Find the image of the strip R = { z in complex with 0 < Re(z) <\(\pi \) } under f(z)=sin(z).

fiora
Oct 25, 2018

#1**+1 **

It looks like it maps to the strip

\(\{z: 0 \leq Re[z] \leq 1,~Im[z] \in \mathbb{R}\}\)

Rom
Oct 26, 2018

#3**+1 **

Thanks,Rom and Guest. Hey Rom,I think your answer is partially correct.

The mapping f(z)=sin(z)=sin(x+iy)=sin(x)cosh(y)+icos(x)cosh(y) map z to Real(f(z)) in positive real number and Im(f(z)) to real, which map to first and fourth quaduant and real axis of complex plane. That is my simpilfie answer and this probelm was one of test question,and I am doing test correction. Does anyone have any thoughts on this?

fiora
Oct 26, 2018

#4**+1 **

I dont understand, are you saying that the image of the strip R under sin(z) is the first quadrant+the fourth quadrant+the real axis?

This answer is wrong (If you know what the open mapping theorem is, you can see how the theorem shows that this can't be the image, and that the image has to be an open set)

Also keep in mind that sin(x+iy)=sin(x)cosh(y)+icos(x)sinh(y), not sin(x)cosh(y)+icos(x)cosh(y)

Guest Oct 26, 2018

#5**+1 **

Thanks guest fort spoting my mistake. I am sorry, I was doing another homework and didn't double check my typing.Yeah , you are right，the correct formula for sin(z) is sin(z)=sin(x+iy)= sin(x)*cosh(y)+i*cos(x)sinh(y). The image of strip R under mapping f(z)=sin(z) is quadrant+the fourth quadrant+**the positive real axis of complex plane.** Let w=f(z),then Re(w) is the set positive number and Im(w) is real number.Then the image is an open set,isn't it? Where the definition of open set is **A set is open if all its points are interior points,**and the definition of interior point is **A point a ∈ G is an interior point of G if some open disk with center a is a subset of G.** And **open disk is the inside of this circle with center a and radius r; we use the notation D[a,r] := {z ∈ C : |z − a| < r} .**

See a similar problem posted on stack exchange

I consider the last answer as correct answer in the preceeding link.

See my instuctor's answer to his sample test problem

follow the link, see problem 5

fiora
Oct 26, 2018

#6**+1 **

Yes I think your answer is correct (The image is the right half plane)

I followed your link but question 5 is different, am I missing something?

Guest Oct 27, 2018

#7**+1 **

Thanks for verification,guest. My professor like to give the student some wired question.He is differently something else and very infamous in my University. He even taught the co-chair of Department of Math & Computer Sicence of my University,so you can imagine his age. I am pretty sure that sample problem in the link is the best and the only sample I can found.Thanks again for your suggestion.

fiora
Oct 29, 2018