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?

  • 1
    $\begingroup$ Try not to just post questions, as people won't know where to start to be able to help you (and you will get downvotes and close votes, as has happened here). Try to explain what you have tried and where and why you got stuck/need help. It would also help if you said where you found this question (week X of course Y, where we covered topic Z). $\endgroup$
    – user1729
    Oct 17, 2018 at 11:22

3 Answers 3


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.

  • $\begingroup$ Thanks! Now this make perfect sense. $\endgroup$
    – Sam Tilly
    Oct 19, 2018 at 3:52

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.


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.

  • $\begingroup$ As complex numbers we see as $\pm 2+0i$ with zero imaginary parts . $\endgroup$ Oct 17, 2018 at 22:58
  • $\begingroup$ Graph $y=x^2+x-7$ and $y=x-3$ and find the intersection points. You should get $x=\pm 2$ $\endgroup$ Oct 18, 2018 at 2:42
  • $\begingroup$ real numbers $\pm 2$ are solutions and they are the same as $\pm 2 +0i$ $\endgroup$ Oct 18, 2018 at 3:26
  • $\begingroup$ Correct, you can not see the roots from the square root formula. Because you hit negatives under radical sign. That is why I asked you to graph $ y=x^2+x−7$ and $y=x−3$ $\endgroup$ Oct 18, 2018 at 3:39

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