Images of complex function

[Steven G]

I just paid my admission fee for help, by helping with posted problems.
Suppose w=f(z)= (i-1)/(z+i).
How can I find the image of
a) |z|=1 (the unit circle)
b) |z|<1 (inside the unit circle)
c) the imaginary axis.

Of course I can plot points that that is not the way to graph
For c) I got the the image was the line that crosses 1+i and -1-i but does not include the point 0

 

[mario99]

What have you done so far for (a)?

Hint: Parametrize the curve and let Desmos draw it.

 

[mario99]

For (a), I got this line. I am wondering if professors Dave and blamocur will get something else!

 

[mario99]

View attachment 38653

For (a), I got this line. I am wondering if professors Dave and blamocur will get something else!

This is for (b) if my calculations were correct ([imath]r = 0.7[/imath]). Both results, (a) and (b) are expected because the image of circles under [imath]\displaystyle\frac{i - 1}{z + i}[/imath] are either lines or circles.

The result of (c) is more interesting than (a) and (b). I will leave it for you as an exercise.

 

[khansaheb]

I can plot points that that is not the way to graph

Why not?

Plotting points is a time honored method of analyzing a function. May be not fast - but most of the time correct. I would rather be slow & correct than fast & wrong!!!

 

[khansaheb]

How can I find the image of

c) the imaginary axis.

The same way you would find the image of y-axis in real domain - trouble-maker.

 

[Dr.Peterson]

View attachment 38653

For (a), I got this line. I am wondering if professors Dave and blamocur will get something else!

Did you check the images of a couple points to make sure this is right? For instance, what is the image of -i?

View attachment 38654

This is for (b) if my calculations were correct ([imath]r = 0.7[/imath]). Both results, (a) and (b) are expected because the image of circles under [imath]\displaystyle\frac{i - 1}{z + i}[/imath] are either lines or circles.

How can (b) be so different from (a)??
If this is meant to be (c), check some points again. What is the image of 0?

What have you done so far for (a)?

Hint: Parametrize the curve and let Desmos draw it.

Of course, if one doesn't want to depend on technology, the parametric equation can be rewritten by eliminating the parameter ...

 

[blamocur]

Did you check the images of a couple points to make sure this is right? For instance, what is the image of -i?

How can (b) be so different from (a)??
If this is meant to be (c), check some points again. What is the image of 0?


Of course, if one doesn't want to depend on technology, the parametric equation can be rewritten by eliminating the parameter ...

I get a similar plot for [imath]|z|=1[/imath] but a different one for [imath]|z| = 0.7[/imath]:

 

[mario99]

Did you check the images of a couple points to make sure this is right? For instance, what is the image of -i?

How can (b) be so different from (a)??
If this is meant to be (c), check some points again. What is the image of 0?


Of course, if one doesn't want to depend on technology, the parametric equation can be rewritten by eliminating the parameter ...

I get a similar plot for [imath]|z|=1[/imath] but a different one for [imath]|z| = 0.7[/imath]:

Oops, I have made a mistake. Corrected it and got the image as professor blamocur.

I did the conjugate as [imath](re^{i\pi} + i)(re^{i\pi} - 1) = r^2 + 1[/imath]

when it was supposed to be [imath](re^{i\theta} + i)(re^{i\theta} - 1)[/imath]

I was surprised the simplification happened so fast. Now I know why?! Because it was wrong.



Comparison

 

[mario99]

Did you check the images of a couple points to make sure this is right? For instance, what is the image of -i?

Yes, I did. And I usually do what you do, [imath]1, -1, i, \ \text{and} \ -i[/imath]. That point is [imath](2,-i)[/imath]

How can (b) be so different from (a)??

Because the radius is changing.

If this is meant to be (c), check some points again. What is the image of 0?

No. It is (b) and I was not surprised to get a circle because I am familiar with [imath]\displaystyle \frac{i - 1}{z + i}[/imath]. There is nothing at zero but I think that as [imath]r[/imath] goes to [imath]\infty[/imath] the dot in the origin will be the circle. You already know that Desmos allows us to use [imath]r > 1[/imath], just to play around with the curve.

Of course, if one doesn't want to depend on technology, the parametric equation can be rewritten by eliminating the parameter ...

Believe it or not, it is more fun to draw the curve without Technology. It takes time and will not be so precise, but it helps when you are in a test where nothing else is allowed!

 

[mario99]

Correction.

I did the conjugate as [imath](re^{i\pi} + i)(re^{i\pi} - i) = r^2 + 1[/imath]

when it was supposed to be [imath](re^{i\theta} + i)(re^{i\theta} - i)[/imath]

It is too late to edit my post.

 

[Dr.Peterson]

It turns out I'd copied the problem wrong. To my eyes, i looks too much like 1.

For (b), I didn't realize that "This is for (b)" didn't mean it was the graph of the inequality, but just one curve within that region.

In any case, there's a reason I only asked questions, and didn't show what I got ...

 

[blamocur]

I have an impression that the OP wanted to get some answers about the result without having to compute too many points. Below is an attempt, at least for a):

If I remember correctly rational complex functions always map straight lines and circles to straight lines and circles. But [imath]\lim_{z\rightarrow i} f(z) = \infty[/imath], which means that the image of [imath]f[/imath] must be not a circle but a straight line. Now one can compute a couple of points and figure out the exact line.

 

[mario99]

I have an impression that the OP wanted to get some answers about the result without having to compute too many points. Below is an attempt, at least for a):

If I remember correctly rational complex functions always map straight lines and circles to straight lines and circles. But [imath]\lim_{z\rightarrow i} f(z) = \infty[/imath], which means that the image of [imath]f[/imath] must be not a circle but a straight line. Now one can compute a couple of points and figure out the exact line.

I am sure that you mean [imath]\lim_{z\rightarrow -i}f(z)[/imath]

And when you analyze this limit you will get

[imath]\lim_{z\rightarrow -i^{+}}f(z) = +\infty[/imath]

And

[imath]\lim_{z\rightarrow -i^{-}}f(z) = -\infty[/imath]

which is an indication that you have a line.

But that idea will not be so obvious if you are not familiar with that specific mapping.

It is like when you are in a calculus test and you are given this integral:

[imath]\displaystyle \int_{-\pi}^{\pi} x^3\ln\left(x^2\ln x^4\right)\sin^2x \ dx[/imath]

How many students will fail to answer this integral and say we did not study this?!

And

How many students will think outside the box and will realize that the integrand is odd?

which is an indication that:

[imath]\displaystyle \int_{-\pi}^{\pi} x^3\ln\left(x^2\ln x^4\right)\sin^2x \ dx = 0[/imath]

But you have to know that stuff. (odd thing.)

 

[blamocur]

I am sure that you mean \(\lim_{z\rightarrow -i}f(z)\)

You are right -- thank you.

 

[Steven G]

For the record, I figured this problem out. If anyone wants to see the solution I will write it up and post it here. It is quite long.

 

[mario99]

If anyone wants to see the solution I will write it up and post it here. It is quite long.

Of course I want to see your solution, professor Steven. And please in details.

For the record, I figured this problem out.

I do not have a single doubt you did that!