Question

Let z be a complex number where z doesn't equal +/- i.
Prove that z/(1+z^2) is real if and only if z is real or |z| = 1.

I have an idea on how to solve this, but I want to know if the conjugate of z/(1+z^2) is
(1+(conjugate of z)^2)/(conjugate of z)

Answer 1

is that the proper conjugate?

Answer 2

because I know that in order for it to be real z/(1+z^2) = (z/(1+z^2))*

Answer 3

[ z/(1+z^2) ]* = z*/(1+z*^2)

Answer 4

Actually your idea is a smart one. A complex number is real if and only if it equally its conjugate, yes.

Answer 5

alright I will make an attempt with this

Answer 6

Yes, this is a very nice approach.

Answer 7

things are getting very large when I set z = (x+yi)
can anything meaningful be obtained by keeping it as z and z* or should I keep going with this expansion?

Answer 8

if z=0 then both sides are satisfied...so assume \(z\ne 0\)

then if \[\frac{z}{1+z^2}\] is real then \[\frac{1+z^2}{z}=\frac{1}{z}+z\] is real

Answer 9

let \[z=a+bi\]

then you get \[\frac{a}{a^2+b^2}+a+\left(b-\frac{b}{a^2+b^2}\right)i\]

this is real if b=0 or \(|z|=1\)

Answer 10

what did you put the z =a + bi into to get that?

Answer 11

\[\frac{1}{z}+z\]

Answer 12

I'm finding how you got there hard to follow.

Answer 13

wait I think I see where you multiplied the bottom by the conjugate

Answer 14

at the beginning when you say both sides are satisfied what are you referring to?
also does this satisfy an iff?

Answer 15

alright I see what you are referring to now

Answer 16

I'm saying if z=0 then z/(1+z^2) is real iff z is real

Answer 17

I need to get that case out of the way since I reciprocated the function

Answer 18

does this satisfy as the proof for an if and only if or do I need to show another case?

Answer 19

what I wrote above is really an iff...so yes

Answer 20

I find it amazing how you can just transform every problem into something so simple to interpret

Answer 21

thanks a ton for the help on this one as well

Answer 22

Not every problem ... I wish I could :)

Answer 23

it just blows my mind every time you do it lol
seems so complicated until you apply the zarkon magic

Answer 24

Here's an alternative. If z/(z^2 + 1) is equal to its conjugate then
\[ \frac{z^*}{1+z^{*2}} = \frac{z}{1+z^2} \]
i.e.,
\[ z^* + z^2z^* = z + z^{*2}z \]
i.e.,
\[ z^* - z = z^{*2}z - z^2z^* = ( z^* - z)zz^* \]

Now that is true if and only
\[ z^* - z = 0 \ \ \ or \ \ \ zz^* = 1 \]
i.e.,
\[ z = z^* \ \ \ or \ \ \ |z|^2 = 1 \]
i.e.,
either z is real or |z|^2 = 1

Answer 25

nice solution

Answer 26

Yes, it's a nice example of using the 'conjugate property' of real numbers.

Answer 27

why does z* - z need to be 0?

Answer 28

oh wait I see it

Answer 29

z* - z = (z* - z)(zz*)

hence zz* = 1, or if not, it must be that z* - z = 0.

I.e., if a = ab, then b = 1 or a = 0

Answer 30

that is a really nice looking proof right there, thanks!

Answer 31

could you maybe also help me with this problem too?
http://openstudy.com/#/updates/4ed5b89ae4b0bcd98ca8bb86