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$\text{If } |z_1| = |z_2|, \text{ show that } \frac{z_1 + z_2}{z_1-z_2} \text{is imaginary.} $

The first thing I tried to do was to multiply both top and bottom by the conjugate of the denominator...

$$ \frac{z_1 + z_2}{z_1-z_2} \left( \frac{z_1 + z_2}{z_1+z_2} \right) \\ = \frac{z_1^2 + 2z_1z_2 + z_2^2}{z_1^2-z_2^2} $$

Then I $\text{Let }z_1,z_2 = x_1+iy_1,x_2+iy_2$ and expanded.. but then the equation was too big to work with. What I wanted to do was to simplify as much as I can, such as I did with $\frac{1-z}{1+z}$ which just equaled $\frac{-i\sin\theta}{1+\cos\theta}$ (after being written in Mod-Arg form, of course). So, what should I do from here on? Thanks in advance.

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    $\begingroup$ Another approach would be to compute the real part and show that it vanishes. $\endgroup$
    – Semiclassical
    Sep 25, 2014 at 15:59
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    $\begingroup$ If $z_1=z_2=1$, the fraction isn't defined, much less imaginary. $\endgroup$
    – vadim123
    Sep 25, 2014 at 15:59
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    $\begingroup$ Multiplying top and bottom by the conjugate of the denominator sounds OK. The conjugate is $\overline{z_1}-\overline{z_2}$, and not what you wrote. $\endgroup$
    – André Nicolas
    Sep 25, 2014 at 16:00
  • $\begingroup$ @vadim123: Presumably the value of $\frac{z_1+_2}{z_1-z_2}$ goes to infinity along the imaginary axis in that case. $\endgroup$
    – Semiclassical
    Sep 25, 2014 at 16:00
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    $\begingroup$ You are welcome. It is not the most efficient way to solve the problem, but it works. $\endgroup$
    – André Nicolas
    Sep 25, 2014 at 16:04

4 Answers 4

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How can you use the condition $| z_1 | = | z_2 |$? This invites to use the representation of complex numbers in polar coordinates.

So if one writes $z_1 = r e^{i \theta_1}$ and $z_2 = r e^{i \theta_2}$, the question amounts to show that $\frac{e^{i \theta_1} + e^{i \theta_2}}{e^{i \theta_1} - e^{i \theta_2}}$ is imaginary. Diviving the numerator and denominator by $e^{i(\theta_1 - \theta_2)/2}$ leads to the result.

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Multiply instead by $\dfrac{\bar{z_1} + \bar{z_2}}{\bar{z_1} + \bar{z_2}}$ to get $\displaystyle \frac{|z_1|^2 + z_1\bar{z_2} + \bar{z_1}{z_2} + |z_2|^2}{|z_1|^2 + z_1\bar{z_2} - \bar{z_1}{z_2} - |z_2|^2}$. The denominator is imaginary and the numerator is real.

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  • $\begingroup$ Thanks! Forgot about that conjugate notation, thanks again! $\endgroup$
    – Sam Chahine
    Sep 25, 2014 at 16:05
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$$\frac{z_1+z_2}{z_1-z_2}+\frac{\overline z_1+\overline z_2}{\overline{z_1-z_2}}$$

$$=\frac{2\overline z_1z_1-2\overline z_2z_2}{|z_1-z_2|^2}=0$$

as $z\overline z=|z|^2=|\overline z|^2 $

and as $x+iy+\overline{x+iy}=2x$

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  • $\begingroup$ Wait, why are you adding the fractions instead of multiplying them? @lab bhattacharjee $\endgroup$
    – Sam Chahine
    Sep 25, 2014 at 16:20
  • $\begingroup$ @SamirChahine, Have you followed the last line? Where have you found it mandatory to multiply fractions? $\endgroup$
    – lab bhattacharjee
    Sep 25, 2014 at 16:22
  • $\begingroup$ Isn't rationalising the denominator multiplying the whole fraction by its conjugate? Or can I add when dealing with complex numbers? @labbhattacharjee $\endgroup$
    – Sam Chahine
    Sep 25, 2014 at 16:26
  • $\begingroup$ @SamirChahine, It's another way of rationalizing the denominator. Please follow the last line of the answer $\endgroup$
    – lab bhattacharjee
    Sep 25, 2014 at 16:32
  • $\begingroup$ I just did, I am amazed, thank you aha, makes much more sense! $\endgroup$
    – Sam Chahine
    Sep 25, 2014 at 16:35
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Setting $z_1=a+ib,z_2=c+id$

$$\frac{z_1+z_2}{z_1-z_2}=\frac{a+c+i(b+d)}{a-c+i(b-d)}$$

$$=\frac{\{a+c+i(b+d)\}\{a-c-i(b-d)\}}{(a-c)^2+(b-d)^2}$$

Clearly, the real part of the numerator is $(a+c)(a-c)+(b+d)(b-d)=a^2+b^2-(c^2+d^2)=|z_1|^2-|z_2|^2$

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