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The Square and Square Root Functions of a Complex Variable ... Gamelin, Ch. 1, Section 4

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,891
I am reading Theodore W. Gamelin's book: "Complex Analysis" ...

I am focused on Chapter 1: The Complex Plane and Elementary Functions ...

I am currently reading Chapter 1, Section 4: The Square and Square Root Functions ... and need some help in verifying a remark by Gamelin ... ...

The relevant section from Gamelin is as follows:


Gamelin - 1 - Remarks, Ch. 1, Section 4 - PART 1... .png
Gamelin - 2 - Remarks, Ch. 1, Section 4 - PART 2 ... .png


In the above text by Gamelin we read the following .... ...

" ... ... Every value \(\displaystyle w\) in the slit plane is the image of exactly two \(\displaystyle z\) values. one in the (open) right half-plane [Re \(\displaystyle z \gt 0\)], the other in the left half-plane [Re \(\displaystyle z \lt 0\)]. ... ... "


Now, I wanted to demonstrate via an example that a value of \(\displaystyle w\) was given by two values of \(\displaystyle z\) ... so I let \(\displaystyle w = 1 + i\) ... and proceeded as follows ...

\(\displaystyle w = 1 + i\)

so that

\(\displaystyle w = 2^{ \frac{1}{2} } e^{ i \frac{ \pi }{ 4} }\)


So then we have ... ...

\(\displaystyle z_1 = f_1(w) = w^{ \frac{1}{2} } = 2^{ \frac{1}{4} } e^{ i \frac{ \pi }{8} }\)

... and ...

\(\displaystyle z_2 = f_2(w) = - f_1(w) = -w^{ \frac{1}{2} } = 2^{ \frac{1}{4} } e^{ - i \frac{ \pi }{8} } \)


(Note that Gamelin uses \(\displaystyle f_2(w)\) for the second branch of \(\displaystyle w^{ \frac{1}{2} }\) ... and, further, notes that \(\displaystyle f_2(w) = - f_1(w)\) ... ... ... ... )


My problem is that I do not believe my value or \(\displaystyle z_2\) is correct ... but I cannot see where my process for calculating \(\displaystyle z_2\) is wrong ...

Can someone please explain my mistake and show and explain the correct process for calculating \(\displaystyle z_2\) ... ...



Help will be appreciated ...

Peter
 
Last edited:

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
What you have is correct. [tex]\omega= 2^{\frac{1}{2}}e^{i\frac{\pi}{4}}[/tex] so we have [tex]z= 2^{\frac{1}{4}}e^{I\frac{\pi}{8}}[/tex]. We can also write [tex]\omega= 2^{\frac{1}{2}}e^{i(\frac{\pi}{4}+ 2\pi)}= 2^{\frac{1}{2}}e^{i\frac{9\pi}{4}}[/tex] so that [tex]z= 2^{\frac{1}{4}}e^{i\frac{9\pi}{8}}[/tex]. Since [tex]\frac{9\pi}{8}= \pi+ \frac{\pi}{8}[/tex], [tex]e^{i\frac{9\pi}{8}}= e^{i\pi}e^{i\frac{\pi}{8}}= -e^{i\frac{\pi}{8}}[/tex] because [tex]e^{i\pi}= -1[/tex].
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,891
What you have is correct. [tex]\omega= 2^{\frac{1}{2}}e^{i\frac{\pi}{4}}[/tex] so we have [tex]z= 2^{\frac{1}{4}}e^{I\frac{\pi}{8}}[/tex]. We can also write [tex]\omega= 2^{\frac{1}{2}}e^{i(\frac{\pi}{4}+ 2\pi)}= 2^{\frac{1}{2}}e^{i\frac{9\pi}{4}}[/tex] so that [tex]z= 2^{\frac{1}{4}}e^{i\frac{9\pi}{8}}[/tex]. Since [tex]\frac{9\pi}{8}= \pi+ \frac{\pi}{8}[/tex], [tex]e^{i\frac{9\pi}{8}}= e^{i\pi}e^{i\frac{\pi}{8}}= -e^{i\frac{\pi}{8}}[/tex] because [tex]e^{i\pi}= -1[/tex].





Thanks for the help, HallsofIvy ....

However I still think my calculation of \(\displaystyle z_2\) ... that is \(\displaystyle z_2 = 2^{ \frac{1}{4} } e^{ - i \frac{ \pi }{8} }\) is incorrect ...


Note that \(\displaystyle e^{ - i \frac{ \pi }{8} } \neq - e^{ i \frac{ \pi }{8} }\) ...


But ... your post showed me the way ... as follows ..


\(\displaystyle z_2 = f_2(w) = - f_1(w) = -w^{ \frac{1}{2} } = -2^{ \frac{1}{4} } e^{ i \frac{ \pi }{8} }\)


So ...


\(\displaystyle z_2 = 2^{ \frac{1}{4} } ( - e^{ i \frac{ \pi }{8} } ) = 2^{ \frac{1}{4} } e^{ i \frac{ 9 \pi }{8} }\)


or ... if you want the argument to be between \(\displaystyle - \pi\) and \(\displaystyle + \pi\) ...


\(\displaystyle z_2 = 2^{ \frac{1}{4} } e^{ - i \frac{ 7 \pi }{8} }\)



Is that correct now ... ?

Peter
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,681
\(\displaystyle z_2 = 2^{ \frac{1}{4} } e^{ - i \frac{ 7 \pi }{8} }\)

Is that correct now ... ?

Peter
That is correct. Notice that $e^{ - i \frac{ 7 \pi }{8} } = e^{ -i \pi + i\frac{ \pi }{8}} = e^{-i\pi}e^{i\frac\pi8} = -e^{i\frac\pi8}$. Thus $z_2 = -z_1$, as must always be the case with square roots.