if $z= x+iy$ where $y \neq 0$ and $1+z^2 \neq 0$, show that the number $w= z/(1+z^2)$ is real only if $|z|=1$

solution :

$$1+z^2 = 1+ x^2 - y^2 +2xyi$$

$$(1+ x^2 - y^2 +2xyi)(1+ x^2 - y^2 -2xyi)=(1+ x^2 - y^2)^2 - (2xyi)^2$$

real component
$$(1+ x^2 - y^2)x - yi(2xyi) = x + x^3 + xy^2$$

imaginary component

$$-2yx^2 i +yi + x^2 yi - y^3 i=0i$$

$$-2yx^2 +y + x^2 y - y^3 =0$$


can't solve this question


$$\dfrac Z{1+Z^2}=\dfrac{x+iy}{1+x^2-y^2+2xyi}$$


We need $y(1+x^2-y^2)-x(2xy)=0\implies y(1+x^2-y^2-2x^2)=0\iff x^2+y^2=1$

  • $\begingroup$ I see =, the last part is tricky! $\endgroup$ – lolisme Jun 8 '15 at 8:33
  • $\begingroup$ You are not solving the original question ! $\endgroup$ – Yves Daoust Jun 8 '15 at 9:06
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    $\begingroup$ @YvesDaoust, Not sure why it has been accepted ! The question has been actually changed many a times $\endgroup$ – lab bhattacharjee Jun 8 '15 at 9:18
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    $\begingroup$ Mh, I guess that the division has disappeared by magic. $\endgroup$ – Yves Daoust Jun 8 '15 at 9:42
  • $\begingroup$ @ThomasAndrews: why do you tell me that ?? $\endgroup$ – Yves Daoust Jun 8 '15 at 13:08

A another approach is to utilise (as @mann and myself concluded) the fact that $$ w = \bar{w} $$ where bar represents the complex conjugate. $$ \frac{z}{1+z^2} = \frac{\bar{z}}{1+\bar{z}^2} $$ re-arrange to yeild $$ z + z\bar{z}^2 = \bar{z} + \bar{z}z^2 \implies z-\bar{z} = \bar{z}z^2-z\bar{z}^2=z\bar{z}\left(z-\bar{z}\right) $$ thus we get $$ 1=z\bar{z} = |z|^2\implies |z| = 1 $$

  • $\begingroup$ wow that's another way but don't really think it is a good idea solving this way, Good job anyway :) $\endgroup$ – lolisme Jun 8 '15 at 13:55
  • $\begingroup$ @lolisme not to worry, and its not the best method. It was just an idea that I thought about. So now you have a bunch of methods to attack these sorts of problems at least :). $\endgroup$ – Chinny84 Jun 8 '15 at 13:59
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    $\begingroup$ Not the best method or not a good idea?!? If you read that the answer is real, your first attempt should be to set $w = \bar w$ and solve. This is the most succinct way to do this problem. $\endgroup$ – user121330 Jun 8 '15 at 17:30
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    $\begingroup$ this is much better and more elegant method. never mind the op can't appreciate it yet.(+1). $\endgroup$ – abel Jun 9 '15 at 0:24
  • $\begingroup$ @user121330 and Abel. Cheers!! Lucky I do this for my own benefit ha. But as long as the OP has an answer that he is satisfied by then its all good :)!! $\endgroup$ – Chinny84 Jun 9 '15 at 6:56

If I started with your solution $$-2yx^2 +y + x^2 y - y^3 =0\implies y(x^2+y^2-1)=0$$

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    $\begingroup$ @Travis I think this is meant as a hint. The implication does give a significant hint. $\endgroup$ – DRF Jun 8 '15 at 10:04
  • $\begingroup$ i figured it out, thanks anyway $\endgroup$ – lolisme Jun 8 '15 at 13:56
  • $\begingroup$ If I were to start with @lolisme's work, this is the exact way I'd do this problem. $\endgroup$ – user121330 Jun 8 '15 at 17:31

It's easier to work with $w=\dfrac{1+z^2}z=\dfrac1z+z$ (that doesn't change the claim).

