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- Jun 22, 2012

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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:

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

" ... ... Since \(\displaystyle \mid 1/ \sqrt{z} \mid \ = 1/ \sqrt{ \mid z \mid } \to \infty\) as \(\displaystyle z \to 0\) ... ... "

Can someone please explain exactly how/why \(\displaystyle \ \mid 1/ \sqrt{z} \mid \ = 1/ \sqrt{ \mid z \mid }\) ...

Help will be appreciated ...

Peter

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:

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

" ... ... Since \(\displaystyle \mid 1/ \sqrt{z} \mid \ = 1/ \sqrt{ \mid z \mid } \to \infty\) as \(\displaystyle z \to 0\) ... ... "

Can someone please explain exactly how/why \(\displaystyle \ \mid 1/ \sqrt{z} \mid \ = 1/ \sqrt{ \mid z \mid }\) ...

Help will be appreciated ...

Peter

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