Equating Complex Numbers


by KrayzBlu
Tags: complex, complex algebra, equating, exponential, numbers
KrayzBlu
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#1
Feb10-14, 10:31 AM
P: 11
Hi,

We know that if we have two complex numbers in polar format (i.e., magnitude and exponential), that for two complex vectors

z1 = A*exp(iB)
z2 = C*exp(iD)

If z1 and z2 are equal, then A = C and B = D. However, this is assuming these values are all real. What if they are complex? I.e. can we say if we have two complex numbers

z3 = (a+ib)*exp(c+id)
z4 = (e+if)*exp(g+ih)

If z3 and z4 are equal, can we say that (a+ib) = (e+if) and (c+id) = (g+ih)?

Thanks
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SteveL27
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#2
Feb10-14, 12:35 PM
P: 799
Quote Quote by KrayzBlu View Post
Hi,

We know that if we have two complex numbers in polar format (i.e., magnitude and exponential), that for two complex vectors

z1 = A*exp(iB)
z2 = C*exp(iD)

If z1 and z2 are equal, then A = C and B = D.
Not quite true. B and D must differ by an integer multiple of 2pi. You'll need to take that into account when working out the rest of this.
KrayzBlu
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#3
Feb10-14, 12:53 PM
P: 11
Quote Quote by SteveL27 View Post
Not quite true. B and D must differ by an integer multiple of 2pi. You'll need to take that into account when working out the rest of this.
Thanks for pointing this out, SteveL27, I should have said B = D +/- n*2*π, where n is any integer.

AlephZero
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#4
Feb10-14, 02:16 PM
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Equating Complex Numbers


Quote Quote by KrayzBlu View Post
If z1 and z2 are equal, then A = C and B = D. However, this is assuming these values are all real.
You can have A = -C, if B and D are different by an odd multiple of π

z3 = (a+ib)*exp(c+id)
z4 = (e+if)*exp(g+ih)
If z3 and z4 are equal, can we say that (a+ib) = (e+if) and (c+id) = (g+ih)?
It should be easy to see why that is false. For example take
a = 1, b = c = d = 0, e = 0, f = 1, and find g and h to make z3 = z4.

If you convert z3 = x3 + i y3 and z3 = x4 + i y4, you only have 2 equations (x3 = x4 and y3 = y4) but 8 unknowns (a through h). You need 6 more equations before you can hope there is a unique solution.


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