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

• vcsharp2003
In summary, the criss-cross approach, which involves multiplying both sides of an equation by the same number, is a valid step in solving equations with complex numbers. This approach can be used because of the field axioms, which state that any number (except for 0) multiplied by another number will result in an equivalent equation. However, it is important to note any restrictions, such as z≠−1, in order to avoid any invalid operations. It is recommended to test solutions in the original equation to ensure accuracy.f

#### vcsharp2003

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?

$$\frac {z-1} {z+1} = ni$$
$$\implies \frac {z-1} {z+1} = \frac {ni} {1}$$
$$\implies {(z-1)} \times 1= {ni} \times {(z+1)}$$

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It's easy to prove that ##{ab\over cd}=1\Leftrightarrow ab=cd## for complex numbers as well (cd##\ne 0##)

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

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So, the criss-cross approoach will work because of the result you mentioned.

So, the criss-cross approoach will work because of the result you mentioned.
Yes.

• vcsharp2003
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.

• vcsharp2003
"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.

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.

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.

• vcsharp2003
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.

• vcsharp2003