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(210)
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Formally, what this means is that for any number > 0 we can nd a > 0 such that f ( z ) a < whenever 0 < z a < For the limit to exist, it must be independent of the direction in which we approach z = a Note that a limit only exists if the limit is independent of the way that we approach the point in question, a point which is illustrated in Example 214 Limits in the theory of complex variables satisfy the same properties that limits do in the real case Speci cally, let us de ne lim f ( z ) = A
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lim f ( z ) = B
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lim { f ( z ) + g( z )} = lim f ( z ) + lim g( z ) = A + B
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(211) (212) (213)
lim { f ( z ) g( z )} = lim f ( z ) lim g( z ) = A + B
z a z a z a
lim { f ( z ) g( z )} = lim f ( z )
z a z a
}{lim g(z)} = AB
z a
z a
f ( z ) lim f ( z ) A = z a = g ( z ) lim g ( z ) B
z a
(214)
Property in Eq (214) holds as long as B 0 Limits can be calculated in terms of real and imaginary parts Let f = u + iv, z = x + iy, and z0 = x 0 + iy0 , w = u0 + iv0 Then lim f ( z ) = w0
z z0
If and only if lim u( x , y) = u0
x x0 y 0
lim v ( x , y) = v0
x x0 y 0
(215)
OPEN DISKS
Frequently, in complex analysis we wish to consider a circular region in the complex plane We call such a region a disk Suppose that the radius of the disk is a If the points on the edge of the disk, that is, the points lying on the circular curve de ning the border of the disk are not included in the region of consideration, we say that the disk is open Consider a disk of radius one centered at the origin We indicate this by writing z <1 This is illustrated in Fig 26 If the disk of radius r is instead centered at a point a, then we would write z a <r
Functions, Limits, and Continuity
Figure 26 The disk z < 1 is centered at the origin The boundary is indicated with a dashed line, which is sometimes done to indicate it is not included in the region of de nition
For example: z 3 <5 describes a disk of radius ve centered at the point z = 3 This is shown in Fig 27 EXAMPLE 211 Compute lim(iz 1) / 2 in the open disk z < 3 z 3 SOLUTION Notice that the point z = 3 is on the boundary of the domain of the function Just plugging in we nd lim
z 3
iz 1 1 3 = +i 2 2 2
Disk centered at point z = 3 on x axis 2 8 x
Figure 27 The disk described by z 3 < 5
Complex Variables Demysti ed
Let s con rm this by applying the formal de nition of the limit Notice that f (z) = iz 1 iz 1 1 3 = + i 2 2 2 2 i = ( z 3) 2 z 3 = 2
So we ve found that given any > 0 1 3 f (z) + i < 2 2 whenever 0 < z 3 < 2 EXAMPLE 212 Compute lim( z 2 )( z + i )
z i
SOLUTION For illustration purposes, we compute the limits of z 2 and z + i independently and then apply Eq (213) First we have lim z 2 = i 2 = 1
z i
Secondly lim z + i = i + i = 2i
z i
Hence lim( z 2 )( z + i ) = ( 1)(2i ) = 2i
z i
EXAMPLE 213 Using the theorems on limits from Eqs (211) (214) evaluate lim f ( z ) when z 2 i f ( z ) = z 2 + 2 z + 5
SOLUTION We have
Functions, Limits, and Continuity
lim z 2 + 2 z + 5 = lim z 2 + lim 2 z + lim 5
z 2 i
= lim z lim z + 2 lim z + lim 5
z 2 i z 2 i z 2 i z 2 i
= ( 2i ) ( 2i ) + 2 ( 2i ) + 5 i = 4 + 4i + 5 = 1 + 4i EXAMPLE 214 Show that the limit lim z /z does not exist
z 0
z 2 i
z 2 i
z 2 i
SOLUTION A limit only exists if the limit is independent of the way that we approach the point in question For this limit, let s calculate it in two different ways The rst way we ll calculate it is by approaching the origin along the x axis This means that we set y = 0, so z x iy x = =1 z x + iy y=0 x Hence lim
z 0
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