z = i(z*) 

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

Complex Numbers
Gubiv Gubiv
7
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1 Answer

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

Erdos Erdos
4.7K
  • Dynkin Dynkin
    0

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

    • Erdos Erdos
      0

      That was a silly mistake. I revised my solution.

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