# Thread: Mapping - Hyperbolic Functions

1. ## Mapping - Hyperbolic Functions

For the mapping f(z) = sinh(z), find and sketch the image of im(z) = d

z = x + iy
w = u + iv

This is the solution given:
$\left(\frac{v}{sin d}\right)^2 - \left(\frac{u}{cos d}\right)^2 = 1$ << got this one
$\frac{v}{sin(d)}\geq1\ if\ d\neq k\pi/2 \ for\ k\in\mathbb{Z}$
$v=0\ if\ d=n\pi\ for\ n\in\mathbb{Z}$
$\{iv:v\geq1\}\ for\ d=(2n+1)\pi/2\ and\ n\in\mathbb{Z}\ even$
$\{iv:v\leq-1\}\ for\ d=(2n+1)\pi/2\ and\ n\in\mathbb{Z}\ odd$

2. ## Re: Mapping - Hyperbolic Functions

Originally Posted by xaz
For the mapping f(z) = sinh(z), find and sketch the image of im(z) = d

z = x + iy
w = u + iv

This is the solution given:
$\left(\frac{v}{sin d}\right)^2 - \left(\frac{u}{cos d}\right)^2 = 1$ << got this one
$\frac{v}{sin(d)}\geq1\ if\ d\neq k\pi/2 \ for\ k\in\mathbb{Z}$
$v=0\ if\ d=n\pi\ for\ n\in\mathbb{Z}$
$\{iv:v\geq1\}\ for\ d=(2n+1)\pi/2\ and\ n\in\mathbb{Z}\ even$
$\{iv:v\leq-1\}\ for\ d=(2n+1)\pi/2\ and\ n\in\mathbb{Z}\ odd$

$w=\sinh(z)=\sinh(x+\imath y)=\sinh(x)\cos(y) + \imath \cosh(x)\sin(y)$

$u=\sinh(x)\cos(y)$

$v=\cosh(x)\sin(y)$

The image of $\Im(z)=d$ is just that of $y=d$ or

$u=\sinh(x)\cos(d)$

$v=\cosh(x)\sin(d)$

you can play with sketching the image of the line $y=d$