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

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

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.

• Another approach would be to compute the real part and show that it vanishes. – Semiclassical Sep 25 '14 at 15:59
• If $z_1=z_2=1$, the fraction isn't defined, much less imaginary. – vadim123 Sep 25 '14 at 15:59
• 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. – André Nicolas Sep 25 '14 at 16:00
• @vadim123: Presumably the value of $\frac{z_1+_2}{z_1-z_2}$ goes to infinity along the imaginary axis in that case. – Semiclassical Sep 25 '14 at 16:00
• You are welcome. It is not the most efficient way to solve the problem, but it works. – André Nicolas Sep 25 '14 at 16:04

$$\frac{z_1+z_2}{z_1-z_2}+\frac{\overline z_1+\overline z_2}{\overline{z_1-z_2}}$$

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

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

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

• Wait, why are you adding the fractions instead of multiplying them? @lab bhattacharjee – S.E. Chahine Sep 25 '14 at 16:20
• @SamirChahine, Have you followed the last line? Where have you found it mandatory to multiply fractions? – lab bhattacharjee Sep 25 '14 at 16:22
• Isn't rationalising the denominator multiplying the whole fraction by its conjugate? Or can I add when dealing with complex numbers? @labbhattacharjee – S.E. Chahine Sep 25 '14 at 16:26
• @SamirChahine, It's another way of rationalizing the denominator. Please follow the last line of the answer – lab bhattacharjee Sep 25 '14 at 16:32
• I just did, I am amazed, thank you aha, makes much more sense! – S.E. Chahine Sep 25 '14 at 16:35

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.

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.

• Thanks! Forgot about that conjugate notation, thanks again! – S.E. Chahine Sep 25 '14 at 16:05

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 $(a+c)(a-c)+(b+d)(b-d)=a^2+b^2-(c^2+d^2)=|z_1|^2-|z_2|^2$