For \(a,b\in\mathbb C\), regard \(a\) and \(b\) as vectors in \(\mathbb R^2\). Express the complex product \(ab\) in terms of the dot product and the planar cross product of those vectors.
Solution
1. With the standard oriented planar cross product, the dot and cross products of the vectors represented by \(a\) and \(b\) determine \(\overline a b\), not \(ab\).
Write \[ a=x+iy, \qquad b=u+iv, \] and use \[ a\cdot b=xu+yv, \qquad a\times b=xv-yu. \] Then \[ \overline a b =(x-iy)(u+iv) =(xu+yv)+i(xv-yu), \] so \[ \boxed{\overline a b=a\cdot b+i(a\times b)}. \] Equivalently, \[ a\overline b=a\cdot b-i(a\times b). \]
2. Consequently the source exercise, if interpreted as asking for \(ab\) from only the two scalar quantities \(a\cdot b\) and \(a\times b\), is false as written.
Take first \[ a=b=1. \] Then \[ a\cdot b=1, \qquad a\times b=0, \qquad ab=1. \] Now rotate both vectors through \(\pi/2\) and take \[ a=b=i. \] The dot and cross products are still \[ a\cdot b=1, \qquad a\times b=0, \] but now \[ ab=-1. \] Thus no function of the pair \((a\cdot b,a\times b)\) alone can equal \(ab\) for all \(a,b\in\mathbb C\).
3. If one reflects one of the two vectors, then \(ab\) can be written using the same operations.
The vector represented by \(\overline b=u-iv\) is the reflection of the vector represented by \(b\) across the real axis. Hence \[ a\cdot\overline b=xu-yv, \qquad a\times\overline b=-xv-yu. \] Therefore \[ \boxed{ab=a\cdot\overline b-i(a\times\overline b)}. \] This uses the reflected vector \(\overline b\), whereas 1 is the direct identity involving the original two vectors.