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[SOLVED] conformal mapping

dwsmith

Well-known member
Feb 1, 2012
1,673
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
Jan 26, 2012
4,193
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
Mar 1, 2012
275
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)
think about the x-axis y=0
we will have [tex]\frac{x}{x+i} = \frac{x(x-i)}{x^2+1} = \frac{x^2}{x^2+1} - \frac{xi}{x^2+1} [/tex] 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
Feb 7, 2012
2,707
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 $w=1$.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
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
Feb 7, 2012
2,707
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
Feb 1, 2012
1,673
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
Feb 1, 2012
1,673
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.