Complex number problem, could the problem be wrong?

[mariush]

Hi!

I was presented with this problem:

Let z +$\neq -1$ be a complex number with |z|=1.
Show that $\frac{z-1}{z+1}$ 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 $\frac{z-1}{z+1}$ = $\frac{0}{2}$ = 0, which is not an imaginary number..

Is the problem incorrect?

Thanks!

 

[Mark44]

Hi!

I was presented with this problem:

Let z +$\neq -1$ be a complex number with |z|=1.
Show that $\frac{z-1}{z+1}$ 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 $\frac{z-1}{z+1}$ = $\frac{0}{2}$ = 0, which is not an imaginary number..

Is the problem incorrect?

Thanks!

What you have is a valid counterexample to the statement.

 

[jackmell]

I think you have the question wrong. Let me try and make it more interesting:

If $|z|=1$, then show $\frac{z-1}{z+1}$ is pure imaginary except when $z=\pm 1$

 

[mariush]

Thanks for the replies.

The problem itself seems pretty straightforward on second look.

Let z = a + ib.
$\frac{(a-1) +ib}{(a+1) +ib}$

We multyply by the complex conjugate and get

$\frac{((a-1)+ib)((a+1)-ib)}{(a+1)^2+b^2}$,

$\frac{(a-1)(a+1) -ib(a-1) +ib(a+1) +b^2}{(a+1)^2 +b^2}$

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

$\frac{i*2b}{(a+1)^2+b^2}$,

 

[tiny-tim]

hi mariush!

this a + ib stuff is very undignified

try treating complex numbers as numbers, not number-pairs

rewrite |z| = 1 as $z\bar{z} = 1$

then what is the condition Re(f(z)) = 0 as an equation in z and $\bar{z}$ ? ​

 

[mariush]

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?

 

[tiny-tim]

hi mariush!

(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 $f(\bar{z})$ ?

 

[SteveL27]

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?