A$.

That is $e^{z + w} = e^z \times e^w = z_0 \times 1 = z_0$.

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  • $\begingroup$ You have written $(\pi,\pi]$, which is surely not what you meant. $\endgroup$ May 14 at 13:08
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    $\begingroup$ @GerryMyerson Thanks for the assist, answer edited, intended $\theta \in (-\pi, \pi]$. $\endgroup$ May 14 at 13:10

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