Why complex = real in this case

[shahar]

Somebody notice me that the expression in the picture have a solution (a=0) or (b=0) and (a and b [together] > 0)...
My question is why the solution that the expression is the same in the complex number.
How can I should know when the solution of an expression will be the same in real and complex number?

 

[HallsofIvy]

The crucial point in your previous post was that, after squaring both sides of \(\displaystyle \sqrt{a^2+ b^2}= a+ b\), to get \(\displaystyle x^2+ b^2= a^2+ 2ab+ b^2\), the "\(\displaystyle a^2+ b^2\)" terms cancel. Since, in fact, "squaring" is not relevant to this problem, the question of "complex" versus "real" numbers simply does not arise.

 

[shahar]

The crucial point in your previous post was that, after squaring both sides of \(\displaystyle \sqrt{a^2+ b^2}= a+ b\), to get \(\displaystyle x^2+ b^2= a^2+ 2ab+ b^2\), the "\(\displaystyle a^2+ b^2\)" terms cancel. Since, in fact, "squaring" is not relevant to this problem, the question of "complex" versus "real" numbers simply does not arise.

Why the question of "complex" versus "real" numbers simply does not arise​?

 

[Deleted member 4993]

Why the question of "complex" versus "real" numbers simply does not arise​?

Complex number - according to my definition - can be expressed as "a + i * b" , where:

i = sqrt[-1] and

a = any real number (no imaginary part) and

b = any real number (no imaginary part)

So:

What is a "complex" number - according to you?

 

[mmm4444bot]

Why complex = real

All Real numbers are Complex numbers.

Not all Complex numbers are Real because some Complex numbers contain the imaginary component i.

A Complex number is Real when it has no imaginary component.

a + b͏∙i

When b is zero (or both a and b are), then there is no imaginary part, and a+b͏∙i is a Real number.

If b is not zero, then a+b͏∙i is not Real.

 

[JeffM]

Somebody notice me that the expression in the picture have a solution (a=0) or (b=0) and (a and b [together] > 0)...
My question is why the solution that the expression is the same in the complex number.
How can I should know when the solution of an expression will be the same in real and complex number?

If a problem has a solution in the real numbers, it also has a solution in the complex numbers.

If a problem has a solution in the complex numbers, it may or may not have a solution in the real numbers.

It is exactly the same situation as all cows are mammals, but not all mammals are cows.

 

[Dr.Peterson]

Somebody notice me that the expression in the picture have a solution (a=0) or (b=0) and (a and b [together] > 0)...
My question is why the solution that the expression is the same in the complex number.
How can I should know when the solution of an expression will be the same in real and complex number?

The equation is sqrt(a^2 + b^2) = a + b. (It is not an "expression", but an "equation".)

I don't understand the claimed solution. It is true that the equation requires a=0 or b=0, but what does "(a and b [together] > 0)" mean? If they are both positive, then the equation is not true. Or perhaps you mean the solution is that either a=0 and b≥0, or b=0 and a≥0? That makes some sense. But what you quoted doesn't say that.

What do you mean by "the solution that the expression is the same in the complex number"? This makes no sense grammatically. I suppose you have to mean, "Why is the solution of the equation the same over the complex numbers?".

But, more important, if a and b are taken to be complex numbers (that is, not necessarily real), then it is really inappropriate to write the equation at all, as there is no principal square root to be represented by the radical! When working with complex numbers, the only proper meaning for the radical is both roots at once, which is not proper in an equation. For example, sqrt(i) = ±(1 + i)/sqrt(2); neither root can be taken as "the" root, as neither is "positive", and there is no other way to chose a principal root. But we don't write non-functions like this in equations.

So my answer is that the equation can't even be written in the context of complex numbers. Moreover, the solution can't say a>0, because comparisons don't apply to complex numbers.

But if I grant that the equation makes some sort of sense over the complex numbers, then the solution is not really the same as over the reals, because the solution is "either a=0 and b is any complex number, or b=0 and a is any complex number". Over the reals, the solution is "either a=0 and b is any positive real number, or b=0 and a is any positive real number". These are not the same solution.

Who told you this? Do you really trust them? Much of what you are asking about is nonsense.

Have you solved the equation yourself, or seen a proof? Don't just assume it, do it -- and then try applying the same reasoning to complex numbers, to see the difference.

 

[shahar]

The equation is sqrt(a^2 + b^2) = a + b. (It is not an "expression", but an "equation".)

I don't understand the claimed solution. It is true that the equation requires a=0 or b=0, but what does "(a and b [together] > 0)" mean? If they are both positive, then the equation is not true. Or perhaps you mean the solution is that either a=0 and b≥0, or b=0 and a≥0? That makes some sense. But what you quoted doesn't say that.

What do you mean by "the solution that the expression is the same in the complex number"? This makes no sense grammatically. I suppose you have to mean, "Why is the solution of the equation the same over the complex numbers?".

But, more important, if a and b are taken to be complex numbers (that is, not necessarily real), then it is really inappropriate to write the equation at all, as there is no principal square root to be represented by the radical! When working with complex numbers, the only proper meaning for the radical is both roots at once, which is not proper in an equation. For example, sqrt(i) = ±(1 + i)/sqrt(2); neither root can be taken as "the" root, as neither is "positive", and there is no other way to chose a principal root. But we don't write non-functions like this in equations.

So my answer is that the equation can't even be written in the context of complex numbers. Moreover, the solution can't say a>0, because comparisons don't apply to complex numbers.

But if I grant that the equation makes some sort of sense over the complex numbers, then the solution is not really the same as over the reals, because the solution is "either a=0 and b is any complex number, or b=0 and a is any complex number". Over the reals, the solution is "either a=0 and b is any positive real number, or b=0 and a is any positive real number". These are not the same solution.

Who told you this? Do you really trust them? Much of what you are asking about is nonsense.

Have you solved the equation yourself, or seen a proof? Don't just assume it, do it -- and then try applying the same reasoning to complex numbers, to see the difference.

Thanks. You give me what I want.