$$w=f(z)=z^2$$ $$S=\{z\mid Re(z)=a\}$$

I don't get what this is asking and what that second part means. What is the concept here? If I let $z=x+iy$, then I can square it. The real term is then $x^2-y^2$. The second part of the problem says that the real part of $z$ is $a$. I can totally replace the $x^2-y^2$ with $a$ but that doesn't seem like it means anything. Plus the question seems to be asking what would be happening on a graph, so I don't see how that relates.


If you replace the $x^2-y^2$ by $a$, that would be saying that the real part of $z^2$ is $a$. But this is not what the question says, it says the real part of $z$ is $a$.

If $z\in S$ then $z=a+it$, $t\in \Bbb R$, and so $$z^2=(a^2-t^2)+2ati\ .$$ If we write $z^2=u+iv$ then $$u=a^2-t^2\ ,\quad v=2at\ ,$$ and you should recognise this as a parametric representation of a parabola. (Alternative: eliminate $t$ to get $4a^2u=4a^4-v^2$.)

  • $\begingroup$ The $t$ was arbitrarily chosen correct? The standard form uses $a+bi$ $\endgroup$ – whatwhatwhat Jan 26 '16 at 23:25
  • $\begingroup$ It's just a letter..... $\endgroup$ – David Jan 26 '16 at 23:26
  • 2
    $\begingroup$ I used $t$ because people often use that for parametric equations of curves. You could leave it as $y$, or any letter you like. $\endgroup$ – David Jan 26 '16 at 23:27
  • $\begingroup$ I don't get how I'm supposed to recognize this as a parabola. Could you help me out with that bit? $\endgroup$ – whatwhatwhat Jan 26 '16 at 23:43
  • 1
    $\begingroup$ Ahhh I see it now. Thanks for the assist! $\endgroup$ – whatwhatwhat Jan 26 '16 at 23:56

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