Sentence examples for maps the disc from inspiring English sources

Exact(3)

maps the disc U r onto the domain ϕ ( U r ), contained in the unit disc.

If B = 1, then the function (10) maps the disc D onto the half-plane { w ∈ C : ℜ [ w ] > 1 + A 2 }.

maps the disc U r onto a domain contained in the unit disc and tangent to the unit circle at the points F ( z 1 ) and at F ( z 2 ).

Similar(57)

For local galaxies, it will be possible to map the discs and the interstellar medium through the low ionization ionic lines and a variety of molecular tracers, such as OH, H2O and high-J CO.

Let us denote by h k the univalent function, which maps the unit disc U onto the conic domain Ω k with h k ( 0 ) = 1.

This is far from a fix; the KryoFlux only creates a magnetic map of the disc, but the technology does not yet exist to render the data found within, which means that the accessibility of Clifton's words depends upon marbl's ability to keep the old VideoWriter working and the floppy disks from disintegrating further.

This is far from a fix; the KryoFlux only creates a magnetic map of the disc, but the technology does not yet exist to render the data found within, which means that the accessibility of Clifton's words depends upon marbl's ability to keep the old VideoWriter working and the floppy disks from disintegrating further.

Of course the functions from map the unit disc onto convex domains.

In considering the map F ( z ) = z − f ( z ), although it is assumed that F maps the exterior of the disc to the exterior of the disc, the proof allows the image of the interior of the disc under F to lie anywhere in R 2. □.

The function q ( z ) = 1 + z / 2 1 − z / 2, z ∈ U. maps the unit disc onto the disc D ( C, R ) with the center C = 5 / 3 and the radius R = 4 / 3. Hence, putting A = 1 / 2, B = − 1 / 2, b = 4 / 9 in Theorem 3.2, we obtain the following corollary.

Notice that for B ∈ ( − 1, 0 ] the function on the right hand side of (2.13) maps the open unit disc onto the disc D ( C, R ) = { w ∈ C : | w − C | < R }, where C = B ( A − B ) 1 − B 2, R = | A − B | 1 − B 2. This function is also univalent in whence, by (1.1), Theorem 2.3 can be written in the following form.

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