Differentiating Complex Square Root Function: Bruce P. Palka, Ex. 1.5, Ch. III

[Math Amateur]

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 need help with an aspect of Example 1.5, Section 1.2, Chapter III ...

Example 1.5, Section 1.2, Chapter III, reads as follows:
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At the start of the above example we read the following:

" ... ... Write \(\displaystyle \theta (z) = \text{Arg } z\). Then \(\displaystyle \sqrt{z} = \sqrt{ \mid z \mid } e^{ i \theta(z)/2 }\). ... "My question is as follows:

How exactly is \(\displaystyle \sqrt{z} = \sqrt{ \mid z \mid } e^{ i \theta(z)/2 }\)In particular ... surely it should be \(\displaystyle \sqrt{z} = \mid \sqrt{ z } \mid e^{ i \theta(z)/2 }\) ... ...

Peter

 

[HallsofIvy]

In particular ... surely it should be \(\displaystyle \sqrt{z} = \mid \sqrt{ z } \mid e^{ i \theta(z)/2 }\) ... ...

Peter

It wouldn't make much sense to write \(\displaystyle \sqrt{z}\) in terms of \(\displaystyle |\sqrt{z}|\)!

For z a complex number, \(\displaystyle |z|\) is a positive real number and so is \(\displaystyle \sqrt{|z|}\). The purpose here is to write \(\displaystyle \sqrt{z}\) in the form \(\displaystyle re^{i\theta}\) with r and \(\displaystyle \theta\) real numbers.

For example, taking z= 4i, r= 4 and \(\displaystyle \theta= \pi/2\) so that |z|= 4 and \(\displaystyle \sqrt{|z|}= 2\). The two square roots of 4i are \(\displaystyle 2 e^{i\pi/4}= \sqrt{2}+ i\sqrt{2}\) and

\(\displaystyle 2 e^{i(\pi+ 2pi)/4}= 2e^{3i\pi/4}= \sqrt{2}- i\sqrt{2}\).

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[Math Amateur]

It wouldn't make much sense to write \(\displaystyle \sqrt{z}\) in terms of \(\displaystyle |\sqrt{z}|\)!

For z a complex number, \(\displaystyle |z|\) is a positive real number and so is \(\displaystyle \sqrt{|z|}\). The purpose here is to write \(\displaystyle \sqrt{z}\) in the form \(\displaystyle re^{i\theta}\) with r and \(\displaystyle \theta\) real numbers.

For example, taking z= 4i, r= 4 and \(\displaystyle \theta= \pi/2\) so that |z|= 4 and \(\displaystyle \sqrt{|z|}= 2\). The two square roots of 4i are \(\displaystyle 2 e^{i\pi/4}= \sqrt{2}+ i\sqrt{2}\) and

\(\displaystyle 2 e^{i(\pi+ 2pi)/4}= 2e^{3i\pi/4}= \sqrt{2}- i\sqrt{2}\).

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THanks so much for the help ...

.. it is much appreciated...

Peter