Understanding multivalued complex function

Question

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 ?

Answer 1

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