Does this equation have a complex number solution? Does this equation have any solutions:
$$\sqrt{z^2+z-7}=\sqrt{z-3}?$$
I know it does not have any real number solutions, but how about complex number solutions?
I understand that when you solve this problem algebraically, you get $z=\pm 2$ as solutions. But when you input $2$ into the original equation you get $i=i$, indicating that the real number $2$ is not a solution. But since $i=i$ is a true statement, this indicates that the complex number $2$ is a solution. My question is how are $2$ (the real number, which is not a solution) and $2$ (the complex number in the complex plane, which is a solution) different?
 A: We can immediately square both sides.
$$z^2+z-7=z-3$$
$$z^2=4$$
$$z=\pm2$$
When the square roots are taken to be principal roots, the equation is satisfied at these $z$. It is just that the expression under the root is negative.
A: You must consider what is meant by $\sqrt\;$. There are several possible square root functions here: one has domain $\mathbb R^+$ and codomain $\mathbb R$, one has domain $\mathbb R$ and codomain $\mathbb C$, and one has domain $\mathbb C$ and codomain $\mathbb C$.
If you use the first function, then it is undefined when $z=\pm2$, so there is no solution.
If you use the second function, it is defined everywhere and $z=\pm2\in\mathbb R$ are solutions.
If you use the third function, it is defined everywhere and $z=\pm2\in\mathbb C$ are solutions.
A: Upon squaring  both sides of  $$ \sqrt{z^2+z-7}=\sqrt{z-3}$$
and solving for $z$, we get $z=\pm 2$
For z=2, we have $$ \sqrt{-1}=\sqrt{-1}$$ and for $z=-2$ we get  $$ \sqrt{-5}=\sqrt{-5}$$ 
So in complex plane we have both $z=\pm 2$ acceptable. 
Note real numbers are also complex numbers with imaginary part equal to zero, so working in complex plane we have two solutions. 
