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If $|z|=|w|=1$, and $1+zw \neq 0$, then $ {{z+w} \over {1+zw}} \in \Bbb R $

i found one link that had a similar problem. Prove if $|z| < 1$ and $ |w| < 1$, then $|1-zw^*| \neq 0$ and $| {{z-w} \over {1-zw^*}}| < 1$

Community
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vordep
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6 Answers6

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We have $$\frac{z-w}{1-zw}=\frac{(z-w)(1-\overline{zw})}{(1-zw)(1-\overline{zw})}$$ Note that $(1-zw)(1-\overline{zw})=1-(zw+\overline{zw})+|zw|^2$ is real, so we need only consider the numerator (as Berci hinted).

We calculate $(z-w)(1-\overline{zw})=z-w-|z|^2\overline{w}+|w|^2\overline{z}=(z+\overline{z})-(w+\overline{w})$, which is real.

vadim123
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HINT:

Let $z=\cos A+i\sin A, w=\cos B+i\sin B$

$\implies z\cdot w=\cos(A+B)+i\sin(A+B)$

Using $\cos C-\cos D=-2\sin\frac{C-D}2\sin\frac{C+D}2$

and $\sin C-\sin D=2\sin\frac{C-D}2\cos\frac{C+D}2$

$$\frac{z-w}{1-zw}=\frac{\cos A-\cos B+i(\sin A-\sin B)}{1-\{\cos(A+B)+i\sin(A+B)\}}$$

$$=\frac{-2\sin\frac{A-B}2\sin\frac{A+B}2+i2\sin\frac{A-B}2\cos\frac{A+B}2}{2\sin^2\frac{A+B}2-i2\sin\frac{A+B}2\cos\frac{A+B}2}$$

$$=\frac{2i\sin\frac{A-B}2\left(i\sin\frac{A+B}2+\cos\frac{A+B}2\right)}{-2i\sin\frac{A+B}2\left(i\sin\frac{A+B}2+\cos\frac{A+B}2\right)}$$

$$=\frac{\sin\frac{A-B}2}{-\sin\frac{A+B}2}$$

EDIT: The changed question can be addressed in the same way using

$\cos C+\cos D=2\cos\frac{C-D}2\cos\frac{C+D}2$

and $\sin C+\sin D=2\sin\frac{C+D}2\cos\frac{C-D}2$

$$\frac{z+w}{1+zw}=\frac{\cos A+\cos B+i(\sin A+\sin B)}{1+\{\cos(A+B)+i\sin(A+B)\}}$$

$$=\frac{2\cos\frac{A+B}2\cos\frac{A-B}2+i2\sin\frac{A+B}2\cos\frac{A-B}2}{2\cos^2\frac{A+B}2+i2\sin\frac{A+B}2\cos\frac{A+B}2}$$

$$=\frac{2\cos\frac{A-B}2\left(\cos\frac{A+B}2+i\sin\frac{A+B}2\right)}{2\cos\frac{A+B}2\left(\cos\frac{A+B}2+i\sin\frac{A+B}2\right)}$$

$$=\frac{\cos\frac{A-B}2}{\cos\frac{A+B}2}$$

lab bhattacharjee
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Hint: For complex numbers $a,b$ we have $a/b\in\Bbb R\iff a\bar b\in\Bbb R$.

Berci
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We have $|\zeta|=1 \Leftrightarrow \bar{\zeta}=\frac{1}{\zeta}$.

$|z|=|w|=1 \Leftrightarrow \bar{z}=\frac{1}{z} \text{and } \bar{w}=\frac{1}{w} \Leftrightarrow \frac{\bar{z}+\bar{w}}{1+\bar{z}\bar{w}}=\frac{\frac{1}{z}+\frac{1}{w}}{1+\frac{1}{z} \frac{1}{w}} \Leftrightarrow \overline{ \left(\frac{z+w}{1+zw}\right)}=\frac{z+w}{1+zw} \Leftrightarrow \frac{z+w}{1+zw} \in \mathbb R$

user1337
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Draw vectors on the complex plane from the origin to a point on the unit circle. Notice that the argument of the sum of two such vectors is the average of the arguments of the two vectors (observe that the latter is defined only modulo a multiple of $\pi$). This follows from the properties of a rhombus (=parallelogram with equal sides).

Let $\arg z=\alpha$ and $\arg w=\beta$. By the above observation $\arg(z+w)=(\alpha+\beta)/2\pmod\pi$. It is well known that $\arg zw=\alpha+\beta$. As $\arg1=0$ another application of the above observation gives that $\arg(1+zw)=(\alpha+\beta)/2\pmod \pi$.

So $z+w$ and $1+zw$ share the same argument modulo an integer multiple of $\pi$. This means that the argument of their ratio is a multiple of $\pi$, and hence the ratio itself is real.

Jyrki Lahtonen
  • 133,153
  • So what i learned from this is that $arg (z +w)= \frac {a+b} {2} $

    ,$arg (1 + zw)= \frac {0+a+b} {2} $ then,

    $ \frac {z+w}{1+zw}$ =$\frac{\frac {a+b}{2}}{\frac{0+a+b}{2}}$

    this is the ratio you were talking about

     – vordep Jun 23 '13 at 20:16
  • @vordep: In the last step: the argument of a ratio is the difference of the arguments (analogously to the argument of a product being the sum of the arguments). Also, it is important that the pairs, $z$ and $w$, as well as $1$ and $zw$ have the same absolute values. Otherwise the rule: "argument of a sum is the average of arguments" does not hold. The point here was the geometric viewpoint. – Jyrki Lahtonen Jun 24 '13 at 06:31
  • oh yeah, thats what i meant, thanks – vordep Jun 24 '13 at 13:17
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HINT:

As $z \bar z=|z|^2=1$ and similarly for $w,$

$$\frac{z+w}{1+zw}=\frac{\frac1{\bar z}+\frac1{\bar w}}{1+\frac1{\bar z}\frac1{\bar w}}=\frac{\bar w+\bar z}{1+\bar w\bar z}$$

$$\text{Using }\frac ab=\frac cd=\frac{a+c}{b+d},$$

$$\frac{z+w}{1+zw}=\frac{\bar w+\bar z}{1+\bar w\bar z}=\frac{z+w+\bar w+\bar z}{1+zw+1+\bar w\bar z}=\frac{(z+\bar z)+(w+\bar w)}{2+(zw+\overline{wz})}$$ as $\bar z\cdot \bar w=\overline{z\cdot w}$

Observe that each pair within parentheses is real

lab bhattacharjee
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