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

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

Peter

 

[HallsofIvy]

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

Peter

It wouldn't make much sense to write [math]\sqrt{z}[/math] in terms of [math]|\sqrt{z}|[/math]!

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

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

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

​

 

[Math Amateur]

It wouldn't make much sense to write [math]\sqrt{z}[/math] in terms of [math]|\sqrt{z}|[/math]!

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

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

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

​

THanks so much for the help ...

.. it is much appreciated...

Peter