Solving Equality Problem in Complex Numbers

• kelvin490
Therefore, in this case, x1 and x2 are equivalent. In summary, in order for x1 and x2 to be exactly equivalent all the time, A and B must satisfy the condition that A=B* and B=A*. However, if we do not follow this approach and set (A-B*)eiωt=(A*-B)e-iωt, we end up with a constant on one side and a time-dependent function on the other, which is only possible if A=B* and B=A*. Therefore, even if A and B are not equal to their complex conjugates, x1 and x2 can still be considered equivalent. The undefined result of 0/0 in this case is due to the assumption that

kelvin490

Gold Member
Suppose we have two equation x1=Aeiωt + Be-iωt and x2=A*e-iωt + B*eiωt . Where A and B are complex number and A* B* are their conjugate correspondingly.

Now if we want to make x1 and x2 exactly equivalent all the time, one way to do it is to have A=B* and B=A* so that x1 and x2 are equivalent. However, if we don't do it by this approach but instead set (A-B*)eiωt=(A*-B)e-iωt, then we have ei2ωt=(A*-B)/(A-B*). I would like to ask if the A and B chosen can satisfy this criteria (even A≠B* and B≠A*), can we still say that x1 ≡ x2 ?

Another thing trouble me is if A=B* and B=A* , then ei2ωt=(A*-B)/(A-B*)=0/0 which is undefined. What causes this problem?

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kelvin490 said:
Suppose we have two equation x1=Aeiωt + Be-iωt and x2=A*e-iωt + B*eiωt . Where A and B are complex number and A* B* are their conjugate correspondingly.

Now if we want to make x1 and x2 exactly equivalent all the time, one way to do it is to have A=B* and B=A* so that x1 and x2 are equivalent. However, if we don't do it by this approach but instead set (A-B*)eiωt=(A*-B)e-iωt, then we have ei2ωt=(A*-B)/(A-B*). I would like to ask if the A and B chosen can satisfy this criteria (even A≠B* and B≠A*), can we still say that x1 ≡ x2 ?
What you are doing is saying that if $Ae^{i\omega t}+ Be^{-i\omega t}= A^*e^{-i\omega t}+ B^*e^{-i\omega t}$ then we must have $e^{2i\omega t}= \frac{A^*- B}{A- B^*}$. However, the right side of that equation is a constant, independent of t. In order for that to be true, the left side must also be a constant- so what you are doing is equivalent to assuming that $\omega= 0$, a very restrictive condition!

Another thing trouble me is if A=B* and B=A* , then ei2ωt=(A*-B)/(A-B*)=0/0 which is undefined. What causes this problem?
If $A= B^*$ then $B$ must equal $A^*$- that is not a distinct condition. And in that case you are saying $x_1= Ae^{i\omega t}+ A*e^{-i\omega t}$ and $x_2= A*e^{-i\omega t}+ Ae^{i\omega t}$, that is, that $x_1= x_2$ for all t. That is exactly what you said you wanted. There is no "problem". If you have two identical equation, say "px= qy" and "px= qy" and subtract them you get 0x= 0y, of course. It would make no sense to try to "solve for y" in that case, you would just get 0/0 as happens here.

The two orignal expressions are complex conjugates of each other. Therefore to be equal, the imaginary part must be 0.

I think the concept of linear independence can help. Since e-iωt and eiωt are two linearly independent variables, the coefficient of x1 and x2 must be equal so that x1 and x2 are equivalent.

Also from (A-B*)ei2ωt=(A*-B) we can see that the left hand side is a time dependent function (let's say t is time) and the other side is a constant. The only way to make both sides equal all the time is to have A=B* and B=A*.

I would approach this problem by first clarifying the definitions and assumptions being made. In this case, it seems that x1 and x2 are being defined as equivalent if they have the same values at all times. Additionally, it appears that A and B are being treated as complex numbers with specific properties, such as being conjugates of each other.

Based on these definitions, it is not clear to me how the equation ei2ωt=(A*-B)/(A-B*) would be satisfied if A and B are not equal to each other's conjugate. This is because the numerator and denominator would not have the same values, unless A and B have specific values that satisfy this equation. Therefore, I would say that x1 and x2 cannot be considered equivalent if A and B are not equal to each other's conjugate.

The issue with ei2ωt=(A*-B)/(A-B*)=0/0 being undefined is due to the fact that division by 0 is undefined in mathematics. This problem arises because the equation is assuming that A and B are equal to each other's conjugate, which may not always be the case. If we want to avoid this problem, we would need to be careful in our selection of A and B to ensure that they are not equal to each other's conjugate.

1. What are complex numbers?

Complex numbers are numbers that are expressed in the form a + bi, where a and b are real numbers and i is the imaginary unit (which is equal to the square root of -1).

2. Why do we need to solve equality problems in complex numbers?

Complex numbers allow us to work with and solve equations that involve both real and imaginary numbers. Many important mathematical concepts, such as the quadratic formula, involve complex numbers.

3. How do you solve equality problems in complex numbers?

To solve an equality problem in complex numbers, we use the same principles as solving equations with real numbers. We can add, subtract, multiply, and divide complex numbers, and solve for the unknown variable by isolating it on one side of the equation.

4. What are some common mistakes when solving equality problems in complex numbers?

One common mistake is forgetting to distribute the imaginary unit i when using the distributive property. Another mistake is not simplifying complex expressions before solving for the variable. It is also important to remember the order of operations when working with complex numbers.

5. Are there any special techniques for solving equality problems in complex numbers?

Yes, one technique is to convert complex numbers into polar form and use the properties of polar numbers to simplify the equations. Another technique is to use the conjugate of a complex number to eliminate the imaginary part of an equation. Additionally, graphing complex numbers on the complex plane can help visualize solutions to equality problems.