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How to relate complex multiplication to Cartesian products?

  1. Jan 23, 2017 #1
    1. The problem statement, all variables and given/known data
    "ℝ×ℝ and ℂ are very similar in many ways. How do you realize ℂ as a Cartesian product of two sets? Consider how complex numbers are multiplied; by grouping real and imaginary parts, show how the pattern of complex multiplication can be used to define multiplication in ℝ×ℝ. Using this multiplication, find the multiplicative inverse of (1,1) in ℝ×ℝ."

    2. Relevant equations
    ##ℝ×ℝ={(x,y): x, y ∈ ℝ}##
    ##ℂ={x+yi: x, y ∈ ℝ}##

    3. The attempt at a solution
    For the first part, my teacher demonstrated that the imaginary component has its own axis, as does the real component of a complex number. He told us that they formed a plane on the real-imaginary plane, and that's analogous to the Cartesian plane. So I figure that I should somehow relate ##x## to the real part, and ##y## to the imaginary.

    Let ##a,b,c,d∈ℝ##.
    Then ##(a+bi)(c+di)=(ac+adi)+(cbi-bd)=(ac-bd)+(cb+ad)i##.
    Setting these equal to the given ##x## and ##y## values, I tried to fix the variables so that ##(a+bi)(c+di)=1+1i##, but I'm getting many possible solutions. I'm trying to reverse-engineer the process, in other words.

    For example, it works when ##a=1,b=0,c=1,d=1##, but also when ##a=0,b=-1,c=-1,d=1##.

    So I can't really form a Cartesian product from a complex product. And I'm just confused on the concept he is trying to explain to us.
     
  2. jcsd
  3. Jan 24, 2017 #2

    Mark44

    Staff: Mentor

    You're trying to find the multiplicative inverse of (1, 1). IOW, what must (a, b) be so that (1, 1) x (a, b) = (1, 0)? There is a unique solution.
     
  4. Jan 24, 2017 #3
    Let's see... ##a=\frac{1}{2},b=-\frac{1}{2}##, so ##(\frac{1}{2},-\frac{1}{2})##. I'll still need to explain the connection between the Cartesian product and complex multiplication, though.
     
  5. Jan 24, 2017 #4

    Mark44

    Staff: Mentor

    Complex multiplication is easiest to understand by considering the complex numbers in polar form. If you have two complex numbers ##z_1 = r_1e^{i\theta_1}## and ##z_2 = r_2 e^{i \theta_2}##, then ##z_1z_2 = r_1r_2 e^{i (\theta_1 + \theta_2)}##. IOW, the magnitudes of the two complex numbers multiply to make the magnitude of the product, and the angles of the two complex numbers add to make the angle of the product. Note that ##z_1## could also be written as ##r_1(\cos \theta + i \sin \theta)##, and similar for ##z_2##. What I'm calling the "angle" is also called the argument, or arg, for short.

    From your example (1, 1) as a complex number in polar form is ##\sqrt 2 e^{i \pi/4}##. (1/2, -1/2) as a complex number in polar form is ##\frac 1{\sqrt 2}e^{-i\pi/4}##. If you multiply these you will get a complex number whose magnitude is 1 and whose angle is 0; that is, the complex number 1 + 0i. This is why (1, 1) and (1/2, -1/2) are multiplicative inverses of each other.

    I'm not sure whether this is what you were looking for.
     
  6. Jan 24, 2017 #5

    FactChecker

    User Avatar
    Science Advisor
    Gold Member

    Yes. Now put this definition of multiplication in terms of (a,b) x (c,d) = (?,?) and you have multiplication in RxR.
    Before you can talk about finding the multiplicative inverse of (1,1), you need to know what the multiplicative identity is. What values of (Ix,Iy) will give (Ix,Iy)x(c,d) = (c,d) for all values of c and d in R? (Hint: 1 is the multiplicative identity in R.)
    Once you know what the multiplicative identity, (Ix,Iy) is in RxR, you can then find the multiplicative inverse, of (1,1). It is (a,b) where (a,b)x(1,1) = (Ix,Iy). Solve that for a and b.

    I can not really say more in a homework question.

    PS. I see that I am way too slow at writing these. There have been several posts since I began.
     
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