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I'm currently learning for an algebra exam and I have some examples of questions from few last years. And I can't find a solution to this one:

Give three examples of complex numbers where z = -z

The only complex number I can think of is 0. Because it is a complex number, isn't it? Like 0 + 0i.

What two other complex numbers can be given as examples?

Edit: Well, I'm pretty sure it's z = -z. I have only this low-resolution picture, but you can see it in the first task: http://i.imgur.com/2EuugPZ.jpg. Yeah, I know it's all in Polish, but you have to believe me it says to find three examples.

Edit 2: Okay, now I see that it might actually say $\bar{z} = -z$.

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    $\begingroup$ $0$ is indeed the only complex number $z$ for which $z=-z$. Is it possible that you read $1/z = -z$? In that case there would be two of them. ${}\qquad{}$ $\endgroup$ – Michael Hardy Jan 18 '15 at 18:52
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    $\begingroup$ ... or maybe $z=-\bar z$? $\endgroup$ – Hagen von Eitzen Jan 18 '15 at 18:53
  • $\begingroup$ I think you are right. $0$ should be the only solution. $\endgroup$ – AvZ Jan 18 '15 at 18:57
  • $\begingroup$ Sorry, but I have only this low-resolution picture and I'm pretty sure it says z = -z, please take a look (it's in the first task): i.imgur.com/2EuugPZ.jpg And it's all in Polish, so you have to believe me it says about three examples. $\endgroup$ – Damian Jan 18 '15 at 18:58
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    $\begingroup$ Looks like it might be $\bar{z} = -z$. $\endgroup$ – Simon S Jan 18 '15 at 18:59
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If the problem is actually $\bar{z} = -z$, then writing $z = a + bi$ for real $a, b$, we have

$$a - bi = -a - bi$$

Hence $a = 0$ and $b$ can take on any value.

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  • $\begingroup$ Yes, I think that might be the solution. Thank you! $\endgroup$ – Damian Jan 18 '15 at 19:02
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If $$ z = -z $$ then we can add $z$ to both sides of the equation, getting $$ 2z = 0 $$ and then we can divide both sides by $2$, getting $$ z=0. $$ My suspicion is that you've misread the question and it said something other than $z=-z$.

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