0, under the reciprocal map defined on the extended
complex plane is the circle

w

1
2
k

=

1
2
k

The vertical line
x
=
k
consists of all points
z
=
k
+
iy
such that
x
,
y
∈
R
where
k
6
=
0. So we may write
w
=
1
k
+
iy
=
k
k
2
+
y
2

y
k
2
+
y
2
i
Thus our real and imaginary parts of
w
=
f
(
u
,
v
)
are
u
=
k
k
2
+
y
2
,
v
=

y
k
2
+
y
2
,
y
∈
R
But observe that
v
=

yu
k
, which implies
y
=

vk
u
(1)
And so, upon substituting into our initial expression for
u
, we have
u
=
k
k
2
+ (

vk
u
)
2
It is here advantageous to us to simplify and complete the square as follows
u
=
k
k
2
+ (

vk
u
)
2
k
=
u
k
2
+ (

vk
u
)
2
k
=
k
2
u
+
v
2
k
2
u
uk
=
k
2
u
2
+
v
2
k
2
0
=
u
2
+
v
2

u
k

1
2
k
2
=
u
2

u
k
+

1
2
k
2
+
v
2
1
4
k
2
=
u

1
2
k
2
+
v
2
Finally, by recognition, we have

w

1
2
k

=

1
2
k

Note that we never made the restriction in (1) that
u
6
=
0. This is because here we are working in the
extended complex plane
.
3
2.6
Limits and Continuity
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
lim
z
→
2
i
(
z
2

z
)
Let us simply substitute in
z
=
2
i
as follows:
(
2
i
)
2

2
i
=
4
i
2
+
2
i
=

4
+
2
i
3.) Use Theorem 2.6.1 and the properties of real limits on page 104 to evaluate
lim
z
→
1

i
(

z

2

i
¯
z
)
Again, let us try substitution

1

i

2

i
(
1

i
) =
2

i
+
i
2
=
2
+
1

i
=
3

i
5.) Use Theorem 2.6.1 and the properties of real limits on page 104 to evaluate
lim
z
→
π
i
(
e
z
)
Again, we need only substitute.
e
i
π
=

1
9.) Use Theorem 2.6.2 and the basic limits (15) and (16) to compute
lim
z
→
2

i
(
z
2

z
)
We try substitution as follows
(
2

i
)
2

2
+
i
=
4

4
i
+
i
2
+
i

2
=
4

2

1

3
i
=
1

3
i
11.) Use Theorem 2.6.2 and the basic limits (15) and (16) to compute
lim
z
→
e
i
π
/4
(
z
+
1
z
)
Yet again, we need only substitution for the following limit. In particular,
lim
z
→
e
i
π
/4
(
z
+
1
z
) =
e
i
π
/4
+
e

i
π
/4
=
cos
(
π
/4
) +
i
sin
(
π
/4
) +
cos
(

π
/4
) +
i
sin
(

π
/4
) =
√
2
4
13.) Use Theorem 2.6.2 and the basic limits (15) and (16) to compute
lim
z
→
i
z
4

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
z
4

1
=
(
z
2

1
)(
z
2
+
1
) = (
z

1
)(
z
+
1
)(
z
+
i
)(
z

i
)
, then we have that
lim
z
→
i
z
4

1
z
+
i
=
lim
z
→
i
(
z
+
1
)(
z

1
)(
z

i
) = (

i
+
1
)(

i

1
)(

2
i
) =
4
i
19.) Consider the limit
lim
z
→
0
z
z
2
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
+
0
y
=
x
, thus the limit is
lim
x
→
0
x
x
2
=
lim
x
→
0
1
=
1
b.) Along the imaginary axis we have
x
=
0, implying that
z
=
iy
, thus our limit becomes
lim
y
→
0
yi

yi
2
=
lim
y
→
0
(

1
)
2
=
1
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.