z = i(z*) 

I think z = 0 is the only solution. But the answer says the equation represents a line. How is this possible?

1 Answer

Let $z=a+bi$. Then $z=iz^*$ becomes 
\[\Rightarrow a=b.\]
Hence for any number $z=a+ia$, $a\in \mathbb{R}$ we have

Erdos Erdos
  • Dynkin Dynkin

    $z^*$ is the conjugate of $z$. So $z^*=a- ib$, not $a+ib$.

    • Erdos Erdos

      That was a silly mistake. I revised my solution.

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