# Equating Complex Numbers

by KrayzBlu
Tags: complex, complex algebra, equating, exponential, numbers
 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
P: 799
 Quote by KrayzBlu 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.
P: 11
 Quote by SteveL27 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.

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Thanks
P: 6,380

## Equating Complex Numbers

 Quote by KrayzBlu 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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