Prove that the diagonals of a rhombus are orthogonal.

Question

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!

Answer 1

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.

Answer 2

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. $${z+w\over z-w}= -\overline{\Big({z+w\over z-w}\Big)}$$

Answer 3

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.