Conformal Mapping of Strip -1 < Im(z) < 1

[Dustinsfl]

Describe the image of the strip $\{z: -1 < \text{Im} \ z < 1\}$ under the map $z\mapsto\dfrac{z}{z + i}$

So I know that $-\infty < x < \infty$ and $-1 < y < 1$.

Then
$$
\frac{x + yi}{x + i(y + 1)}
$$

Now if I take the the line y = -1, I have
$$
\frac{x-i}{x}
$$

Then find out what happens when y = 1.
Is this the correct way to do this type of problem?

 

[Ackbach]

Describe the image of the strip $\{z: -1 < \text{Im} \ z < 1\}$ under the map $z\mapsto\dfrac{z}{z + i}$

So I know that $-\infty < x < \infty$ and $-1 < y < 1$.

Then
$$
\frac{x + yi}{x + i(y + 1)}
$$

Now if I take the the line y = -1, I have
$$
\frac{x-i}{x}
$$

Then find out what happens when y = 1.
Is this the correct way to do this type of problem?

I don't know if it's the only way to do this problem, but it's the way I would do it.

 

[Amer]

you can't choose that line for two reasons it is not in the strip which you are trying to map, second it had the denominator zero which is (0,-1)
think about the x-axis y=0
we will have \frac{x}{x+i} = \frac{x(x-i)}{x^2+1} = \frac{x^2}{x^2+1} - \frac{xi}{x^2+1} for all real numbers
the real part it is between (0,1)
I do not know if that help but i think that is the only way you take take some points and see where they will go

 

[Opalg]

Describe the image of the strip $\{z: -1 < \text{Im} \ z < 1\}$ under the map $z\mapsto\dfrac{z}{z + i}$

If $w = \dfrac z{z+i}$ then $z = \dfrac{iw}{1-w}$. So we want to find conditions on $w$ to ensure that $-1 < \text{Re}\,\dfrac w{1-w} < 1.$

Let $w = u+iv$. Then $$\frac w{1-w} = \frac{u+iv}{1-u-iv} = \frac{(u+iv)(1-u+iv)}{(1-u-iv)(1-u+iv)}.$$ The real part of that is $$\frac{u(1-u)-v^2}{(1-u)^2+v^2}.$$ You need to find the region in which that fraction lies between –1 and +1.

[sp]I make the answer to be the region outside the circle of radius 1/4 centred at $w=3/4$, and to the left of the vertical line through $w=1$.[/sp]

 

[Dustinsfl]

If $w = \dfrac z{z+i}$ then $z = \dfrac{iw}{1-w}$. So we want to find conditions on $w$ to ensure that $-1 < \text{Re}\,\dfrac w{1-w} < 1.$

Let $w = u+iv$. Then $$\frac w{1-w} = \frac{u+iv}{1-u-iv} = \frac{(u+iv)(1-u+iv)}{(1-u-iv)(1-u+iv)}.$$ The real part of that is $$\frac{u(1-u)-v^2}{(1-u)^2+v^2}.$$ You need to find the region in which that fraction lies between –1 and +1.

[sp]I make the answer to be the region outside the circle of radius 1/4 centred at $w=3/4$, and to the left of the vertical line through $w=1$.[/sp]

I think you lost your i in the numerator. $iw = ui - v$

So I have
$$
\frac{-v+i(u-u^2-v^2)}{(1-u)^2+v}
$$
So for
$$
-1<\frac{-v}{(1-u)^2+v}\Leftrightarrow \frac{1}{4}<\left(v-\frac{1}{2}\right)^2+(1-u)^2
$$
and for
$$
1<\frac{-v}{(1-u)^2+v}\Leftrightarrow \frac{1}{4}<\left(v+\frac{1}{2}\right)^2+(1-u)^2
$$

So it is the two regions outside of the the circles of radius 1/4 centered at $\left(1,\pm\frac{1}{2}\right)$

 

[Opalg]

I think you lost your i in the numerator.

I don't think so. The problem asked for the image of the strip $\{z: -1 < \text{Im} \, z < 1\}$. Instead of taking the imaginary part of $z$, I took the real part of $iz$, which amounts to the same thing.

 

[Dustinsfl]

I don't think so. The problem asked for the image of the strip $\{z: -1 < \text{Im} \, z < 1\}$. Instead of taking the imaginary part of $z$, I took the real part of $iz$, which amounts to the same thing.

Since you multiplied z by i, should we have had $\dfrac{-w}{1-w}$ then too?

 

[Dustinsfl]

I have the $u<1$ and $\frac{1}{16}<\left(u-\frac{3}{4}\right)^2+v^2$. I don't see how you obtained the radius to be 1/4.