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

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?

• Well, for starters, don't you know the value of $\sinh(y)$ for all the points on the given line? – zipirovich Sep 26 '18 at 4:01
• 2? is there a rule that will help me narrow down the shape it should be? should i test points? – 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 $$\displaystyle \frac{u^2}{5}+\frac{v^2}{4}=1$$.