Complex number problem, could the problem be wrong?

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Discussion Overview

The discussion revolves around a problem involving complex numbers, specifically examining whether the expression \(\frac{z-1}{z+1}\) is an imaginary number given that \(|z|=1\) and \(z \neq -1\). Participants explore algebraic manipulations and interpretations of the problem, questioning its validity and the nature of complex numbers.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant presents a counterexample using \(z=1+0b\), concluding that \(\frac{z-1}{z+1} = 0\), which they argue is not an imaginary number, suggesting the problem may be incorrect.
  • Another participant proposes that the question might be misinterpreted and suggests that \(\frac{z-1}{z+1}\) is pure imaginary except when \(z=\pm 1\).
  • A different participant provides an algebraic approach, expressing \(z\) in terms of \(a\) and \(b\) and manipulating the expression to show it simplifies under the condition \(|z|=1\).
  • Some participants discuss the representation of complex numbers, with one suggesting to treat them as numbers rather than pairs, while another finds it helpful to think of them as vectors.
  • There is a debate about whether 0 can be considered a pure imaginary number, with one participant asserting that it is, while another challenges this perspective.

Areas of Agreement / Disagreement

Participants express differing views on the validity of the original problem and the interpretation of imaginary numbers, indicating that multiple competing views remain and the discussion is unresolved.

Contextual Notes

There are unresolved assumptions regarding the definitions of imaginary numbers and the conditions under which the original statement holds true. The algebraic manipulations presented may depend on specific interpretations of complex numbers.

mariush
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Hi!

I was presented with this problem:

Let z +[itex]\neq -1[/itex] be a complex number with |z|=1.
Show that [itex]\frac{z-1}{z+1}[/itex] is an imaginary number.
I am getting nowhere with the algebra, but i did notiec one thing:

let z=1+0b and the equation says [itex]\frac{z-1}{z+1}[/itex] = [itex]\frac{0}{2}[/itex] = 0, which is not an imaginary number..

Is the problem incorrect?

Thanks!
 
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mariush said:
Hi!

I was presented with this problem:

Let z +[itex]\neq -1[/itex] be a complex number with |z|=1.
Show that [itex]\frac{z-1}{z+1}[/itex] is an imaginary number.
I am getting nowhere with the algebra, but i did notiec one thing:

let z=1+0b and the equation says [itex]\frac{z-1}{z+1}[/itex] = [itex]\frac{0}{2}[/itex] = 0, which is not an imaginary number..

Is the problem incorrect?

Thanks!

What you have is a valid counterexample to the statement.
 
I think you have the question wrong. Let me try and make it more interesting:

If [itex]|z|=1[/itex], then show [itex]\frac{z-1}{z+1}[/itex] is pure imaginary except when [itex]z=\pm 1[/itex]
 
Thanks for the replies.

The problem itself seems pretty straightforward on second look.

Let z = a + ib.
[itex]\frac{(a-1) +ib}{(a+1) +ib}[/itex]

We multyply by the complex conjugate and get

[itex]\frac{((a-1)+ib)((a+1)-ib)}{(a+1)^2+b^2}[/itex],

[itex]\frac{(a-1)(a+1) -ib(a-1) +ib(a+1) +b^2}{(a+1)^2 +b^2}[/itex]

And since |z| = [itex]\sqrt{a^2 +b^2}[/itex]=1 [itex]\Rightarrow[/itex] a^2 + b^2 = 1
, the a^2 + b^2 -1 casel and all that is left is

[itex]\frac{i*2b}{(a+1)^2+b^2}[/itex],
 
hi mariush! :smile:

this a + ib stuff is very undignified :redface:

try treating complex numbers as numbers, not number-pairs

rewrite |z| = 1 as [itex]z\bar{z} = 1[/itex]

then what is the condition Re(f(z)) = 0 as an equation in z and [itex]\bar{z}[/itex] ? :wink:
 
Tim :)

I agree it looks horrible. I have found it helpful to think of the number-pairs as vectors (hencec thinking in number pairs) and since it's my first day whit these complex numbers I haven't gotten to the section that discusses de Moivres formula ++ yet.

I don't think i fully understand what you ment by the last. Maybe you could explain?
 
hi mariush! :smile:

(no, I'm not suggesting you use de moivre)

if a function f(z) is purely imaginary (ie Re(f(z)) = 0), then what is the relation between f(z) and [itex]f(\bar{z})[/itex] ? :wink:
 
Mark44 said:
What you have is a valid counterexample to the statement.

I don't think so. 0 is a pure imaginary number. Its real part is zero, right?
 

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