Complex Cross Product and Area of a parallelogram

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SUMMARY

The area of the parallelogram formed by two complex numbers, z and z', can be calculated using the formula |z x z'|, where z x z' represents the cross product. In a 2D context, this can be expressed as Im(~zz'), where Im denotes the imaginary part and ~ denotes the conjugate of z. The computed scalar from the cross product should be treated as a magnitude, which is equivalent to its absolute value. Thus, for complex numbers z = 1 + i and z' = 1 + 2i, the area is |1|, confirming that the absolute value is indeed 1.

PREREQUISITES
  • Understanding of complex numbers and their representation as vectors
  • Knowledge of vector cross products, particularly in 3D
  • Familiarity with complex conjugates and their properties
  • Basic grasp of imaginary numbers and their significance in calculations
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  • Explore the properties of complex conjugates and their applications in vector calculations
  • Learn about the geometric interpretation of the cross product in both 2D and 3D
  • Study the relationship between complex numbers and their corresponding vector representations
  • Investigate the implications of using absolute values in mathematical contexts
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mattmns
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Hello, just a quick question.

I have two complex numbers (say z and z'), and I want to find the area of the parallelogram that is generated by the two complex numbers (written as vectors, ie, if z = x + iy is a complex number, then the vector is (x,y)).

Now the area of the parallelogram generated by z and z' is |z x z'|

However, when I compute z x z' I get what I would consider a "scalar" and then I am asked to take the "magnitude" (or is it "absolute value") of this "scalar." Do I just take the "absolute value"?

For example.

say z = 1 + i, and z' = 1 + 2i
then z x z' = (1)(2) - (1)(1) = 2 - 1 = 1.
The area is |z x z'| = |1|. Is this the absolute value of 1 (which would equal 1) ?

Thanks.
 
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The cross product is really only defined in 3D. But you can apply it in 2D by just taking the third coordinate as 0. So (x,y,0) X (x',y',0) = (0,0,xy'-x'y). Now if z=x+iy and z'=x'+iy', can you think of a way of expressing xy'-x'y in terms of z and z'?
 
Yes, xy' - x'y = Im(~zz') [That is the Imaginary part of (conjugate of z times z')]. (I am not sure how to put a bar over z) However, I must be missing how this is connected to whether or not I am supposed to just take the "absolute value" of z x z' to get the area of the parallelogram spanned by z and z'. Could you elaborate? Thanks.

edit... hmmm... Maybe you are getting at that our new vector is (0, xy' -x'y) since z x z' = Im(~zz'), then we can take the "magnitude" (the modulus) which would be just the absolute value as I suspected.

Sorry about loosely using the terms magnitude, and absolute value.
 
Last edited:
"Area" is, by definition, a non-negative real number. Taking the cross product as you are, with the z-component of each vector 0, gives a vector with both x- and y-components 0. The area is the length of that vector, the absolute value of the z-component.
 

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