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I have the following question.

Let $z_1$ and $z_2$ be complex numbers.


  1. $|z_1|=|z_2|=1$

  2. $ z_1z_2 \neq -1$

What I have to prove is that:


is real.

My thoughts: First, I multiplied the numerator and denominator by the conjugate of the denominator. Therefore, the denominator is real for sure, because of the proof I know that:

$$z\bar z = |z|^2 $$

Therefore it is real, and I can ignore the denominator.

The result is now:

$$(z_1 + z_2)(1 + \overline{z_1}\,\overline{z_2})$$

Another thought I had is to use trigonometric identities, but that do very well either.

Any help is appreciated.

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It might be more helpful to use conjugation rules to show that $\frac{z_1 + z_2}{1+z_1 z_2}$ is itself. – grantfgates May 9 '14 at 9:47
up vote 3 down vote accepted

Hint: continue the expansion $$ (z_1+z_2)(1+\overline{z_1}\,\overline{z_2})= z_1+z_2+z_1\overline{z_1}\,\overline{z_2}+\overline{z_1}\,z_2\overline{z_2} $$

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I got to the stage where I have: $$z1 + \bar z1 + z2 + \bar z2$$. Is it right that the sum of these numbers is real becuase each complex number with its conjuction is real? – Alan May 9 '14 at 10:02
@Alan $z+\overline z = 2\operatorname{Re}(z)$, can you show and then use this? – Christoph May 9 '14 at 10:07
Yes, I can. Thank you all, I think I have proved it. @Christoph – Alan May 9 '14 at 10:08
@Alan Yes, $z+\bar{z}$ is equal to its conjugate, so it's real. – egreg May 9 '14 at 10:08
@egreg thank you. – Alan May 9 '14 at 10:09

Since $z_1$ and $z_2$ lie on the unit circle they can be written as

$$ z_1=e^{i\alpha}, z_2=e^{i\beta} $$

where $\alpha$ and $\beta$ are real. So

$$ \frac{z_1+z_2}{1+z_1 z_2}=\frac{(z_1+z_2)(1+\bar z_1\bar z_2)}{|1+z_1 z_2|^2}=\frac{(e^{i\alpha}+e^{i\beta})(1+e^{-i(\alpha+\beta)})}{|1+z_1 z_2|^2}= $$

$$ \frac{e^{i\alpha}+e^{i\beta}+e^{-i\alpha}+e^{-i\beta}}{|1+z_1 z_2|^2}=\frac{2(\cos\alpha+\cos\beta)}{|1+z_1 z_2|^2} \in \mathbb R $$

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Best answer in my opinion. – recursive recursion May 9 '14 at 20:23

Hint: $$\text{Im}(z) = \frac{1}{2i}(z - \bar{z})$$

If you can prove that the imaginary part of a number is $0$, then it is real.

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Let be $\displaystyle w=\frac{z_{1}+z_{2}}{1+z_{1}z_{2}}$ and $w=a+ib$

We will prove that $w-\bar{w}=0$.

$$w-\bar{w}=\frac{z_{1}+z_{2}}{1+z_{1}z_{2}}-\frac{\bar z_{1}+\bar z_{2}}{1+\bar z_{1}\bar z_{2}}=\frac{z_{1}+\bar z_{2}+z_{2}+\bar z_{1}-\bar z_{1}-\bar z_{2}-z_{1}- z_{2}}{\left|1+z_{1}z_{2}\right|^{2}}=0$$

$$a+ib-a+ib=0 \Leftrightarrow b=0.$$ So $w$ real.

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