# Prove that the diagonals of a rhombus are orthogonal.

I'm trying to solve some of the problems in Ahlfors' Complex Analysis book. On the section about analytic geometry, the following problem is stated:

Prove that the diagonals of a rhombus are orthogonal.

Since the idea was to use complex analysis tools to solve it, I came up with the following.

I take any arbitrary rhombus and draw it on the complex plane. After this, I re-orient it such that one of the corners is lying on the intersection of the real and imaginary axis and one of the diagonals is on the imaginary axis. Relating one of the sides of the rhombus with the imaginary number $$z$$, I obtain the following scenario:

Recalling that the reflection about the imaginary axis is given by $$z \to -\overline{z}$$.

Using this construction, if indeed the diagonals are orthogonal this would mean that $$\frac{z -\overline{z}}{z +\overline{z}}$$ is purely imaginary (since multiplying by $$i$$ results in a $$90^\circ$$ rotation in the complex plane). To show this, I do $$\overline{\left(\frac{z -\overline{z}}{z +\overline{z}}\right)} = \frac{\overline{z} -\overline{\overline{z}}}{\overline{z} +\overline{\overline{z}}} = \frac{\overline{z} -z}{\overline{z} + z} = - \left(\frac{z -\overline{z}}{z +\overline{z}}\right)$$ And since \begin{align} z = a+ib \text{ is purely imaginary } \iff a=0 \iff a =-a \iff a -ib = -a-ib \iff \overline{z} = -z \end{align} this concludes the solution.

Is my solution correct? Thank you very much!

• Hmm, since everything is hapening in upper plane there is no conjugation. Also don't understand how you got those sides. Commented Jul 25, 2020 at 18:49
• What is this quantity $\frac{z -\overline{z}}{z +\overline{z}}$? Commented Jul 25, 2020 at 18:50
• @Aqua, I got the sides by using that the reflection about the imaginary axis of a complex number $z$ is given by the negative of the conjugate. I understand this is because if $z = a+ib$, then $-a+ib = -\overline{z}$ gives the same number but with opposite sign real part. Commented Jul 25, 2020 at 18:52
• And by $\frac{z -\overline{z}}{z +\overline{z}}$ I denote the complex number such that when I multiply it with $z + \overline{z}$ it results in $z -\overline{z}$. Commented Jul 25, 2020 at 18:53
• I don't think this proof is OK since you already actually use the fact that diagonals are perpendicular by using reflections. Commented Jul 25, 2020 at 19:03

## 3 Answers

This is essentially right. There's one problem with

if indeed the diagonals are orthogonal this would mean that ... is purely imaginary

Here what you want is the converse:

if ... is purely imaginary then the diagonals are orthogonal

You might try an alternative proof using $$z$$ and $$z+e^{i\theta}z$$ for the first two edges of the rhombus.

• Oh, you're right. I'm assuming $\iff$ when I only know $\implies$. But can I prove that this statement is an "if and only if", or does this not go both ways? Commented Jul 25, 2020 at 18:56
• It goes both ways, so long as the ratio isn't $0$, that is, so long as $z$ isn't real, which you can clearly assume. Commented Jul 25, 2020 at 18:59
• Is this really correct? @saulspatz Commented Jul 25, 2020 at 19:10
• @Aqua Well, I think so, certainly. What do you think is wrong? Commented Jul 25, 2020 at 19:14
• He use reflection for sides which implictly means he used orthogonalty. @saulspatz Commented Jul 25, 2020 at 19:15

You are to much relaying on a picture. You assume that $$D$$ and $$B$$ are symmetric to imaginary axsis (since you wrote $$w= -\overline{z}$$) which authomaticly means $$AC\bot BD$$ (since $$C$$ lies on imaginary axsis) which is to be prove. So your proof is not correct.

