Question

"Consider w = z/(z^2+1), where z=x+iy, y \neq 0 and {z^2+1} \neq 0. Given that Im (w) = 0, show that | z | = 1."

I would appreciate it if someone pointed me towards the right direction and showed one or two of the first steps, but preferably not the entire answer.

Answer 1

Written out again because it looks all messed up:

\[\frac{ z }{ z^2+1 } , z=x+iy , y \neq 0 , z^2+1 \neq 0 \]

Answer 2

@terenzreignz

Answer 3

I usually decompose the entire thing... how about expressing w as
\[\Large \frac{x+\color{blue}iy}{(x+\color{blue}iy)^2 + 1}\] and separate the real and imaginary parts?

Answer 4

so you mean expanding the brackets?

Answer 5

\[\frac{ x+iy }{ x^2+2xyi-y^2+1 }\]

Answer 6

Yeah, that's it, and then, well, the real part and the imaginary parts aren't quite separated yet...
\[\large \frac{x+\color{blue}iy}{x^2 -y^2 +1 + 2xy\color{blue}i}\]
Let's try multiplying the numerator and denominator by the conjugate of the denominator?

Answer 7

Mind you, I don't really know what I'm doing, I'm just hoping that something might come out :3

Answer 8

hahah alright then :D

Answer 9

so the conjugate would be \[x^2-y^2+1-2xyi\]
?

Answer 10

That's right...

Answer 11

So it gives: \[\frac{(x+iy)(x^2-y^2+1-2xyi) }{(x^2-y^2+1+2xyi) (x^2-y^2+1-2xyi) }\]
I'm not sure how to proceed as the powers will get pretty big if we multiply it out

Answer 12

Don't mind the denominator, that's sure to be some real number anyway :)
Focus on the numerator, multiply the polynomials, and separate the real and the imaginary part...

Answer 13

so:\[x^3-y^2x+x-x^2yi-y^3i+yi+2xy^2\]

Answer 14

But to know Im (w) don't we need the denominator also?

Answer 15

Ah, I see now... the magic...

Answer 16

Here, we have this...
\[\Large \frac{x^3-y^2x+x+2xy^2-x^2yi-y^3i+yi }{(x^2-y^2+1+2xyi) (x^2-y^2+1-2xyi) }\]

Answer 17

I think the 2xy^2 should be 2xy^2i

Answer 18

No it shouldn't, because it was the result of you multiplying \(\large y\color{blue}i\) by \(\large -2xy\color{blue}i\)

Answer 19

So, ready to proceed? ^_^

Answer 20

Aaa I see

Answer 21

Shall we?

Answer 22

Yep :)

Answer 23

So, the denominator is simply \(z^2 + 1\) multiplied by its conjugate, yes?
so, it becomes just \(\large |z^2 +1|^2\)
a real number, which isn't zero since \(|z^2+1|\ne0\)

Answer 24

You following me so far?

Answer 25

yes that seems to be correct

Answer 26

you make it pretty clear so it's easy to follow :D

Answer 27

So, the entire thingy is just\[\Large \frac{x^3-y^2x+x+2xy^2-x^2yi-y^3i+yi }{|z^2+1|^2 }\]
I actually just did that to lessen the scary-factor of the denominator :3

Answer 28

there is a much simpler way of solving this

Answer 29

I will defer :)
Complex analysis is not my forte...
but maybe you guys could just let me finish this?
Pretty please? ^_^

Answer 30

I am taking account of the fact that:\[\log{w}=0\implies\log{\frac{z}{z^2+1}}=0\implies z=z^2+1\]

Answer 31

Could you explain that part?

Answer 32

log?
How does log factor into all of this? D:

Answer 33

if \(\log(x)=0\implies x=e^0=1\)

Answer 34

aren't we given \(\log(w)=0\)

Answer 35

No... we're given Im(w) = 0
D:

Answer 36

I don't think we're given that, although I could be mistaken

Answer 37

I thought 'lm' was 'ln' mistyped :)

Answer 38

:D Would you happen to have other ideas or should terenzreignz continue from where we left off?

Answer 39

strangely enough if it was 'ln' then it leads to the same answer! :)

Answer 40

Yes it does, I just realized that :D Maybe there's something to it....

Answer 41

It's Terence (or TJ) @Teinis
>.>
Back to reality, I suppose :)

We may now neatly separate the real part from the imaginary part, like this...
\[\Large \frac{\color{red}{x^3-y^2x+x+2xy^2}+\color{blue}{(-x^2y-y^3+y)}\color{green}i }{|z^2+1|^2 }\]

Answer 42

So, in effect,
\[\Large Im(w) = \frac{-x^2y-y^3+y}{|z^2+1|^2}=0\]

Answer 43

Catch me so far?

Answer 44

Yep sounds good

Answer 45

@terenzreignz has it spot on! sorry for my mis understanding of your question. :)

Answer 46

Well, we have established that \(\large |z^2 + 1|^2\) is a nonzero real number, so it may be multiplied to both sides of the equation to give... (and cancel out)
\[\Large -x^2y-y^3+y = 0\]
aye?

Answer 47

aaaa Im starting to see where you're going with this

Answer 48

Yep that seems good

Answer 49

You want to continue it?

Answer 50

I could give it a try :)

Answer 51

Hint: Factor ;)

Answer 52

We can factorize it as \[y(-x^2-y^2+1) = 0\]

Answer 53

mhmm <nods>

Answer 54

and since we know:
\[y \neq 0, that means -x^2-y^2+1 = 0\]

Answer 55

yes...

Answer 56

Stuck?

Answer 57

Actually 'I got stuck again, not quite sure how to continue now...

Answer 58

NO WAIT!

Answer 59

waiting...

Answer 60

Omg I just had a light bulb moment

Answer 61

Do elaborate ;)

Answer 62

\[-x^2-y^2+1 = x^2+y^2-1\]

Answer 63

huh?
That makes no sense...

Answer 64

I meant if you multiply both sides by -1 you can get it into that form

Answer 65

^THAT makes sense

Answer 66

\[x^2+y^2 = 1\]

Answer 67

<drum roll>

Answer 68

and since \[\left| z \right| = \sqrt{x^2+y^2}\]

Answer 69

<drum roll gets faster>

Answer 70

\[\sqrt{x^2+y^2} = 1\]

Answer 71

so \[\left| z \right| = 1\]

Answer 72

!!!

Answer 73

<applause>

Answer 74

Heh... not bad for a kid that didn't know what he was doing just about a half-hour ago, huh? ^_^

Answer 75

I would like to thank my mother and father.....

Answer 76

and tj ;)

Answer 77

hehe :D

Answer 78

yeah I guess, now that I think about it it doesn't seem that hard, you just need someone to walk you through some of the process

Answer 79

Thanks a lot for helping me with that :)

Answer 80

Yes... and I didn't really tell you the answer, now did I? ;)

Answer 81

hahah nope ;)

Answer 82

You actually did most of the dirty work yourself... I detest algebra XD

Answer 83

yeah those letters can be quite messy

Answer 84

Well, I guess it's time to wrap this question up, thanks again!

Answer 85

Signing off now.
---------------------------
Terence out