0
$\begingroup$

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

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.

$\endgroup$
2
$\begingroup$

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 $4a^2u=4a^4-v^2$.)

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.