|
|
Post by mondo on Jun 5, 2023 7:08:16 GMT
On the bottom of page 462 author says "Consider the standard example $g(z) = e^{1/z}$... if we write $z = re^{i\theta}$ then $|g(z)| = e^{\frac{\cos(\theta)}{r}}$"
My question is how was the last formula for modulus of $g(z)$ obtained?
Thank you.
|
|
|
|
Post by Admin on Jun 5, 2023 14:06:31 GMT
Mondo
This is all on page 366! I don't understand where you got 462 from!
However:
If we write $\displaystyle z=re^{i\theta}$ then
$\displaystyle\frac{1}{z}=\frac{1}{r}e^{-i\theta}=\frac{1}{r}(\cos\theta-i\sin\theta)$
So $\displaystyle g(z)=e^{\frac{1}{z}}=e^{\frac{1}{r}e^{-i\theta}}=e^{\frac{1}{r}(\cos\theta-i\sin\theta)}=e^{\frac{\cos\theta}{r}}\cdot e^{\frac{-i\sin\theta}{r}}$
Therefore $\displaystyle |g(z)|=e^{\frac{\cos\theta}{r}}$
Just a bit of algebra.
Vasco
|
|
|
|
Post by mondo on Jun 5, 2023 20:49:56 GMT
Mondo This is all on page 366! I don't understand where you got 462 from! However: If we write $\displaystyle z=re^{i\theta}$ then $\displaystyle\frac{1}{z}=\frac{1}{r}e^{-i\theta}=\frac{1}{r}(\cos\theta-i\sin\theta)$ So $\displaystyle g(z)=e^{\frac{1}{z}}=e^{\frac{1}{r}e^{-i\theta}}=e^{\frac{1}{r}(\cos\theta-i\sin\theta)}=e^{\frac{\cos\theta}{r}}\cdot e^{\frac{-i\sin\theta}{r}}$ Therefore $\displaystyle |g(z)|=e^{\frac{\cos\theta}{r}}$ Just a bit of algebra. Vasco Thank you Vasco, I only wonder why can we drop the $e^{-i\sin\theta}$? It suggests me that this is because when we do the module it becomes symmetric with $e^{\frac{\cos\theta}{r}}$ but is it really the case? |
|
|
|
Post by Admin on Jun 5, 2023 21:08:02 GMT
Mondo
We are not dropping anything.
When we write a complex number as $Ue^{i\phi}$ where $U$ and $\phi$ are real numbers then $U$ is the modulus.
Vasco
|
|
|
|
Post by Admin on Jun 5, 2023 21:49:20 GMT
Mondo
$e^{\frac{\cos\theta}{r}}\cdot e^{\frac{-i\sin\theta}{r}}=Ue^{i\phi}$ where $U=e^{\frac{\cos\theta}{r}}$ and $e^{i\phi}=e^{\frac{-i\sin\theta}{r}}$
Vasco
|
|
|
|
Post by mondo on Jun 5, 2023 22:49:46 GMT
Right, sorry I got blinded for a moment. Okay I will likely have more questions to this section. Will update soon.
Thank you.
|
|
|
|
Post by mondo on Jun 6, 2023 7:11:00 GMT
Two more questions I do have:
1. In the middle of page 366, below the definition of essential singularities author writes "If $f$ were bounded in the vicinity of $s$ then $s$ would not be a singularity at all, but on the other hand $f(z)$ cannot approach $\infty$ az $z$ approaches $s$ from all directions, for then $s$ would only be a pole." - this is quite confusing to me. First of all why a bound on the function argument eliminates the chance of singularity in that point (or its vicinity)?
2. "Cannot approach $\infty$..would only be a pole" author didn't define "essential singularity" well so it is hard to tell what kind of alternative he is referring to here but what does he mean by "only a pole" what else could it be?
|
|
|
|
Post by Admin on Jun 6, 2023 9:32:14 GMT
Mondo
1. The bound is not on the function argument, but on the function $f$ itself - "If $f$ were bounded...".
2. An essential singularity is a point where the function is not bounded, but where the point is not a pole.
He has used the example $f=e^{1/z}$ to illustrate that this function has an essential singularity at the origin. This is illustrated in figure 19.
For a function to be described as bounded at a point $s$, then it cannot approach $\infty$ as we approach the point $s$, from any direction in the complex plane.
If a function at a point $s$ is not bounded but $s$ is not a pole then $s$ is an essential singularity.
Vasco
|
|
