Show that question regarding complex numbers

**Glitch** · August 3rd 2010, 03:07 AM

The question:

Show that $Re(\frac{1 - z}{1 + z}) = 0$ for any complex number z with |z| = 1.

My attempt:
I tried finding the the result of the division, hoping to get a value of 0 as the real part, to confirm the problem. This is what I got:

Let z = a+ib
$\frac{1 - (a + ib)}{1 + (a + ib)} = \frac{(1 - a) - ib)}{(a + 1) + ib}$
$\frac{(1 - a) - ib)}{(a + 1) + ib} . \frac{(a + 1) - ib)}{(a + 1) - ib} = \frac{[(1 - a) - ib)][(a + 1) - ib]}{(a + 1)^2 + b^2}$
$\frac{(1 - a^2) - 2ib -b^2)}{(a + 1)^2 + b^2} = \frac{(1 - a) - 2ib -b^2}{(a + 1) + b^2}$
$\frac{(1 - a) - b^2 - 2ib}{(a + 1) + b^2}$
Therefore the real part is:
$\frac{(1 - a) - b^2}{(a+1) + b^2}$

Now I'm not sure where to go from here. I think what I've done so far is correct. Any help would be great, thanks!

**Archie Meade** · August 3rd 2010, 03:26 AM

Originally Posted by Glitch

The question:

Show that $Re(\frac{1 - z}{1 + z}) = 0$ for any complex number z with |z| = 1.

My attempt:
I tried finding the the result of the division, hoping to get a value of 0 as the real part, to confirm the problem. This is what I got:

Let z = a+ib
$\frac{1 - (a + ib)}{1 + (a + ib)} = \frac{(1 - a) - ib)}{(a + 1) + ib}$
$\frac{(1 - a) - ib)}{(a + 1) + ib} . \frac{(a + 1) - ib)}{(a + 1) - ib} = \frac{[(1 - a) - ib)][(a + 1) - ib]}{(a + 1)^2 + b^2}$
$\frac{(1 - a^2) - 2ib -b^2)}{(a + 1)^2 + b^2}$

Why are you dropping terms here?

$= \frac{(1 - a) - 2ib -b^2}{(a + 1) + b^2}$
$\frac{(1 - a) - b^2 - 2ib}{(a + 1) + b^2}$
Therefore the real part is:
$\frac{(1 - a) - b^2}{(a+1) + b^2}$

Now I'm not sure where to go from here. I think what I've done so far is correct. Any help would be great, thanks!

Note that since |z|=1, then

$a^2+b^2=1$

(Pythagoras' Theorem)

**Vlasev** · August 3rd 2010, 03:31 AM

I think you have a mistake either in your calculation or your writing. the final expression is:

[LaTeX ERROR: Convert failed]

Now, since the magnitude of z is 1, you have [LaTeX ERROR: Convert failed] . Thus the numerator simplifies to 0 and hence the statement is proven!

Edit: Didn't see you post Archie Meade.

**Glitch** · August 3rd 2010, 03:36 AM

Whoops, I cancelled the terms by accident. Now it makes sense! Thanks.

**HallsofIvy** · August 3rd 2010, 03:37 AM

Originally Posted by Glitch

The question:

Show that $Re(\frac{1 - z}{1 + z}) = 0$ for any complex number z with |z| = 1.

My attempt:
I tried finding the the result of the division, hoping to get a value of 0 as the real part, to confirm the problem. This is what I got:

Let z = a+ib
$\frac{1 - (a + ib)}{1 + (a + ib)} = \frac{(1 - a) - ib)}{(a + 1) + ib}$
[tex]\frac{(1 - a) - ib)}{(a + 1) + ib} . \frac{(a + 1) - ib)}{(a + 1) - ib} = \frac{[(1 - a) - ib)][(a + 1) - ib]}{(a + 1)^2 + b^2}[/MATH
$\frac{(1 - a^2) - 2ib -b^2)}{(a + 1)^2 + b^2} = \frac{(1 - a) - 2ib -b^2}{(a + 1) + b^2}$ ]

Looks like an error here, perhaps a typo. " $1- a^2$ " in the denominator of the first fraction has become " $1- a$ ". Also " $(a+ 1)^2$ " in the denominator has become " $(a+ 1)$ " but since a fraction is 0 if and only if the numerator is 0, the denominator is irrelevant.

$\frac{(1 - a) - b^2 - 2ib}{(a + 1) + b^2}$
Therefore the real part is:
$\frac{(1 - a) - b^2}{(a+1) + b^2}$

Now I'm not sure where to go from here. I think what I've done so far is correct. Any help would be great, thanks!

The numerator shoud be $1- a^2- b^2- 2ib$ - and $a^2+ b^2= 1$ .

**Glitch** · August 3rd 2010, 03:42 AM

It wasn't a typo, I foolishly tried to cancel the terms. Ahh well, I'll learn with practise!

**Plato** · August 3rd 2010, 03:57 AM

$\displaystyle\frac{{1 - z}}{{1 + z}} = \frac{{\left( {1 - z} \right)\left( {1 + \overline z } \right)}}{{\left| {1 + z} \right|^2 }} = \frac{{1 + \overline z - z - \left| z \right|^2 }}{{\left| {1 + z} \right|^2 }} = \frac{{\text{Im} (z)i}}{{\left| {1 + z} \right|^2 }}$
It is clear that the real part is zero.

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