# [SOLVED]conformal mapping

#### dwsmith

##### Well-known member
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

##### Indicium Physicus
Staff member
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

##### Active member
you cant 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)
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

##### MHB Oldtimer
Staff member
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.

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 $w=1$.

• Amer and dwsmith

#### dwsmith

##### Well-known member
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.

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 $w=1$.

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 $\left(1,\pm\frac{1}{2}\right)$

Last edited:

#### Opalg

##### MHB Oldtimer
Staff member
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.

#### dwsmith

##### Well-known member
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?

#### dwsmith

##### Well-known member
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.