Can we use criss-cross approach with complex number equations?

AI Thread Summary
The discussion centers on the validity of using the criss-cross approach in solving complex number equations, specifically the equation $$\frac {z-1} {z+1} = ni$$. Participants agree that this method is essentially multiplying both sides by a non-zero term, which is a valid operation in algebra. The importance of ensuring that the variable does not take on values that would lead to division by zero is emphasized, particularly noting that $$z \ne -1$$. While the term "criss-cross" may be less relevant, the underlying principle of maintaining equality through multiplication is affirmed. Overall, the approach is deemed acceptable as long as care is taken with the conditions of the variables involved.
vcsharp2003
Messages
906
Reaction score
178
Homework Statement
If ##\frac {z-1} {z+1} = ni## and ## Z\ne -1##, then can we simply criss-cross the denominators of the two sides so that LHS is multiplied by 1 and RHS is multiplied by ##z+1##? This is part of a bigger problem, but I just wanted to clear my doubts on whether this approach is possible.
Relevant Equations
None
I am not sure why criss-cross approach would work here, but it seems to get the answer. What would be the reason why we could use this approach?

$$\frac {z-1} {z+1} = ni$$
$$\implies \frac {z-1} {z+1} = \frac {ni} {1}$$
$$\implies {(z-1)} \times 1= {ni} \times {(z+1)}$$
 
Last edited:
Physics news on Phys.org
It's easy to prove that ##{ab\over cd}=1\Leftrightarrow ab=cd## for complex numbers as well (cd##\ne 0##)

##\ ##
 
  • Like
Likes vcsharp2003
BvU said:
It's easy to prove that ##{ab\over cd}=1\Leftrightarrow ab=cd## for complex numbers as well (cd##\ne 0##)

##\ ##

So, the criss-cross approoach will work because of the result you mentioned.
 
vcsharp2003 said:
So, the criss-cross approoach will work because of the result you mentioned.
Yes.
 
  • Like
Likes vcsharp2003
vcsharp2003 said:
Homework Statement:: If ##\frac {z-1} {z+1} = ni## and ## Z\ne -1##, then can we simply criss-cross the denominators of the two sides so that LHS is multiplied by 1 and RHS is multiplied by ##z+1##? This is part of a bigger problem, but I just wanted to clear my doubts on whether this approach is possible.
Relevant Equations:: None

I am not sure why criss-cross approach would work here, but it seems to get the answer. What would be the reason why we could use this approach?
"Criss-cross" approach, as a term, seems less useful to me than what it is that you're actually doing. IOW, you're multiplying both sides of the given equation by ##z + 1##. This is guaranteed to be a valid step because it is given that ##z \ne -1##.
Multiplication of both sides of an equation, whether real or complex, is a property of equations that results in an equivalent equation, provided that what you multiply by is nonzero.
 
  • Like
Likes vcsharp2003
Mark44 said:
"Criss-cross" approach, as a term, seems less useful to me than what it is that you're actually doing. IOW, you're multiplying both sides of the given equation by ##z + 1##. This is guaranteed to be a valid step because it is given that ##z \ne -1##.
Multiplication of both sides of an equation, whether real or complex, is a property of equations that results in an equivalent equation, provided that what you multiply by is nonzero.
Yes, I think your explanation is very logical. If two equals are multiplied by the same number (##\ne 0##), then they are still equals.
 
Mark44 said:
This is guaranteed to be a valid step because it is given that ##z \ne -1##.
This multiplication is always allowed, even in the case ##z=-1.## Whether it makes sense is a different question. The important parts of the field axioms in this context are ##1\neq 0## and all ##z\ne 0## have a multiplicative inverse.
 
fresh_42 said:
This multiplication is always allowed, even in the case z=−1. Whether it makes sense is a different question.
Certainly multiplication of both sides of an equation by zero is allowed, but this operation isn't useful and doesn't produce a new equation that is equivalent to the one you started with. I made this clear in the next paragraph of what I wrote.
 
Yes, with complex numbers, you can do all the mathematical manipulation that you are used to with real numbers. In fact, you can do more. But the consequences are profound and worth a lot of study.
As always, when you cross-multiply you should be careful to keep track of things like ##z \ne -1## where the original equation was not allowed to divide by zero. Don't lose track of those things.
 
  • Like
Likes vcsharp2003
  • #10
In fact, it is often best to just go back and test any solutions in the original equations. Keeping perfect track of which values are illegal can get overwhelming. But in simple cases, it is best to keep track as the calculation proceeds.
 
  • Like
Likes vcsharp2003
Back
Top