Math 450 Worked Problems in Limits and Continuity of Complex Functions Pay careful attention to the solution to these th ese problems.
Problem 1:
Find the following limits. lim ( z 2 − 4 z + 2 + 5i )
z →2 +i
substitution itution of z = 2 + i in the quadratic z 2 − 4 z + 2 + 5i . Solution. This requires subst 2
lim ( z 2 − 4 z + 2 + 5i ) = ( 2 + i ) − 4 ( 2 + i ) + 2 + 5i
z →2 + i
= ( 3 + 4i ) − 8 − 4i + 2 + 5i = 3 − 8 + 2 + 4i − 4i + 5i = − 3 + 5i
Problem 2: Let f ( z ) = 3 z − 1 . We prove that lim f ( z ) = 2 + 3i . z →1+ i
for δ > 0 such that f ( z ) − (2 + 3i) < ε . Solution: Let ε > 0 be given. We look for
We take any
such that 0 < δ < ε / 3 .
Problem 3:
Prove the Theorem lim f ( z ) = ∞ if and only if lim z → z0
z → z0
1 f ( z )
=0
Solution:
We start the proof by noting that the reciprocal of f ( z ) is
1 f ( z )
and the reciprocal of
infinity or 1/0 is 0/1 or 0. Thus we use the reciprocal of epsilon to tie it all together. Hence, we say f ( z ) − ∞ >
1
ε
whenever 0 < z − z0 < δ . This says that the point w = f ( z )
neighborhood w >
1
ε
lies in the ε
of ∞ whenever z lies in the deleted neighborhood 0 < z − z0 < δ of
z0 . Now. since we have chosen the epsilon well, we now try to change the left side into
the right by taking the reciprocal. The reciprocal of f ( z ) − ∞ >
1
ε
is
1 f ( z )
− 0 < ε , which is in the form we desire because
this means that lim f ( z ) = ∞ if and only if lim
z → z0
z → z0
1 f ( z )
=0.
Problem 4:
xy 2 x2 y +i 6 2 Let f ( z ) = x 2 + 2 y 6 x +y 0
when z ≠ 0 when z = 0
Show that f(z) is not continuous at the point z = 0.
Solution : Part 1: Approach from the a line y = mx Solution. Consider the line xy
3
x + 2 y 2
6
=
x +y 6
xy
and get:
2
3
3
x + 2 y 2
x (mx )
m x
3
2
x (mx ) 6 2 x + ( mx ) 3
4
2 6 6 x + 2m x
m x
+i
6
4
+i
+i
1 + 2m 6 x 4
x (mx )
+i
2 6 x + 2(mx )
3
=
6
3
2
3
=
x +y
3
x + 2 ( mx ) xm x
x y
+i
6
3
=
6 2 2 x +m x 4
x m 6 2 2 x +m x 2
x m x +m 4
2
We can now calculate the limit for points
lim f ( z ) = z →0
into
3
x y
+i
f ( x, mx ) =
through the origin. Substitute
lim
( x , y )→ (0,0)
, where L is any line y = mx.
f ( x , y)
= lim f ( x, mx ) x →0
3
= lim x →0
=
m x
4
1 + 2m 6 x 4
3 2 m 0
1 + 2m 6 04 = 0 + 0i
+i
+ i lim x →0
m0
2
x m x +m 4
2
2
04 + m 2
=0
Part 2: Approach from the curve x = y3 (We could use any line or curve of approach)
xy 2
Substitute x = y into 3
f ( y , y ) =
x + 2 y 2
3
=
= =
3
2
y 6
1 3
6
y
+i
x +y 6
+i
+ 2 y6
y + 2 y
and get:
3
3
6
x6 + y 2
x y
+i
6
y y
( y )
+i
x 2 + 2 y 6
3
xy
3
x2 y
+i
2
( y
3
(y ) 3
y
6
y
+ y2
10
y +y 18
)
3
2
8
y + 1 16
We can now calculate the limit for points z in the plane as they approach 0 along the curve x = y3. lim f ( z ) = z →0
lim
( x , y )→( 0,0)
f ( x, y )
= lim f ( y 3 , y ) y →0
1
y
8
= lim + i lim 16 y →0 3 y →0 y +1 = = =
1 3 1 3 1
+i
08 016 + 1
+ 0i
3
Since this value is not the same as the limit along lines through the origin that we obtained in Part 1, we conclude that f ( z ) is not continuous at the point z = 0 .
1
Problem 5:
Let f1 ( z ) = z e
i
Arg ( z )
3
3
θ θ = r 3 cos + i sin , 3 3 1
where z = reiθ , r = z ≠ 0, θ = Arg ( z ), and r > 0, − π < θ ≤ π
Here f1 ( z ) denotes the principal cube root function. 1 3
Show that f1 ( z ) is a branch of the multivalued cube root f ( z ) = z . Solution : 1 3
f1 ( z ) = z e
i
Arg ( z )
3
, so
1 i Arg3( z ) ( f1 ( z ) ) = z 3 e
3
3
3
Arg ( z )
1 i3 = z 3 e
3
= z ei Arg( z ) = z This shows that f1 ( z ) is indeed a branch of the cube root function.
1
Problem 6: Let f1 ( z ) = z 3 e
i
Arg ( z )
3
θ θ = r 3 cos + i sin , 3 3 1
where z = reiθ , r = z ≠ 0, θ = Arg ( z ), and r > 0, − π < θ ≤ π Here f1 ( z ) denotes the principal cube root function.
What is the range of
?
Solution:
Use polar coordinates z = reiθ in the z-plane and w = ρ eiφ in the w-plane. 1
Then w = f1 ( z ) = z 3 e
i
Arg ( z )
3
θ θ = r 3 cos + i sin 3 3 1
13 θ maps the point ( r , θ ) in the xy-plane onto the point ( ρ ,φ ) = r , in the uv-plane, 3 1 3
and we get the equations ρ = r and φ =
θ . 3
Using the equations r = ρ 3 and θ = 3φ , we find that the image of
is π π which can be written as ρ eiφ : ρ > 0 and − < φ ≤ . 3 3 1
Problem 7: Let f1 ( z ) = z e
i
Arg ( z )
3
3
θ θ = r 3 cos + i sin , 3 3 1
where z = reiθ , r = z ≠ 0, θ = Arg ( z ), and r > 0, − π < θ ≤ π
Here f1 ( z ) denotes the principal cube root function.
Where is
continuous ?
Solution: 1
f1 ( z ) = z 3 e
i
Arg ( z )
3
θ θ = r 3 cos + i sin , 3 3 1
where z = reiθ , r = z ≠ 0, θ = Arg ( z ), and r > 0, − π < θ ≤ π is continuous for all z except z = 0 and along the negative x-axis where r > 0 and θ = π .
is continuous for all z except
and along the negative x-axis where
.
