Complex Analysis simple Mapping question

In summary, the conversation discussed how to find the image of a rectangle with four vertices under the function f(z)=e^x. The solution involved mapping the corners of the rectangle to points on the complex plane and considering how the exponential function changes along the four straight lines between the corners. It was determined that the edges of the rectangle would be mapped to circular arcs and straight lines on the complex plane. The conversation also highlighted the importance of being precise in notation and thinking when working with complex numbers.
  • #1
RJLiberator
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Homework Statement



Find the image of the rectangle with four vertices A=0, B= pi*i, C= -1+pi*i, D = -1 under the function f(z)=e^x

2. The attempt at a solution
So, the graph of the original points is obvious.
Now I have to map them to the new function.
Seems easy enough, but I am not getting one step. Can someone help me get started?

We have f(z) = e^x

So is it as simple as letting A = 0 = z = e^0 aka 1.
B = pi*i so z = pi+i so f(z) = e^(pi*i) aka cos(pi)+i*sin(pi) = -1
C = -1*pi*i so z = -1+pi*i and thus e^(-1+pi*i) = (1/e)*e^(pi*i) = -1/e
D = -1 so z = -1 and e^(-1) = 1/e

Thus we'd have points (0, 1) = A, (pi*i, -1) = B, (-1+pi*i, -1/e) = C, (-1, 1/e) = D

Not sure how i'd graph C tho :/
 
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  • #2
Yes, you have mapped the corners correctly. Now consider how the exponential changes along the 4 straight lines between the corners A,B,C,D. Example: For z going from A to B only the angle of ez changes. The modulus remains constant. So that line AB gets mapped in a circular arc.
 
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  • #3
RJLiberator said:
Thus we'd have points (0, 1) = A, (pi*i, -1) = B, (-1+pi*i, -1/e) = C, (-1, 1/e) = D
Up to this point, I agree. What do you mean here? A = 0 => eA = e0 = 1; You just need to graph the points 1, -1, -1/e, and 1/e and map the lines between them. For instance, the line AB gets mapped to an arc of constant radius=1 that is centered at 0 and goes counter clockwise from 1 to -1. Do the same thinking for the other 3 lines.
 
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  • #4
Hm. Well, I am very happy to hear my early calculations are correct.

I'm not quite sure how to graph it.

From your second post, A goes to (1, 0) ?? instead of (0, 1)?

B goes to (-1, 0) ?? And then it's a circle with radius one connecting them so it's a semi-circle above the x-axis? Then what happens to the other lines?
 
  • #5
RJLiberator said:
And then it's a circle with radius one connecting them so it's a semi-circle above the x-axis?
Yes.
Then what happens to the other lines?
Just remember that ez = ex+iy = exeiy. So if x=Re(z) is constant while y=Im(z) changes, then the modulus is constant and only the angle changes. Similarly, if y=Im(z) is constant while x=Re(z) changes, then the angle is constant while the modulus changes. So the 4 edges of the rectangle should be simple to map. Just keep track of what is changing and if it is increasing or decreasing.
 
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  • #6
So C to D would also be a semi circle from -1/e to 1/e. Hm, this doesn't seem right.
Your explanations are clear, but I am still missing something on a basic level it seems.

Not sure what B to C and D to A would be like.

A to B is decreasing
B to C is increasing
C to D is increasing
D to A is increasing
 
  • #7
A=0+i0 to B=0+iπ : real is constant 0 while imaginary increases from 0 to iπ => maps to an arc of radius 1 and angle from 0 increasing to π
B=0+iπ to C=-1+iπ : imaginary is constant iπ while real decreases from 0 to -1 => maps to a line of constant angle π whose modulus decreases from 1 to e-1

Do similar logic for C to D and for D to A
 
  • #8
Hm. Lights are starting to flicker.

So A to b increased from 0 to pi in angle with an arc of radius 1
B to C is a line that goes down from 0 to -1 in the imaginary axis
C to D is another angle dependent one that goes from pi to 0 degrees
and then D to A is a straight line back to the origin

Similar to polar coordinates it seems.

I really appreciate your help, time for me to go to work now. You have been wonderful and I plan on re-reading what you wrote again later to fully understand the problem.

Cheers.
 
  • #9
Ok. Be sure to be very precise in your notation and thinking. Talk about the real and imaginary parts of A, B, C, and D. Talk about the modulus and argument (angle) of eA, eB, eC, and eD. Don't mix them up. You seem to be doing that in your writing, and maybe in your thinking.
 
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  • #10
Hey mate,

I just wanted to follow up and thank you for your help. I understand now what you were trying to tell me.

So it's a semi-circle above the x-axis that goes from 1 to -1, and then comes into -1/e and back with a new radius to 1/e.
It makes sense to me. Still hard for me to explain mathematically, but it makes sense.

Thanks again for the help.
 
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Related to Complex Analysis simple Mapping question

1. What is complex analysis?

Complex analysis is a branch of mathematics that deals with the study of functions of complex numbers. It involves the analysis of complex-valued functions, derivatives, and integrals.

2. What is a simple mapping in complex analysis?

A simple mapping, also known as a conformal mapping, is a transformation of a complex plane that preserves angles and shapes. It is a one-to-one mapping that preserves the structure of the original function.

3. How is complex analysis used in real life?

Complex analysis has various applications in fields such as engineering, physics, and economics. It is used to model and analyze systems with complex variables, such as electrical circuits, fluid flow, and economic models.

4. What are some common techniques used in complex analysis?

Some common techniques used in complex analysis include Cauchy's integral theorem, power series, residue theorem, and conformal mapping. These techniques are used to solve problems related to complex functions and their derivatives.

5. What are the benefits of studying complex analysis?

Studying complex analysis can improve problem-solving skills, analytical thinking, and mathematical reasoning. It also provides a better understanding of complex systems and their behavior, making it a valuable tool in various fields of science and engineering.

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