Note: In order to save time, the extreme pedantism and wordiness that is seen in the text has here been avoided.
For instance, substitution to evaluate a limit may take one line rather than half a page.
1.) Use Theorem 2.6.1 and the properties of real limits on page 104 to evaluate
z→2ilim(z2−z) Let us simply substitute in z=2i as follows:
(2i)2−2i=4i2+2i= −4+2i
3.) Use Theorem 2.6.1 and the properties of real limits on page 104 to evaluate
z→1−ilim (|z|2−i¯z)
Again, let us try substitution
|1−i|2−i(1−i) =2−i+i2=2+1−i= 3−i
5.) Use Theorem 2.6.1 and the properties of real limits on page 104 to evaluate
z→πilim(ez) Again, we need only substitute.
eiπ = −1
9.) Use Theorem 2.6.2 and the basic limits (15) and (16) to compute
z→2−ilim (z2−z) We try substitution as follows
(2−i)2−2+i=4−4i+i2+i−2=4−2−1−3i= 1−3i
11.) Use Theorem 2.6.2 and the basic limits (15) and (16) to compute lim
z→eiπ/4
(z+1 z)
Yet again, we need only substitution for the following limit. In particular, lim
z→eiπ/4
(z+1
z) =eiπ/4+e−iπ/4=cos(π/4) +i sin(π/4) +cos(−π/4) +i sin(−π/4) = √ 2
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13.) Use Theorem 2.6.2 and the basic limits (15) and (16) to compute
z→−ilim z4−1
z+i
For the first time thus far, substitution will not suffice, for, obviously, it yields an indeterminate form (note that L’Hopital’s rule applies only to the reals). However, if we simply recognize that z4−1 = (z2−1)(z2+1) = (z−1)(z+1)(z+i)(z−i), then we have that
a.) What value foes the limit approach as z approaches 0 along the real axis?
b.) What value does the limit approach as z approaches along the imaginary axis?
c.) Do the answers from (a) and (b) imply that the limit exists? Explain.
d.) What value does the limit approach as z approaches along the line y=x?
e.) What can you say about the limit in general?
a.) Along the x axis we have z=x+0y=x, thus the limit is
b.) Along the imaginary axis we have x=0, implying that z=iy, thus our limit becomes
limy→0
c.) No. The limit must be the same along any of the infinitely (uncountably) many paths in the complex plane for it to exist in general. In other words, two, three, or even a trillion paths, while perhaps suggesting that the limit may exist and equal some value c, do not actually demonstrate that the limit is c. This must be shown using more general methods. However, if the limit along one path does not equal the limit along another, then we may say that the limit does not exist.
d.) If we approach along y=x then z=x+ix and
e.) As discussed in part (c), the limit does not exist because the limit is dependent on path.
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21.) Use (24) or (25), Theorem 2.2, and the basic limits (15) and (16) to compute
z→∞lim
z2+iz−2 (1+2i)z2
Let us attempt to simplify our expression, as its current form yields an indeterminate form and there is no natural idea of the top ”approaching infinity faster” than the bottom as there is inR (this is more formally shown by dividing by the largest power of the variable in question), at least not when we have an imaginary number in question. So, upon simplification we have
(1−2i)z2+ (2+i)z− (2−4i) 5z2
Now that all factors are properly separated with the denominator a real expression of a complex vari-able, we may use our familiar laws of rational functions as apply in R. In particular, it is found that
z→∞lim f(z) = 1 5 −2
5i
23.) Use (24) or (25), Theorem 2.2, and the basic limits (15) and (16) to compute
limz→i
z2−1 z2+1
By substitution we have that, noting the nature of complex infinity (e.g. its lack of sign) z2−1
z2+1 →−2
0 → ∞
27.) Show that f is continuous at z0
f(z) =z2−iz+3−2i ; z0=2−i We have that
f(2−i) =5−8i and similarly
z→2−ilim (z2−iz+3−2i) = (2−i)2−i(2−i) +3−2i=4−4i−1−2i−1+3−2i=5−8i Therefore, by definition f is continuous at z0.
28.) Show that f is continuous at z0
f(z) =z3−1
29.) Show that f is continuous at z0
f(z) = z
3
z3+3z2+z; z0=i
Substitution here works for computing the limit, and substitution is equivalent in computation to eval-uating f(z0), thus the two are equal. Here is the explicit calculation:
z→zlim0 f(z) = i
31.) Show that f is continuous at z0
f(z) =
( z3−1
z−1 :|z| 6=1 3 :|z| =1 where z0=1
Clearly, f(1) =3 and so we simply try the limit as follows (note the expression we chose to evaluate – this was done because it is equivalent to the value obtained along the other possible path)
z→1lim Therefore, it has been demonstrated that f is continuous at z0=1.
35.) Show that f is discontinuous at z0
f(z) = z
2+1
z+i ; z0= −i Evaluating f(−i)yields
(−i)2+1
−i+1 = ∅
Thus, the criteria for continuity are not met and f is discontinuous at z0.
37.) Show that f is discontinuous at z0
f(z) =Arg(z); z0= −1
The problem here is not that f does not exist but that in fact the limit does not exist, thus not meeting the criteria for continuity, and so making f discontinuous at z0.
Let z be a point on the negative real axis, then Arg(z) =πbut we have that there are points{zn}such that they are arbitrarily close to z but have their image under f has some nonzero imaginary part and so the argument becomes arbitrarily close to−π. Thus, we have that the limit does not exist. (An e−δ argument would make this more precise — Also, note that we must be careful with substitution in the case of poorly behaved functions like this)
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39.) Show that f is discontinuous at z0
f(z) =
( z3−1
z−1 :|z| 6=1 3 :|z| =1 where z0=i
If we consider the approach along the imaginary axis then we have
lim z→i f(z) = i
3−i i−1 =i
However, trivially we know that along the unit circle we have that the limit evaluates to 3. But, 36= i, thus f is discontinuous at the point in question.
41.) Use Theorem 2.3 to determine the largest region in the complex plane on which the function f is continuous. z0
f(z) =Re(z)Im(z) Let z=x+iy, then
f(z) =Re(x+iy)Im(x+iy) =xy
But, we know that the function f(z) = xy is continuous for arbitrary z, thus f is continuous on all of C.
43.) Use Theorem 2.3 to determine the largest region in the complex plane on which the function f is continuous.
f(z) = z−1 z¯z−4 Let z=x+iy, then
f(z) = x+iy−1
(x+iy)(x−iy) −4 = x−1
x2+y2−4+i y x2+y2−4 Therefore,
u(x, y) = x−1
x2+y2−4 ; v(x, y) = y x2+y2−4
These expressions are continuous on their domains. In particular, they are continuous for all(x, y) : x2+y26=4, thus f is continuous for all z :|z| 6=2.
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