Corresponding Areas in the z and w Planes for Algebraic Equations

In summary, you need to solve for z when solving for the interior of the unit circle in the w-plane. This is the same thing as solving for r(radius) in the real plane.
  • #1
Wishbone
139
0
The question reads:

"what part of the z-plane corresponds to the interior of the unit circle in the w-plane if

a) w = (z-1)/(z+1) b) w = (z-i)/(z+i)"


I really am having problems understanding what the question is asking. I don't understand what the w plane is, and in which plane is the unit circle, the z plane, or the w plane?
 
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  • #2
w is a function of z. And it's a bijection. So it takes the complex plane and spawns it all over again. This is what we mean by the w-plane: the complex plane, as spawned by the function w. It's no different than the complex plane itself.

What the question is asking is, if you consider the interior of the unit circle in the w-plane (read complex plane), what is its pre-image? I.e. what subset A is such that w(A) = interior of the unit circle. You will have to find the inverse function [itex]w^{-1}[/itex] and apply the unit circle to it. See what it spawns.

Hope that was understandable.
 
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  • #3
Exactly what quasar987 said. Essentially you need to solve the inequalities
a) |(z-1)(z+1)|< 1 b) |(z-1)/(z+1)|< 1

I would recommend starting by solving the corresponding equations.
 
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  • #4
wow hmm...


Are you saying that I need to find the numbers that are within the unit circle in the complex Z plane?
 
  • #5
More like "what numbers in the complex z plane correspond (via the w mapping) to numbers that are in the unit circle in the w-plane."
 
  • #6
one question, about solving for z, when I solve those inequaltities for z, what is that tell me? Is that the same thing as r(radius) in the real plane?
 
  • #7
also I cannot seem to solve those equations for z? is it possible?
 
  • #8
Wishbone said:
one question, about solving for z, when I solve those inequaltities for z, what is that tell me? Is that the same thing as r(radius) in the real plane?

Well, no, z is a complex number and so is not a radius!

also I cannot seem to solve those equations for z? is it possible?

Yes, of course it's possible. Setting z= x+ iy, then |(z-1)(z+1)|= |z2- 1|= |(x2-y2)+i(2xy)|.

|(z-1)(z+1)|=|(x2-y2)+i(2xy)|= 1 is the same as
[tex]\sqrt{(x^2-y^2)^2+ 4x^2y^2}= 1[/tex]

What (x,y) satisfy that?
 

FAQ: Corresponding Areas in the z and w Planes for Algebraic Equations

What is complex algebra?

Complex algebra is a branch of mathematics that deals with the manipulation and solving of equations involving complex numbers. These numbers are a combination of a real number and an imaginary number, represented as a + bi, where a and b are real numbers and i is the imaginary unit.

What is the difference between real and complex numbers?

The main difference between real and complex numbers is that real numbers can be represented on a number line, while complex numbers cannot. Real numbers only have a real part, while complex numbers have both a real and an imaginary part. Additionally, real numbers follow the properties of addition, subtraction, multiplication, and division, while complex numbers have additional properties such as conjugation and modulus.

How do you add and subtract complex numbers?

To add or subtract complex numbers, you simply add or subtract the real parts and the imaginary parts separately. For example, (2 + 3i) + (4 + 2i) = (2 + 4) + (3 + 2)i = 6 + 5i. Similarly, (5 - 2i) - (3 + 4i) = (5 - 3) + (-2 - 4)i = 2 - 6i.

How do you multiply and divide complex numbers?

To multiply complex numbers, you use the FOIL method, where FOIL stands for First, Outer, Inner, Last. This means you multiply the first terms, then the outer terms, then the inner terms, and finally the last terms. For example, (2 + 3i)(4 + 2i) = 2(4) + 2(2i) + 3i(4) + 3i(2i) = 8 + 4i + 12i + 6i^2 = 8 + 16i - 6 = 2 + 16i. To divide complex numbers, you use the conjugate of the denominator to rationalize the expression. For example, (2 + 3i) / (4 + 2i) = [(2 + 3i)(4 - 2i)] / [(4 + 2i)(4 - 2i)] = (8 + 16i - 6) / 20 = 2/5 + (8/20)i = 2/5 + 2/5i.

How do you solve complex algebraic equations?

To solve complex algebraic equations, you use the same methods as solving real algebraic equations, such as combining like terms, isolating the variable, and using the properties of multiplication and division. However, you must also remember to simplify any complex numbers and use the FOIL method or conjugates when necessary. It is also important to check your answer by plugging it back into the original equation to ensure it is correct.

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