1. ## complex = real

When I know that a solution of equation is the same in the real numbers and the complex numbers so they are the same?

2. ## Re: complex = real

This makes no sense as written. You are supposed to have a 5 Bagrut in English. Use it.

3. ## Re: complex = real

Originally Posted by romsek
This makes no sense as written. You are supposed to have a 5 Bagrut in English. Use it.
In this question the complex and the real is the same because some of the solutions are canceled.
(1)So the answer if I understand correctly is when the solution is cancelled. But Now I need a guideline that tell me if I am rights.
The question:
sqrt (a^2 + b^2) = a + b

(2)There is another solutions when they are the same when there no cancelling...?

4. ## Re: complex = real

Originally Posted by policer
In this question the complex and the real is the same because some of the solutions are canceled.
(1)So the answer if I understand correctly is when the solution is cancelled. But Now I need a guideline that tell me if I am rights.
The question:
sqrt (a^2 + b^2) = a + b

(2)There is another solutions when they are the same when there no cancelling...?
What do you mean by "cancelled"? The solution provided by romsek in the original post does not involve any "cancelling". He squared both sides. The problem with squaring both sides (when both sides are real) is that you introduce the possibility that one of the sides was negative while the other positive.

On the other hand, the Complex Square Root finds the first root of unity. So, you still need either $a=0$ or $b=0$. But, you can have $a=x+iy$ with $x> 0$ or $x=0,y\ge 0$ as a valid solution as well. (When you square it, the first root of unity will be the positive real component while the second root of unity will be the negative real component).

5. ## Re: complex = real

Originally Posted by SlipEternal
What do you mean by "cancelled"? The solution provided by romsek in the original post does not involve any "cancelling". He squared both sides. The problem with squaring both sides (when both sides are real) is that you introduce the possibility that one of the sides was negative while the other positive. This is similar to the issue with complex numbers (where squaring may provide more solutions than actually exist). Squaring both sides over the complex numbers is still possible, and any result from that must still hold for the "unsquared" solution. So, squaring both sides gives a necessary condition for a solution, but it does not give a sufficient solution. So, let's look for additional solutions over the complex numbers. We know that either $a$ or $b$ must be zero (that is still a necessary condition for a solution). Now, we want to check if $b=0$, what possible complex values are available for $a$:

$$\sqrt{(x+iy)^2}=x+iy$$

This is true when $x\ge 0$.
Thanks, Thanks...!!!

6. ## Re: complex = real

Originally Posted by policer
Thanks, Thanks...!!!
I updated my answer. If $x>0$, then $y$ can be anything. But if $x=0$, then you need $y\ge 0$.