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This is my homework, which reads as follows: Let $z_1, z_2$ be complex numbers. Prove that when $z_1z_2 \neq -1$ and $|z_1| = |z_2| = 1$, then the imaginary part of $$ \frac{z_1 + z_2}{1 + z_1z_2} $$ is zero.

I've tried to approach this in several ways, but whichever one I try, at some point the expression gets too large and doesn't simplify (at least I cannot find a way to simplify it).

The way I tried to approach this problem is to:

  1. Set $z_1 = a+bi$, $z_2 = c+di$, then $z_1+z_2=(a+c)+(b+d)i$, and in polar form, $\sqrt{(a + c)^2 + (b + d)^2}\times (\cos\theta + i\sin\theta)$, $\tan \theta = \frac{b+d}{a+c}$.
  2. Find $\tan \phi$, where $\phi$ is the angle of $\frac{1}{1 + z_1z_2}$.
  3. Substitute into $\tan(\theta + \phi) = \frac{\tan(\theta)+\tan(\phi)}{1-\tan(\theta)\tan(\phi)} = 0$.

The problem is $\tan(\phi)$ is a huge expression, and I cannot get it to simplify, so I cannot also show that the last formula is actually true.

Since this is a homework, and we are expected to do this with pan and paper (not even a calculator...), I don't believe we are to do this many calculations by hand. Hence, there must be some "trick" to get this to simplify, or a completely different approach. Would you care to give me a hint?

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Hint: show that $$\frac{z_1+z_2}{1+z_1z_2} = \overline{\left(\frac{z_1+z_2}{1+z_1z_2}\right)}$$

Added: Don't write the complex numbers in algebraic or trigonometric form. Work with $z_1$ and $z_2$ and use properties of conjugation: the conjugate of a ratio is the ratio of conjugates, the conjugate of a sum is ..., the conjugate of a product is ... . Also useful is to see that $1 = |z_1|^2 = z_1 \bar{z_1}$.

[The condition $z_1z_2 \neq 1$ should be $z_1z_2\neq -1$.]

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  • $\begingroup$ When I try doing this, first by multiplying the numerator and denominator by denominator's conjugate, I get an expression with several dozens terms, and if I try to expand it further, I eventually might get cubes or even higher powers etc. But it never really simplifies to something manageable. There must be some "trick" I'm missing in simplification... $\endgroup$ – wvxvw Apr 30 '16 at 12:05
  • $\begingroup$ I've extended my initial answer. $\endgroup$ – Catalin Zara Apr 30 '16 at 12:10
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The trick you're looking for is to notice that since $|z| = 1$, we can simply represent numbers by $z = e^{i \theta}$, since they are just points on the unit circle!

Let:

\begin{gather*} z_1 = e^{ia} \\ z_2 = e^{ib} \end{gather*}

Let:

$$ y = \frac{z_1 + z_2}{1 + z_1z_2} = \frac{e^{ia} + e^{ib}}{1 + e^{ia}e^{ib}} = \frac{e^{ia} + e^{ib}}{1 + e^{i(a + b)}} $$

Multiplying by the complex conjugate of the denominator (so we have a real denominator that we can ignore):

$$ y = \frac{e^{ia} + e^{ib}}{1 + e^{i(a + b)}} \cdot \frac{1 + e^{-i(a + b)}}{1 + e^{-i(a + b)}} $$

We know that the denominator will be read, so all we care about is the numerator that we need to simplify

\begin{align*} y_{num} ={}& e^{ia}(1 + e^{-i(a + b)}) + e^{ib}(1 + e^{-i(a + b)}) ={} \\ {}={}& e^{ia} + e^{ia -ia - ib)}+e^{ib} + e^{ib - ia - ib} ={} \\ {}={}& e^{ia} + e^{-ib} + e^{ib} + e^{-ia} ={} \\ {}={}& (e^{ia} + e^{-ia}) + (e^{ib} + e^{-ib}) ={} \\ {}={}& 2 \cos(a) + 2 \cos(b) \end{align*}

Hence, the numerator is purely real. Since we already know that the denominator is real, we know that the question is real as well

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  • $\begingroup$ Thanks a lot for posting your solution. I'm going to accept Catalin Zara's anther though because he was also trying to be more educational. $\endgroup$ – wvxvw Apr 30 '16 at 12:19
  • $\begingroup$ Ah, I understand :) I thought that having the answer in full might help though! $\endgroup$ – Siddharth Bhat May 1 '16 at 11:14
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Since $|z_1|^2=z_1 \overline{z_1}=1$, we have $\frac{1}{z_1}=\overline{z_1}$.

Similarly, $\frac{1}{z_2}=\overline{z_2}$.

So:

\begin{align*} \frac{z_1+z_2}{1+z_1z_2}={}&\frac{1/z_1+1/z_2}{1+1/(z_1z_2)}=\frac{\overline{z_1}+\overline{z_2}}{1+\overline{z_1z_2}}={} \\ {}={}&\overline{\left(\frac{z_1+z_2}{1+z_1z_2}\right)}. \end{align*}

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  • $\begingroup$ This is a very nice way to prove it. Thank you! But I'll accept Catalin Zara's answer because it was trying to be more educational. $\endgroup$ – wvxvw Apr 30 '16 at 12:18
  • $\begingroup$ @wvxvw: Thank you! You did the right thing to tell us that it was a homework question and you only needed a hint - you didn't ask for a complete solution. I am also glad that you see the educational value in my approach to providing guidance in situations like that. $\endgroup$ – Catalin Zara May 1 '16 at 0:09
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Multiplying the numerator by the conjugate of the denominator,

$$(z_1+z_2)(1+\overline{z_1z_2})=z_1+z_2+|z_1|^2\overline{z_2}+\overline{z_1}|z_2|^2=z_1+z_2+\overline{z_2}+\overline{z_1}.$$

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