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1 y 1 y = 2 2 2 2 2 x + y 2 ( z + iy)( z iy) Therefore, there are two simple poles located at z = iy These lie directly on the y axis, one in the upper half plane and one in the lower half plane To get the right answer for the integral we seek, we need to compute using both cases First we consider the pole in the upper half plane The residue corresponding to z = + iy is a 1 = ( z iy) ye ikz 2 2 ( z + iy)( z iy) = yeky 1 ky e = 2 2 (2iy) 4 2i

This arrangement is not new It is provided in Windows NT by what is called a roaming profile Under Windows 2000 the job is taken over by the Group Policy, as described earlier in the chapter This policy dictates how the desktop, menu items, and other application properties are presented to the user at login This promises a comfortable familiarity to the user For the administrator, it means the ability to lock down restricted parts of the OS

11000000101010000000010000111101 11000000101010000000010000111110 11000000101010000000010000111111 - Can t use all ones

Special folders can be set up to replicate to a network drive For example, while you are working with a document in the My Documents folder, it can be replicated to a secure network location When you log out and then in on another machine, the system checks with the special folder, identifies the most recent files, and retrieves your documents in that folder The Offline Folders, Synchronization, and Group Policy tools all play a role in providing this feature

Applying the residue theorem, we nd that I= 1 2 y 1 1 e ky e ikx dx = 2 i 2 eky = x 2 + y2 4 i 2

11000000101010000000010001100000 - Can t use all zeroes 11000000101010000000010001100001 11000000101010000000010001100010

Distributed File System (Dfs)

However, now let s consider enclosing the other singularity, which would be an equally valid approach The singularity is located at z = iy, which is below the x axis, so we would need to use a semicircle in the lower half plane to enclose it This time the residue is ye ikz a 1 = ( z + iy) 2 2 ( z + iy)( z iy) Using this result, we obtain I= 1 2 y 1 1 e ky e ikx dx = 2 i 2 e ky = x 2 + y2 4 i 2

11000000101010000000010001111101 11000000101010000000010001111111 11000000101010000000010001111111 - Can t use all ones

Another feature available now for NT 40 users and one that will be standard in Windows 2000 enhances the scalability of Windows networks The Distributed File System (Dfs) allows you to collect network shares and reshare them as a single share on any Windows server This enables consolidation of many disparate shares into a single namespace Users no longer have to mount each share on a different machine with a new drive letter, because all the shares they need will be under a single share on a Dfs server The Dfs server doesn't actually house the shares; they remain on the computers where they were stored before consolidation Dfs is built into Windows 2000, so there are no additional components to install Currently, Dfs can be installed as a service on NT 40 machines An NT 40 workstation can act as a client to a Dfs server, but the workstation cannot itself host a Dfs share NT 40 servers can host a single Dfs share and can also act as a Dfs client You can download the Dfs software from the following Microsoft address: http://wwwmicrosoftcom/ntserver/nts/downloads/winfeatures/NTSDistrFile/defaultasp

Note that in this case you only get 25 2 = 30 hosts per network ID! These better be small networks! Converting these to dotted decimal we get: 192168432/27 (192168433 192168462) 192168464/27 (192168465 192168494) 192168496/27 (192168497 1921684126) 1921684128/27 (1921684129 1921684158) 1921684160/27 (1921684161 1921684190) 1921684192/27 (1921684193 1921684222) These two examples started with a Class C address There s no reason not to start with any starting network ID Nothing changes from the process you just learned

Combining both results gives the correct answer, which is I= EXAMPLE 711 Compute the integral given by I= SOLUTION This integral is given by I = 2 i residues in upper half plane We nd the residues by considering the complex function z2 f ( z) = 4 z +1 The singularities are found by solving the equation z4 + 1 = 0 This equation is solved by z = ( 1)1/ 4 But, remember that 1 = ei x2 dx 1 + x 4