Complex analysis: Find a function that maps z1 and z2 onto w1 and w2

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Homework Help Overview

The problem involves finding a linear transformation that maps the complex numbers z1 = 2 and z2 = -3i to w1 = 1+i and w2 = 3. Participants are exploring the nature of this transformation and the conditions it must satisfy.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Some participants discuss the need to determine a function that correctly maps the specified inputs to outputs, questioning the validity of the textbook's provided answer. Others suggest using linear transformations and treating complex numbers as vectors in R2 to approach the problem.

Discussion Status

The discussion is ongoing, with participants questioning the accuracy of the textbook's answer and exploring different methods to derive the transformation. There is no consensus yet, as some participants have identified potential errors in the problem statement regarding the values of w2.

Contextual Notes

There is a noted discrepancy regarding the value of w2, with some participants suggesting it should be 1 instead of 3, which affects the approach to solving the problem.

Juwane
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Homework Statement



What is the function (linear transformation) that maps z_{1} = 2 and z_{2} = -3i onto w_{1} = 1+i and w_{2} = 3?

I think it's asking for the function that if you put 2 in it, it should give 1+i, and if you put -3i in the same function, it should give 3.

The answer given at the back of the book is w=f(z)=\frac{3+2i}{13}z + \frac{7+9i}{13}

Homework Equations



Maybe these would help:

x = \frac{z + \overline{z} }{2} and y = \frac{z - \overline{z} }{2i}

The Attempt at a Solution



I have no idea how to even start. The horrible book I am using doesn't give a clue. One possible way is to see what do we have to z_{1} = 2 to get 1+i. The answer is: -1+i , but we can't add this to z_{2} = -3i, since that would give us 1-2i whereas we must get 3. Is there any other way to find out?
 
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Juwane said:

Homework Statement



What is the function (linear transformation) that maps z_{1} = 2 and z_{2} = -3i onto w_{1} = 1+i and w_{2} = 3?

I think it's asking for the function that if you put 2 in it, it should give 1+i, and if you put -3i in the same function, it should give 3.

The answer given at the back of the book is w=f(z)=\frac{3+2i}{13}z + \frac{7+9i}{13}

Homework Equations



Maybe these would help:

x = \frac{z + \overline{z} }{2} and y = \frac{z - \overline{z} }{2i}

The Attempt at a Solution



I have no idea how to even start. The horrible book I am using doesn't give a clue. One possible way is to see what do we have to z_{1} = 2 to get 1+i. The answer is: -1+i , but we can't add this to z_{2} = -3i, since that would give us 1-2i whereas we must get 3. Is there any other way to find out?
The textbook's answer gives f(z) as a linear polynomial with complex coefficients, which is one way to represent this function. It gives the right results for the two given complex numbers.

Edit: Scratch part of what I said. The formula gives the right result for 2 + i, but not the right result for -3i.

Another way to approach this problem is to use linear transformations, and treat complex numbers as vectors in R2.

From the given information, T(2) = T(<2, 0>T) = <1, 1>T, and
T(-3i) = T(<0, -3>T) = <3, 0>T.

Using the properties of linear transformations, it's easy to find T(<1, 0>T) and T(<0, 1>T). That means that you can find T(a + bi) = T(<a, b>T) = aT(<1, 0>T) + bT(<0, 1>T), for any complex number a + bi.
 
Last edited:
I think you mean w2=1. Double check the book. A linear function is f(z)=a*z+b for complex constants a and b. Just put in the given values and solve the two simultaneous equations for a and b.
 
Last edited:
Dick is right. If w2 is 3, then the formula given as an answer is wrong.
 
Dick said:
I think you mean w2=1. Double check the book. A linear function is f(z)=a*z+b for complex constants a and b. Just put in the given values and solve the two simultaneous equations for a and b.

Yes, you're right. w2 is 1, not 3. I'll try to use the method you've given here. Thanks.
 

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