My question is

Find and sketch the image of the line $y=\sinh^{−1}(2)$ under the mapping $w=\sin(z)$.

Write down the image curve as one equation in $u$ and $v$ where $w=u+iv$ for $u,v\in\mathbb{R}$.

I'm not sure how to go about this when there's trigonometry involved. I know I need $\sin(x+iy)=\sin(x)\cosh(y)+i\cos(x)\sinh(y)$, but what happens after that?

  • $\begingroup$ Well, for starters, don't you know the value of $\sinh(y)$ for all the points on the given line? $\endgroup$ – zipirovich Sep 26 '18 at 4:01
  • $\begingroup$ 2? is there a rule that will help me narrow down the shape it should be? should i test points? $\endgroup$ – gigglegirl6 Sep 26 '18 at 8:16

First of all, from $y=\sinh^{-1}(2)$, we know that $\sinh(y)=2$ for all points on this line. Then from the identity $\cosh^2(y)-\sinh^2(y)=1$, we can find that $\cosh(y)=\sqrt{5}$ (because $\cosh$ of a real number is always positive). Now we can put these values into what you already know: $$\sin(x+iy)=\sin(x)\cosh(y)+i\cos(x)\sinh(y)=\sqrt{5}\sin(x)+2i\cos(x)=u+iv,$$ where $$u=\sqrt{5}\sin(x), \quad v=2\cos(x), \quad x\in\mathbb{R}.$$ These are the parametric equations of the ellipse $\displaystyle \frac{u^2}{5}+\frac{v^2}{4}=1$.

  • $\begingroup$ Oh i forgot about the hyperbolic identity cosh2(y)−sinh2(y)=1. thank you :) $\endgroup$ – gigglegirl6 Sep 28 '18 at 10:40

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