If $|z|=1$, show that: $$\mathrm{Re}\left(\frac{1 - z}{1 + z}\right) = 0$$

I reasoned that for $z = x + iy$, $\sqrt{x^2 + y^2} = 1\implies x^2 + y ^2 = 1$ and figured the real part would be:

$$\frac{1 - x}{1 + x}$$

I tried a number of manipulations of the equation but couldn't seem to arrive at any point where I could link the two to show that the real part was = 0.

  • 1
    $\begingroup$ write $z=re^{i\theta}$ and observe that $r=1$ and $e^{i\theta} = cos \theta + i sin \theta$. Now manipulate the given condition. $\endgroup$ – Dilawar Mar 11 '12 at 12:39
  • $\begingroup$ You can not calculate the real part like that (see anon's answer/comment below). $\endgroup$ – AD. Mar 15 '12 at 7:58

The easy method to do this kind of exercise is to notice that $Re(z)=0$ if and only if $z=-\overline z$. Notice that if $|z|=1$ then $\overline z=\frac{1}{z}$.

In your case $$ \overline{\left(\frac{1-z}{1+z}\right)}=\frac{1-\overline z}{1+\overline z}=\frac{1-\frac{1}{z}}{1+\frac{1}{z}}=\frac{z-1}{1+z}=-\frac{1-z}{1+z}$$

Usually, the replacement of the algebraic form $z=x+iy$ makes things much more complicated in problems like this. First you should try applying the standard results from the general case, such as $|z|^2=z\overline z$, or $z \in \Bbb{R}$ if and only if $z=\overline z$, etc.


Let $z=a+bi$, where $a^2+b^2=1$ and $z\neq -1$. Then $$\begin{eqnarray} \Re\left(\frac{1-z}{1+z}\right) &=&\Re\left(\frac{(1-z)\overline{(1+z)}}{|1+z|^2}\right)\\ &=&\Re\left(\frac{(1-a-bi)(1+a-bi)}{(a+1)^2+b^2}\right)\\ &=&\Re\left(\frac{1-(a^2+b^2)-2bi}{(a+1)^2+b^2}\right)\\ &=&\Re\left(\frac{-2bi}{(a+1)^2+b^2}\right)=0\end{eqnarray}$$


Using polar form of a complex number, $z=e^{i\theta},\quad\theta\in\mathbb{R}$.

Hence $\frac{1-z}{1+z}=\dfrac{1-e^{i\theta}}{1+e^{i\theta}}=\dfrac{e^{-\frac{i\theta}{2}}-e^{\frac{i\theta}{2}}}{e^{-\frac{i\theta}{2}}+e^{\frac{i\theta}{2}}}=\dfrac{-2i\sin\frac{\theta}{2}}{2\cos\frac{\theta}{2}}=-i\tan\frac{\theta}{2}$

  • $\begingroup$ How did you get from term 2 to 3 and from term 3 to 4? $\endgroup$ – stariz77 Mar 18 '12 at 22:36
  • $\begingroup$ @stariz77 Multiplying numerator and denominator by $e^{-\frac{i\theta}{2}}$ and then using the formula $e^{ix}=\cos x+i\sin x$. $\endgroup$ – Tapu Mar 19 '12 at 8:02

Since we are adding multiple proofs, here are two more.

1) Geometric proof.

enter image description here

$\displaystyle z$ corresponds to point $\displaystyle B$. $\displaystyle 1+z$ corresponds to $\displaystyle H$ and $\displaystyle 1-z$ to $\displaystyle G$.

$\displaystyle ACGD$ and $\displaystyle ADHB$ are rhombii, with $CD$ parallel to $\displaystyle AH$. In rhombii the diagonals intersect at right angles, and so $\displaystyle AG$ is perpendicular to $\displaystyle AH$.

Thus $\displaystyle \frac{1-z}{1+z} = ci$ for some $\displaystyle c$.

2) Using vectors.

We refer to the above figure.

$\displaystyle H = (1+\cos \theta, \sin \theta)$.

$\displaystyle G = (1- \cos \theta, -\sin \theta)$.

