The complex shoelace formula for the signed area of a triangle with vertices given by the complex numbers $a, b, c$ is $$\frac{i}{4} \begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ \overline{a} & \overline{b} & \overline{c} \\ \end{vmatrix} $$

I have seen the algebraic proof for this formula using elementary row operations and the multilinear nature of the determinant, however that style of proof appears to imply that the simplicity of the final result (i.e. a determinant involving only conjugates) is a coincidence; furthermore it provides little intuition.

I am looking for a way to understand this formula through a geometric/intuitive argument (not necessarily rigorous) in order to gain a deeper understanding rather than just accepting that the algebra works out.

Isaac Greene
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  • It seems a little hard to answer your question as you don't specify what you "allow" or "disallow". –  Nov 14 '18 at 21:55

4 Answers4


Consider the case $\,a=0\,$, first, where the proposed formula reduces to:

$$ S_{OBC} = \frac{i}{4} \begin{vmatrix} 1 & 1 & 1 \\ 0 & b & c \\ \overline{0} & \overline{b} & \overline{c} \\ \end{vmatrix} = \frac{i}{4} \begin{vmatrix} b & c \\ \overline{b} & \overline{c} \\ \end{vmatrix} = \frac{i}{4}\left(b \bar c - \bar b c\right) = -\,\frac{1}{2}\,\operatorname{Im}(b \bar c) \tag{1} $$

Let $\,b = |b| e^{i \beta}\,$, $\,c = |c| e^{i\gamma}\,$, then $\,b \bar c = |b|\cdot|c|\cdot e^{i(\beta-\gamma)}=\color{blue}{|b|}\cdot\color{blue}{|c|}\cdot \left(\cos(\beta-\gamma)+ i \color{red}{\sin(\beta-\gamma)}\right)\,$, so $\,(1)\,$ is equivalent to the familiar triangle (signed) area formula $\,S_{OBC} = \frac{1}{2}\,\color{blue}{OB} \cdot \color{blue}{OC} \cdot \color{red}{\sin \widehat{BOC}}\,$.

But area is, of course, invariant under translations, so it follows that in the general case:

$$ S_{ABC} = \frac{1}{2}\,AB \cdot AC \cdot \sin \widehat{BAC} = -\,\frac{1}{2}\,\operatorname{Im}\left((b-a) \overline{(c-a)}\right) \tag{2} $$

The latter is equivalent to the posted formula via elementary column operations:

$$ \begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ \overline{a} & \overline{b} & \overline{c} \\ \end{vmatrix} \;=\; \begin{vmatrix} 1 & 0 & 0 \\ a & b-a & c-a \\ \overline{a} & \overline{b-a} & \overline{c-a} \\ \end{vmatrix} $$

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  • I appreciate this neat explanation, however I fail to see the intuition it offers as to why the final answer is an extremely simple 3x3 determinant. It's still as if the simplicity of the final formula is a coincidence. – Isaac Greene Aug 19 '18 at 22:50
  • @IsaacGreene $\;(1)\,$ above links the geometric formula $\,S=\frac{1}{2}bc \sin A\,$ to the minor $\,\begin{vmatrix} b & c \\ \overline{b} & \overline{c} \\ \end{vmatrix}\,$ of the determinant. The rest is just the classic shoelace argument, plus recognizing the Laplacian expansion by minors to write everything as one single $3$x$3$ determinant. I am not entirely sure what kind of "intuition" you are looking for, which you would find satisfactory. – dxiv Aug 20 '18 at 00:35
  • It isn't necessary for the final formula to be so 'neat' (i.e. it could be much more complicated) I am seeking some intuition for why it is so neat. – Isaac Greene Aug 20 '18 at 02:08
  • @IsaacGreene I guess you could start with the [Equation of line in form of determinant](https://math.stackexchange.com/questions/2111593/equation-of-line-in-form-of-determinant), then adapt the argument used in [What's an intuitive way to think about the determinant?](https://math.stackexchange.com/questions/668/whats-an-intuitive-way-to-think-about-the-determinant/669#669) to the complex case. Basically, show that the row-linearity and the value of $\,1\,$ for points $\,0,1,i\,$ uniquely define the area of the triangle. – dxiv Aug 20 '18 at 02:38

Here's one way to intuit it. Area should be a degree-$2$ homogeneous polynomial function of $a,\,b,\,c,\,a^\ast,\,b^\ast,\,c^\ast$, construed as $6$ independent variables. Exchanging each variable with its own conjugate is equivalent to imposing $i\mapsto -i$, i.e. flipping the plane, so should change the sign of the signed area. Similarly, exchanging any two of $a,\,b,\,c$ flips the triangle, so that should change the sign of the signed area too. The only natural geometrical way to get this kind of behaviour is a determinant of some square matrix, with $a,\,b,\,c$ in one row and their conjugates in the other. And to ensure exchanging columns doesn't change the unsigned area, we need the final row to consist of the same number $3$ times. It may as well be $1$; now we just need an overall multiplier out front.

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Without loss of generality, you can assume that $c=0$ (either by translating the triangle, or by reducing the $3\times3$ determinant to $2\times2$) and the formula is in fact

$$\frac i4(a\overline b-\overline ab)=\frac i4(a\overline b-\overline {a\overline b}),$$

which is also

$$\frac{\Im(a\overline b)}{2}.$$

On the other hand, it is known that $a\overline b$ is a complex number with components equal to the dot product and the cross product of $a$ and $b$. And the cross product is the area of the parallelogram they form.


I am not very familiar with intuition behind the determinant of a complex matrix, but I have a very nice intuitive explanation that the following very similar formula is correct.

Given a $2$-dimensional triangle with vertices $A$, $B$ and $C$. Define $A':=(A_x,A_y,1)^T$, $B':=(B_x,B_y,1)^T$ and $C':=(C_x,C_y,1)^T$. The area of the triangle $ABC$ equals $$|\det(A',B',C')/2|.$$

Simple algebra shows this formula is the same as yours, but for this formula I have a nice intuitive explanation of its correctness.

By the geometric definition, the determinant is the volume of the corresponding parallelepiped $\mathcal{P}:=\{\alpha A'+\beta B'+\gamma C'|\alpha,\beta,\gamma\in[0,1]\}$. It should be pretty clear that $\mathcal{P}$ lies in the region $0\leq z\leq3$. So we can split the volume of $\mathcal{P}$ in three parts: $0\leq z\leq1$, $1\leq z\leq2$ and $2\leq z\leq3$. Call these parts $\mathcal{P}_0$, $\mathcal{P}_1$ and $\mathcal{P}_2$.

Now comes the important part. We translate $\mathcal{P}_1$ by $-A'$ and we translate $\mathcal{P}_2$ by $-2A'$. Then all three parts are merely touching and they form the parallelepiped $\mathcal{P}'$ of the points $A'$, $B'-A'$ and $C'-A'$ (*). Notice that $A'_z=1$ while the other two points have $z=0$. So the parallelogram of these two points forms the base of $\mathcal{P}'$, and $\mathcal{P}'$ has height $1$. We find that the volume of $\mathcal{P}'$ equals the area of this parallelogram, which equals twice the area of the triangle $ABC$.

(*) This is not obvious. However to prove this, I believe you will need to use algebra. But to get an intuition for this, you can simply try to imagine the parallelepiped pieces. This is not easy though, but you could also go over the top and make an animation or even a real life model.

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