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Complex logarithm

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
I quote an unsolved question from Yahoo! Answers

let D be the domain obtained by deleting the ray {x, x<=0} from the plane and let G(z) be a branch of log z on D. Show that G maps D onto a horizontal strip of width of 2pi,
{ x+iy: belong to R, c<y<c+2pi} and the mapping is one to one on D.

Thanks
I have given a link to the topic there so the OP can see my complete response.
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Firstly consider $G:D\to\mathbb{C}$, $G(z)=\log z=\log |z|+i\arg z$ where $\arg$ is the principal argument of $z$. When $z$ varies on $D$, $|z|$ varies on $(0,+\infty)$ hence, $\log |z|$ varies on $(-\infty,+\infty)$. When $z$ varies on $D$, $\arg z$ varies on $(-\pi,\pi)$. This implies $$G(D)=\mathbb{R}+(-\pi,\pi)i=\{x+iy:x\in\mathbb{R},y\in(-\pi,\pi)\}$$ Another continuous argument has the form $\arg_c z\in(c,c+2\pi)$, and we get the result. On the other hand, $$G(z_1)=G(z_2)\Rightarrow \log |z_1|+i\arg_c z_1=\log |z_2|+i\arg_c z_2\Rightarrow\\ \log |z_1|=\log |z_2|\;\wedge\;\arg_c z_1=\arg_c z_2\Rightarrow |z_1|=|z_2|\;\wedge\;\arg_c z_1=\arg_c z_2\\\Rightarrow z_1=z_2\Rightarrow G\mbox{ is one to one on the domain }D$$