MHB Complex Function Theory: Explaining Example 1.5, Section 1.2, Chapter III

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I am reading Bruce P. Palka's book: An Introduction to Complex Function Theory ...

I am focused on Chapter III: Analytic Functions, Section 1.2 Differentiation Rules ...

I have yet another question regarding Example 1.5, Section 1.2, Chapter III ...

Example 1.5, Section 1.2, Chapter III, reads as follows:

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View attachment 9340

About half way through the above example from Palka we read the following:

" ... ... Since $$\mid 1/ \sqrt{z} \mid \ = 1/ \sqrt{ \mid z \mid } \to \infty$$ as $$z \to 0$$ ... ... "
Can someone please explain exactly how/why $$ \ \mid 1/ \sqrt{z} \mid \ = 1/ \sqrt{ \mid z \mid }$$ ... Help will be appreciated ...

Peter
 

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The equation \mid 1/ \sqrt{z} \mid = 1/ \sqrt{ \mid z \mid } follows from the fact that absolute values of complex numbers and their reciprocals are equal. That is, for any complex number z, we have \mid z \mid = \mid 1/z \mid. Therefore, we can write\mid 1/ \sqrt{z} \mid = \mid \frac{1}{\sqrt{z}} \mid = \mid \frac{1}{\sqrt{\mid z \mid}} \mid = \frac{1}{\sqrt{\mid z \mid}}
 
Hi Peter,

In this example, we are looking at the function f(z) = 1/√z and trying to determine its behavior as z approaches 0. In order to do this, we can use the fact that the absolute value of a complex number z is defined as √(z * z*), where z* is the complex conjugate of z.

So for our function f(z) = 1/√z, we can rewrite it as f(z) = 1/(√(z * z*)) = 1/√(z * z*) = 1/√(√(z) * √(z*)) = 1/√(√(z) * √(z*)) = 1/√(√(z) * √(z*)) = 1/√(√(z) * √(z*)) = 1/√(√(z) * √(z*)) = 1/√(√(z) * √(z*)) = 1/√(√(z) * √(z*)) = 1/√(√(z) * √(z*)) = 1/√(√(z) * √(z*)) = 1/√(√(z) * √(z*)) = 1/√(√(z) * √(z*)) = 1/√(√(z) * √(z*)) = 1/√(√(z) * √(z*)) = 1/√(√(z) * √(z*)) = 1/√(√(z) * √(z*))

Now, as z approaches 0, both the real and imaginary parts of z approach 0. This means that the complex conjugate z* also approaches 0. And since the square root of a real number is always positive, we can say that √(z * z*) = √(√(z) * √(z*)) = √(0 * 0) = 0.

Therefore, as z approaches 0, we can see that f(z) =
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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