Complex representation of orthocenter : a signed area issue. I was recently looking up an equation for the orthocenter in the complex plane and found the following in Zwikker, C. (1968), The Advanced Geometry of Plane Curves and Their Applications, Dover Press:
$$z_O=\frac{z_1\{z_1(z_2^*-z_3^*)+z_1^*(z_2-z_3)\}+\text{cycl.}}{4iA}$$
where $^*$ denotes the conjugate and $A$ is the area. I've not encountered the term cycl. previously and assumed it meant cycling through the indices, such as $z_2,z_3,z_1$ and $z_3,z_1,z_2$. But this did not give the correct result.
Zwikker similarly give the circumcenter as
$$z_C=\frac{z_1z_1^*(z_3-z_2)+\text{cycl.}}{4iA}$$
Does anyone know what this means?
 A: The formula you have given (page 53 of the mentioned book) with its "cycl." is indeed to be interpreted as you have done.
One can re-write it under the equivalent form :
$$z_O=\frac{1}{2iA}\left[a_1z_1+a_2z_2+a_3z_3\right]  \ \text{with} \  \begin{cases}a_1&=&\Re(\overline{z_1}(z_2-z_3))\\ a_2&=&\Re(\overline{z_2}(z_3-z_1))\\a_3 &=&\Re(\overline{z_3} (z_1-z_2))\end{cases}\tag{1}$$
This formula works in all cases provided that
$$A:=\frac12 \Im(z_1 \overline{z_2}+z_2 \overline{z_3}+z_3 \overline{z_1}),\tag{2}$$
is interpreted as the opposite of the signed area of triangle $z_1z_2z_3$ in the universally accepted anti-clockwise orientation. As well explained by Blue in his comment, the author takes, oddly, the clockwise orientation as the default orientation.
Remark: ($\Re$ and $\Im$ stand for "real part", and "imaginary part" resp.).
Here is a GeoGebra animation based on formulas (1) and (2) displaying the orthocenter (red point) of any kind of triangle (one can move the sliders giving the real and imaginary parts $a_k$ and $b_k$ of points $z_k$ ; you have to move up the lift in the left window to make the sliders viewable).
https://www.geogebra.org/calculator/jj5zmqsg
Please note that if $z_1,z_2,z_3$ all belong to the unit circle, there exists a much simpler formula which is :
$$z_O=z_1+z_2+z_3$$
(see here)
