MHB Verify Gamelin's Remark: Complex Square and Square Root Functions

[Math Amateur]

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:View attachment 9288
View attachment 9289In the above text by Gamelin we read the following ... ...

" ... ... Every value $$w$$ in the slit plane is the image of exactly two $$z$$ values. one in the (open) right half-plane [Re $$z \gt 0$$], the other in the left half-plane [Re $$z \lt 0$$]. ... ... "Now, I wanted to demonstrate via an example that a value of $$w$$ was given by two values of $$z$$ ... so I let $$w = 1 + i$$ ... and proceeded as follows ...

$$w = 1 + i$$

so that

$$w = 2^{ \frac{1}{2} } e^{ i \frac{ \pi }{ 4} }$$So then we have ... ...

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

... and ...

$$ z_2 = f_2(w) = - f_1(w) = -w^{ \frac{1}{2} } = 2^{ \frac{1}{4} } e^{ - i \frac{ \pi }{8} } $$(Note that Gamelin uses $$f_2(w)$$ for the second branch of $$w^{ \frac{1}{2} }$$ ... and, further, notes that $$f_2(w) = - f_1(w)$$ ... ... ... ... )My problem is that I do not believe my value or $$z_2$$ is correct ... but I cannot see where my process for calculating $$z_2$$ is wrong ...

Can someone please explain my mistake and show and explain the correct process for calculating $$z_2$$ ... ...
Help will be appreciated ...

Peter

 

[HallsofIvy]

What you have is correct. \omega= 2^{\frac{1}{2}}e^{i\frac{\pi}{4}} so we have z= 2^{\frac{1}{4}}e^{I\frac{\pi}{8}}. We can also write \omega= 2^{\frac{1}{2}}e^{i(\frac{\pi}{4}+ 2\pi)}= 2^{\frac{1}{2}}e^{i\frac{9\pi}{4}} so that z= 2^{\frac{1}{4}}e^{i\frac{9\pi}{8}}. Since \frac{9\pi}{8}= \pi+ \frac{\pi}{8}, e^{i\frac{9\pi}{8}}= e^{i\pi}e^{i\frac{\pi}{8}}= -e^{i\frac{\pi}{8}} because e^{i\pi}= -1.

 

[Math Amateur]

What you have is correct. \omega= 2^{\frac{1}{2}}e^{i\frac{\pi}{4}} so we have z= 2^{\frac{1}{4}}e^{I\frac{\pi}{8}}. We can also write \omega= 2^{\frac{1}{2}}e^{i(\frac{\pi}{4}+ 2\pi)}= 2^{\frac{1}{2}}e^{i\frac{9\pi}{4}} so that z= 2^{\frac{1}{4}}e^{i\frac{9\pi}{8}}. Since \frac{9\pi}{8}= \pi+ \frac{\pi}{8}, e^{i\frac{9\pi}{8}}= e^{i\pi}e^{i\frac{\pi}{8}}= -e^{i\frac{\pi}{8}} because e^{i\pi}= -1.

Thanks for the help, HallsofIvy ...

However I still think my calculation of $$z_2$$ ... that is $$z_2 = 2^{ \frac{1}{4} } e^{ - i \frac{ \pi }{8} }$$ is incorrect ...Note that $$e^{ - i \frac{ \pi }{8} } \neq - e^{ i \frac{ \pi }{8} }$$ ...But ... your post showed me the way ... as follows ..$$z_2 = f_2(w) = - f_1(w) = -w^{ \frac{1}{2} } = -2^{ \frac{1}{4} } e^{ i \frac{ \pi }{8} }$$So ...$$z_2 = 2^{ \frac{1}{4} } ( - e^{ i \frac{ \pi }{8} } ) = 2^{ \frac{1}{4} } e^{ i \frac{ 9 \pi }{8} }$$ or ... if you want the argument to be between $$- \pi$$ and $$+ \pi$$ ...$$z_2 = 2^{ \frac{1}{4} } e^{ - i \frac{ 7 \pi }{8} }$$
Is that correct now ... ?

Peter

 

[Opalg]

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

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.