I want to find the image of the function that defined as $\exp(x+iy)=e^x(\cos(y)+i\sin(y))$ for $z\in L$ where $L=\left\{z\in \mathbb C \mid a\mathrm{Re}(z)+b\mathrm{Im}(z)+c=0 \right\}$ and $a^2+b^2\ne 0$.

I tried to get to an expression for $z$ uing the identities $\mathrm{Re}(z)=\dfrac{z+\bar z}{2}$ and $\mathrm{Im}(z)=\dfrac{z-\bar z}{2i}$ but I got to a dead end.

How should I approach this problem?


marked as duplicate by dxiv, José Carlos Santos, Community Apr 3 '18 at 21:48

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If $b\neq0$, then $ax+by+c=0\iff y=-\frac bax-\frac ca$ and therefore\begin{align}\exp(x+yi)&=\exp\left(x-\left(\frac bax+\frac ca\right)i\right)\\&=e^x\left(\cos\left(\frac bax+\frac ca\right)-\sin\left(\frac bax+\frac ca\right)\right).\end{align}This is a spiral. Can you deal with the case $b=0$?

  • $\begingroup$ thank you for answering, you also assumed that $a\ne 0$? I forgot to mention that $a^2+b^2\ne 0$ in the question. $\endgroup$ – segevp Apr 3 '18 at 11:47
  • $\begingroup$ @segevp I assumed that $a$ and $b$ weren't both $0$. $\endgroup$ – José Carlos Santos Apr 3 '18 at 11:53
  • $\begingroup$ @JoséCarlosSantos Guess the OP also forgot the condition $\,a,b \in \mathbb{R}\,$ from the duplicate question ;-) $\endgroup$ – dxiv Apr 3 '18 at 19:57
  • $\begingroup$ @dxiv It's not just the same question. It uses the same notation! $\endgroup$ – José Carlos Santos Apr 3 '18 at 19:59
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    $\begingroup$ @segevp You may still want to edit the condition $\,a,b \in \mathbb{R}\,$ into the question, before someone points out that the answer would be quite more laborious otherwise ;-) $\endgroup$ – dxiv Apr 3 '18 at 21:31

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