The author wanted to show that the expression $$(z_1-z_2)^2+(z_2-z_3)^2+(z_3-z_1)^2=0 \tag{1}\label{one}$$ is both translation invariant and rotation invariant.

Let us first take translation invariant. What it means is if we shift every point in the complex plane by a fixed vector $\zeta$, then too the expression (given above) doesn't change. This means if $z_i \mapsto z_i+\zeta$ and so on, then too \eqref{one} will not change. So $\zeta$ can be any (fixed) complex number. And $((z_j\,+\,\zeta )\,-\,(z_k\,+\,\zeta ))^2\,=\,(z_j\,-\,z_k)^2$ is for translational invariance not rotational.

For rotational invariance we want to see what will happen if we rotate every point in $\Bbb{C}$ by an angle $\theta$ in the counterclockwise direction, i.e. what happens when we $z_k \mapsto z_k \cdot e^{i \theta}$ (note that multiplication by $e^{i \theta}$ causes the vector represented by $z_k$ to be rotated by an angle $\theta$). As shown by the author $(z_je^{i\theta}\,-\,z_ke^{i\theta})^2\,=\,e^{i2\theta}(z_j\,-\,z_k)^2$, so every term in \eqref{one} will have the same factor $e^{2i\theta}$ so it will get canceled out and hence the expression will be invariant under rotation as well.

Once we have established that the expression \eqref{one} is invariant under both rotation and translation, then without the loss of generality we can work with values of $z_1,z_2,z_3$ that satisfy \eqref{one} but are also convenient to use. So with that the author chose $z_1=0$ and $z_2$ to be on the positive real axis (because we can translate and rotate the segment $z_1z_2$ such that $z_1$ comes to the origin and $z_2$ can be on the positive real axis and still our expression \eqref{one} will remain intact). Now once $z_1=0$ and $z_2=\xi >0$, then we can locate where $z_3$ should be. Once you solve the quadratic $z_3^2-z_3\xi+\xi^2=0$, we get $z_3=\xi e^{i\frac{\pm\pi}{3}}$ (note: this is simply rotating $z_2=\xi$ by $60^{\circ}$ in either clockwise or counterclockwise direction).

Now you realize that these points are the vertices of an equilateral triangle, then that means our general $z_1,z_2,z_3$ that satisfy \eqref{one} will also be forming an equilateral triangle.

The author wanted to show that the expression $$(z_1-z_2)^2+(z_2-z_3)^2+(z_3-z_1)^2=0 \tag{1}\label{one}$$ is both translation invariant and rotation invariant.

Let us first take translation invariant. What it means is if we shift every point in the complex plane by a fixed vector $\zeta$, then too the expression (given above) doesn't change. This means if $z_i \mapsto z_i+\zeta$ and so on, then too \eqref{one} will not change. So $\zeta$ can be any (fixed) complex number. And $((z_j\,+\,\zeta )\,-\,(z_k\,+\,\zeta ))^2\,=\,(z_j\,-\,z_k)^2$ is for translational invariance not rotational.

For rotational invariance we want to see what will happen if we rotate every point in $\Bbb{C}$ by an angle $\theta$ in the counterclockwise direction, i.e. what happens when we $z_k \mapsto z_k \cdot e^{i \theta}$ (note that multiplication by $e^{i \theta}$ causes the vector represented by $z_k$ to be rotated by an angle $\theta$). As shown by the author $(z_je^{i\theta}\,-\,z_ke^{i\theta})^2\,=\,e^{i2\theta}(z_j\,-\,z_k)^2$, so every term in \eqref{one} will have the same factor $e^{2i\theta}$ so it will get canceled out and hence the expression will be invariant under rotation as well.

Once we have established that the expression \eqref{one} is invariant under both rotation and translation, then without the loss of generality we can work with values of $z_1,z_2,z_3$ that satisfy \eqref{one} but are also convenient to use. So with that the author chose $z_1=0$ and $z_2$ to be on the positive real axis (because we can translate and rotate the segment $z_1z_2$ such that $z_1$ comes to the origin and $z_2$ can be on the positive real axis and still our expression \eqref{one} will remain intact). With $z_1=0$ and $z_2=\xi >0$ we can find $z_3$ by solving the quadratic $z_3^2-z_3\xi+\xi^2=0$. From this we get $z_3=\xi e^{i\frac{\pm\pi}{3}}$ (note: this is simply rotating $z_2=\xi$ by $60^{\circ}$ in either clockwise or counterclockwise direction).