Taking the imaginary part,

$$-\frac y{x^2+y^2}+y=y\left(1-\frac1{x^2+y^2}\right)=0,$$ i.e. $$x^2+y^2=1.$$

Or, in polar coordinates,

$$-\frac1r\sin(\phi)+r\sin(\phi)=0\implies r=1.$$

  • $\begingroup$ Exactly the condition i had by! $w=\bar{w}$ , precisely. $|z|^2+z^2 +(\bar{z})^2+1=0$ $\endgroup$ – Mann Jun 8 '15 at 9:30
  • $\begingroup$ @mann write that up as an answer as I was about to use that approach also ;). $\endgroup$ – Chinny84 Jun 8 '15 at 10:49
  • $\begingroup$ @Chinny84 , I have to go out in a few min, you'd write the answer :) $\endgroup$ – Mann Jun 8 '15 at 11:05


Let $z=r(\cos A+i\sin A)$ where $r\ge0,A$ are real

As $y\ne0,\sin A\ne0$



$r(\cos A+i\sin A)\cdot(1+r^2\cos2A-ir^2\sin2A)$ $=\cdots+ir[\sin A(1+r^2\cos2A)-r^2\cos A\sin2A]=\cdots+ir[\sin A-r^2\sin(2A-A)]$

$=\cdots+ir\sin A(1-r^2)$


I tried to solve this with geometry and how the arithmetic operations on complex numbers can be interpreted geometrically.

  1. By dividing two complex numbers, their arguments (angles, $\varphi$) are subtracted.
  2. If $w$ should be real, its argument has to be zero.

That means $\varphi(z) = \varphi(1+z^2)$.

Here's a little sketch of both those complex numbers drawn as arrows in the complex plane: enter image description here

You can see $z$, $z^2$, $1$ and $1+z^2$ in green. I also added the arguments of $z$ and $z^2$ in red, for no real reason (pun not intended) other than to remember that they have a constant ratio of $\varphi(z) = 2\varphi(z^2)$. So you can't do much about that, but $\varphi(1+z^2)$ is a whole 'nother story, because of the $+1$

The true goal is still to line up the green arrow (i.e. $1+z^2$) with $z$. In order to be lined up, one has to be derivable from the other by multiplying with a real number $a$. I guess this is called "being linear dependent", but am not sure. Anyway, the formula for that looks like that:

$$ \begin{align} 1+z^2 &= az\\ z^2 -az +1 &= 0\\ z_{1/2} &= \frac{a}{2} \pm \sqrt{\left(\frac{a}{2}\right) ^2-1}\\ \end{align} $$

This is kind of a good thing, because there's $\pm$, which means this is a two for one. And circles aren't functions because they need two lines. So maybe we are on to something here. Let's see if the above formula can be squeezed into being a circle.

You said $z= x+iy$ where $y \neq 0$ and that means there has to be an imaginary part, so whatever $a$ is (cannot be bothered to calculate it), the term under the square will be negative and by that forces $z_{1/2}$ to have an imaginary part.

$$ \begin{align} z_{1/2} &= \frac{a}{2} \pm \sqrt{(-1)\left(-\left((\frac{a}{2}\right) ^2+1\right)}\\ &= \frac{a}{2} \pm \sqrt{-1} \sqrt{-\left(\frac{a}{2}\right)^2+1}\\ &= \frac{a}{2} \pm i \sqrt{-\left(\frac{a}{2}\right)^2+1}\\ &= x+iy\\ \end{align} $$

Finally, what's $|z|$?

$$\begin{align} |z| = |z_{1/2}| &= \sqrt{x^2 + y^2}\\ &= \sqrt{\left(\frac{a}{2}\right)^2 + \left(\sqrt{-\left(\frac{a}{2}\right)^2+1}\right)^2}\\ &= \sqrt{\left(\frac{a}{2}\right)^2 -\left(\frac{a}{2}\right)^2+1}\\ &=1 \end{align}$$

I'm not sure if this is rigorous enough to be a proof, but you just asked to show it, so I hope this is ok and offers a bit of a different approach.


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