Let's Do Geometry The Complex Way!

Let $2a, 2b, 2c, 2d$ represent the complex numbers running along the edges of the square $ABCD$ .Since these complex numbers form a closed path therefore,

$\boxed{a+b+c+d = 0}$ .... $(1)$

Let $A$ (refer this point in the diagram of the question) be the origin. We know that if $z$ represents a complex number then $iz$ represents another complex number with the same magnitude as $z$ rotated by an angle $90^\circ$ in the anti-clockwise direction with respect to $z$ . Using this analysis we can represent the positions of the centres of the squares as follows:

• $G \equiv a+ia$

• $F \equiv 2a+b+ib$

• $H \equiv 2a+2b+c+ic$

• $E \equiv 2a+2b+2c+d+id$

Now, let the complex number from $G$ to $H$ be $z_1$ . Therefore $z_1 = a+2b+c+i(c-a)$ and $\left\vert z_1 \right\vert = length(GH)$

Also, let the complex number from $F$ to $E$ be $z_2$ . Therefore $z_2 = b+2c+d +i(d-b)$ and $\left\vert z_2 \right\vert = length(FE)$ .

It's easy to observe that $z_1 = iz_2$

(The above statement can be verified as follows: $z_2+iz_1 = (a+b+c+d)+i(a+b+c+d) = 0 \Rightarrow z_2 = -iz_1 \Rightarrow \boxed{z_1=iz_2}$ )

This means:

• $z_1$ is $z_2$ rotated by $90^\circ \Rightarrow \boxed{\theta = 90^\circ}$

• $\left\vert z_1 \right\vert = \left\vert z_2 \right\vert \Rightarrow \boxed{\frac{l(GH)}{l(FE)} = 1}$

This problem is from a wonderful book Visual Complex Analysis By Tristan Needham

I just assumed the squares are equal. Therefore, their ratio is 1 and the angle formed is 90. LOL HAHAHA

I remember Napoleon's theorem and applied it here.. How lucky I am!