Now you realize that these points are the vertices of an equilateral triangle, then that means our general $z_1,z_2,z_3$ that satisfy \eqref{one} will also be forming an equilateral triangle.

The author wanted to show that the expression $$(z_1-z_2)^2+(z_2-z_3)^2+(z_3-z_1)^2=0 \tag{1}\label{one}$$ is both translation invariant and rotation invariant.

Let us first take translation invariant. What it means is if we shift every point in the complex plane by a fixed vector $\zeta$, then too the expression (given above) doesn't change. This means if $z_i \mapsto z_i+\zeta$ and so on, then too \eqref{one} will not change. So $\zeta$ can be any (fixed) complex number. And $((z_j\,+\,\zeta )\,-\,(z_k\,+\,\zeta ))^2\,=\,(z_j\,-\,z_k)^2$ is for translational invariance not rotational.

For rotational invariance we want to see what will happen if we rotate every point in $\Bbb{C}$ by an angle $\theta$ in the counterclockwise direction, i.e. what happens when we $z_k \mapsto z_k \cdot e^{i \theta}$ (note that multiplication by $e^{i \theta}$ causes the vector represented by $z_k$ to be rotated by an angle $\theta$). As shown by the author $(z_je^{i\theta}\,-\,z_ke^{i\theta})^2\,=\,e^{i2\theta}(z_j\,-\,z_k)^2$, so every term in \eqref{one} will have the same factor $e^{2i\theta}$ so it will get canceled out and hence the expression will be invariant under rotation as well.

Once we have established that the expression \eqref{one} is invariant under both rotation and translation, then without the loss of generality we can work with values of $z_1,z_2,z_3$ that satisfy \eqref{one} but are also convenient to use. So with that the author chose $z_1=0$ and $z_2$ to be on the real axis (because we can translate and rotate the segment $z_1z_2$ such that $z_1$ comes to the origin and $z_2$ can be either on the positive or negative real axis and still our expression \eqref{one} will remain intact). Now once $z_1=0$ and $z_2^2=\xi >0$, then we can locate where $z_3$ should be. Once you know $z_3=\xi e^{i\frac{\pm\pi}{3}}$, you realize that these vertices form an equilateral triangle, then that means our general $z_1,z_2,z_3$ that satisfy \eqref{one} will also be forming an equilateral triangle.

The author wanted to show that the expression $$(z_1-z_2)^2+(z_2-z_3)^2+(z_3-z_1)^2=0 \tag{1}\label{one}$$ is both translation invariant and rotation invariant.

Let us first take translation invariant. What it means is if we shift every point in the complex plane by a fixed vector $\zeta$, then too the expression (given above) doesn't change. This means if $z_i \mapsto z_i+\zeta$ and so on, then too \eqref{one} will not change. So $\zeta$ can be any (fixed) complex number. And $((z_j\,+\,\zeta )\,-\,(z_k\,+\,\zeta ))^2\,=\,(z_j\,-\,z_k)^2$ is for translational invariance not rotational.

For rotational invariance we want to see what will happen if we rotate every point in $\Bbb{C}$ by an angle $\theta$ in the counterclockwise direction, i.e. what happens when we $z_k \mapsto z_k \cdot e^{i \theta}$ (note that multiplication by $e^{i \theta}$ causes the vector represented by $z_k$ to be rotated by an angle $\theta$). As shown by the author $(z_je^{i\theta}\,-\,z_ke^{i\theta})^2\,=\,e^{i2\theta}(z_j\,-\,z_k)^2$, so every term in \eqref{one} will have the same factor $e^{2i\theta}$ so it will get canceled out and hence the expression will be invariant under rotation as well.

Once we have established that the expression \eqref{one} is invariant under both rotation and translation, then without the loss of generality we can work with values of $z_1,z_2,z_3$ that satisfy \eqref{one} but are also convenient to use. So with that the author chose $z_1=0$ and $z_2$ to be on the positive real axis (because we can translate and rotate the segment $z_1z_2$ such that $z_1$ comes to the origin and $z_2$ can be on the positive real axis and still our expression \eqref{one} will remain intact). Now once $z_1=0$ and $z_2=\xi >0$, then we can locate where $z_3$ should be. Once you solve the quadratic $z_3^2-z_3\xi+\xi^2=0$, we get $z_3=\xi e^{i\frac{\pm\pi}{3}}$ (note: this is simply rotating $z_2=\xi$ by $60^{\circ}$ in either clockwise or counterclockwise direction).

Now you realize that these points are the vertices of an equilateral triangle, then that means our general $z_1,z_2,z_3$ that satisfy \eqref{one} will also be forming an equilateral triangle.