Correct way is to say $$AD =w$$ and we know $$|w|=|z|$$. Then we need to see that $$\displaystyle{z+w\over z-w}$$ is imaginary number, i.e. $${z+w\over z-w}= -\overline{\Big({z+w\over z-w}\Big)}$$

• I think this last part can then be shown by doing $$-\overline{\left(\frac{z+w}{z-w}\right)}=\frac{\overline{w}+\overline{z}}{\overline{w}-\overline{z}}=\frac{|w|^2+\overline{z}w}{|w|^2-\overline{z}w}= \frac{|z|^2+\overline{z}w}{|z|^2-\overline{z}w}=\frac{\overline{z}(z+w)}{\overline{z}(z-w)}=\frac{z+w}{z-w}$$ right? Assuming $z,w \neq0$ to begin with. Commented Jul 26, 2020 at 16:16
• Also, what I wanted to know is if somehow I could "salvage" my solution using something similar to your method here. Using your diagram as a reference, if I just denote the sides of the rhombus as $z$ and $w$ (like you do), I can choose to put the diagonal on the imaginary axis (without assuming anything else about the position of $w$ respect ot $z$ this time). As you pointed out in the comments earlier, If I know that the diagonal bisects the angle $\arg\left(\frac{w}{z}\right)$, then I can justify that $-\overline{z} = w$ and continue the solution as I did. Correct? Commented Jul 26, 2020 at 16:25
• But my question is, if I know that every rhombus is additionally a parallelogram, is this sufficient to show that in the construction detailed above $-\overline{z}=w$? I believe this is enough (like the case where I previously know the bisection to be true) but I'm not sure. Could you tell me if I'm correct on this line of reasoning? Commented Jul 26, 2020 at 16:27
• No, it make no difference. You would still use reflection. And actually you would not use that it is paralelogram at all. Commented Jul 26, 2020 at 16:48
• Ok, but at least using that the diagonal bisects does work to justify that $-\bar{z}=w$. So if I can show that beforehand, then the original solution should be "saved", right? Commented Jul 26, 2020 at 16:56

Meanwhile preparing solution with "pure" complex properties, I come to one decision, which worthy, imho, to be here.

1) Let's take $$z_1,z_2,z_3,z_4$$ complex numbers in opposite clockwise direction, which gives us rhombus. We can construct rhombus sides by $$a=z_3-z_2$$ and $$b=z_2-z_1$$ and diagonals by $$d_2=b+a$$ and $$d_1=a-b$$. If we consider scalar product for diagonals, we have: $$d_1 \cdot d_2= (a-b) \cdot (a+b)= (a_1-b_1)(a_1+b_1)+(a_2-b_2)(a_2+b_2)=\\= a_1^2-b_1^2 + a_2^2-b_2^2 = |a|^2-|b|^2=0$$ In one hand it's zero because we are in rhombus, but on another hand this give diagonals perpendicularity.

Second solution coming. Hope.

Addition.

2) As promised lets look on slight different solution. Again we take complex numbers $$z_1,z_2,z_3,z_4$$ and construct sides and $$a=z_3-z_2$$ and $$b=z_2-z_1$$ and diagonals by $$d_2=b+a$$ and $$d_1=a-b$$. For 2 complex numbers we know, that division argument is difference between numerator and denominator arguments i.e. $$\Im \frac{d_1}{d_2}=\Im{d_1}-\Im{d_2}$$. So let's calculate now complex numbers division: $$\frac{a+b}{a-b}=\frac{1}{|a-b|^2}\left(a+b \right)\left(\overline{a-b} \right)=\frac{1}{|a-b|^2}\left( |a|^2-|b|^2+\overline{a}b-\overline{b}a \right)=\\=\frac{1}{|a-b|^2}\left( |a|^2-|b|^2+\overline{a}b-\overline{\overline{a}b} \right)=\frac{1}{|a-b|^2}\left( |a|^2-|b|^2+i2\Im(a\overline{b}) \right)$$

As we are in rhombus, then we have $$|a|^2-|b|^2=0$$ and we get, that division of rhombus diagonals is pure imaginary complex number, so angle between diagonals is right angle.