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By comparing complex numbers in complex plane and vectors in real plane, we see that they behave 'same'. For example we can find solutions of $z_1+z_2$, $|z_1-z_2|$ and so on by considering $z_i=(x_i,y_i)$ as a vector in real plane.

But I couldn't find the analogous to the product of two complex numbers as vectors in real plane. The definition $(x_1,y_1)(x_2,y_2) = (x_1x_2 - y_1y_2,y_1x_2 + x_1y_2)$ doesn't represent any type of products in real plane, i.e. in case of dot product $(x_1,y_1)(x_2,y_2)=(x_1x_2,y_1y_2)$ and in case of cross product $(x_1,y_1)(x_2,y_2)=(0,0)$ since there is no component in x-y real plane.

What is the vector product comparison of $(x_1,y_1)(x_2,y_2) = (x_1x_2 - y_1y_2,y_1x_2 + x_1y_2)$ in x-y real plane?

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  • $\begingroup$ You can add the arguments and multiply the magnitudes, if that counts as a geometrical interpretation. $\endgroup$ – Element118 Nov 18 '15 at 2:40
  • $\begingroup$ @Element118 - I meant by geometrical comparison, the type of vector product. Sorry, I edited it. Thanks $\endgroup$ – user231343 Nov 18 '15 at 2:44
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    $\begingroup$ The multiplication of quaternions can be written in terms of products of 3D-vector by scalar, dot product of two 3D-vectors, and the cross product of two 3D-vectors. Complex multiplication is a special case of this, where the 3D-vectors are confined to a single axis. But I don't think this point of view adds much light to your question. Except possibly in the sense that it underlines the fact that the multiplication thinks of the real axis as scalars, and everything else as vectors. The 1+3 dimensional view of quaternions becomes a 1+1 dimensional view of the complex numbers. $\endgroup$ – Jyrki Lahtonen Nov 18 '15 at 8:14
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    $\begingroup$ For easy reference: the quaternion product of $q=(a,\vec{u})$ and $q'=(b,\vec{v})$ is $$qq'=(ab-\vec{u}\cdot\vec{v},a\vec{v}+b\vec{u}+\vec{u}\times\vec{v}).$$ When dealing with complex numbers $\vec{u}$ and $\vec{v}$ are constrained to the $\vec{i}$-axes. $\endgroup$ – Jyrki Lahtonen Nov 18 '15 at 8:18
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The arguments add mod $2\pi$ and the magnitudes multiply.

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