Their dot product = $\displaystyle (1 + \cos \theta)(1- \cos \theta) - \sin^2 \theta = 1 - \cos^2 \theta - \sin^2 \theta = 0$

So $\displaystyle \vec{AH}$ and $\displaystyle \vec{AG}$ are perpendicular.

Incidentally, the converse is also true:

If $\displaystyle \text{Re}\left(\frac{1-z}{1+z}\right) = 0$, then $\displaystyle |z| = 1$.


Hint $\: $ Realizing the denominator produces a purely imaginary numerator, i.e.

$$\displaystyle\ z\bar z\: =\: 1\ \:\Rightarrow\:\ \frac{(1-z)\:(1+\bar z)}{(1+z)\:(1+\bar z)}\: =\: \frac{\bar z-z}{|1+z|^2}\: =\: \frac{r\:i}{s},\ \ r,s\in \mathbb R$$

Note $\ $ This is an instance of the powerful method of rationalizing denominators to simplify arithmetic of algebraic numbers or functions. Generally, $\rm F$-rationalize means coerce to an element of the base ring $\rm F$ of an algebraic extension, which classically is $\mathbb Q = $ rationals. Above we have the base ring $\rm F =\mathbb R\subset \mathbb R[{\it i}] = \mathbb C,\:$ so rationalize means realize.


I want to point out that the problem in your approach is that, generally,

$$\mathrm{Re}\left(\frac{1-z}{1+z}\right) \ne \frac{1-\mathrm{Re}(z)}{1+\mathrm{Re}(z)}.$$

In other words, just because there's a Re on the left of the equation doesn't mean you can replace a complex variable $z$ with its real part $x$; that's tantamount to switching the order of evaluation of two functions composed (as in $f\circ g$ vs. $g\circ f$, here with $f=\mathrm{Re}$ and $g(z)=\frac{1-z}{1+z}$).

The general approach you want to take is replacing $z$ with $x+yi$, then trying to get the whole thing inside the real part in the form $\text{blah}_1+\text{blah}_2i$, where both $\text{blah}$s are entirely real, so you can just chop off $\text{blah}_2$ to finish evaluating. Frequently this process involves slapping the conjugate $x-yi$ in there somewhere to make things real.

  • $\begingroup$ Not feeling foo and bar today? $\endgroup$ – Alex Becker Mar 11 '12 at 15:05

Yet another approach, is to note that $Re(\overline z_1z_2)=z_1 \cdot z_2$, where $\cdot$ denotes the euclidean dot product in $R^2$.

Let $z=x+iy$ for $x,y$ in $R$, then we have $Re(\frac{1-z}{1+z})=\overline{1-z}{\frac{1}{1+z}}=(1-x,y)\cdot (\frac{1+x}{1+2x+x^2+y^2},\frac{-y}{1+2x+x^2+y^2})=\frac{1-(x^2+y^2)}{2+2x}=0$

  • $\begingroup$ Re$(\bar i\bar i)=-1$ but $i\cdot i=1$ if I read you correctly. And $\overline{\frac1{1+z}}$ is proportional to $(1+x,y)$, not $(1+x,-y)$. $\endgroup$ – Did Mar 15 '12 at 9:04
  • $\begingroup$ oh, you are right. I think it is fixed now. Thanks! $\endgroup$ – Henrik Mar 15 '12 at 9:33
  1. The map $$T:z\mapsto \frac{1-z}{1+z}$$ is a Möbius transform.
  2. It is well known that Möbius transforms maps circles into straight lines or circles.
  3. By 1 and 2 it is sufficient to check the image of three points in order to decide the image of $\{z:\,|z|=1\}$.

Now, $$T(1)= 0$$ $$T(-1)= \infty$$ $$T(i)=\frac{1-i}{1+i}= \frac{(1-i)\overline{(1+i)}}{|1+i|^2}=\frac{(1-i)^2}{2} = \frac{-2i}{2}=-i$$ That is the image of the unit circle is the imaginary axis and in particular $$\Re\frac{1-z}{1+z}=0, \qquad \text{when $|z|=1$}$$


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