# November 1, 2010 Lecturer Dmitri Zaitsev Michaelmas Term 2010

```November 1, 2010
Lecturer Dmitri Zaitsev
Michaelmas Term 2010
Course 2325 2010 Complex Analysis I
Sheet 1
Due: at the end of the lecture
Exercise 1
Find logz, Logz and
√
z for
(i) z = 2i;
(ii) z = 1 − i;
(iii) z = 2/(1 +
√
3i).
Solution
Use formulas logz = log|z| + i arg z and Logz = log|z| + iArgz with arg z being the
set of all arguments and −π &lt; Argz ≤ π the principal value.
(i) logz = log|z| + i arg(z) = log2 + iπ/2 + {2πk : k ∈ ZZ},
Logz = log2 + iπ/2,
√
√
z = &plusmn; 2eiπ/4 .
√
(ii) logz = log 2 − iπ/4 + {2πk : k ∈ ZZ},
√
Logz = log 2 − iπ/4,
√
z = &plusmn;21/4 e−iπ/8 .
(iii) z = 2/(2eiπ/3 ) = e−iπ/3
logz = −iπ/3 + {2πk : k ∈ ZZ},
Logz = −iπ/3’
√
z = &plusmn;e−iπ/6 .
Exercise 2
Prove that Im(iz) = Rez, Re(iz) = −Imz, ez = ez , e−z =
1
ez .
Solution
Let z = a + ib, then iz = −b + ia, so clearly Im(iz) = a = Rez and Re(iz) = −b =
−Imz.
z = a − ib. So, ez = ea e−ib = ea (cos(−b) + isin(−b)) = ea (cosb − isinb), whereas
ez = ea (cosb + isinb), so ez = ez .
1
ez
=
1
ea (cosb+isinb)
=
1
ea (cosb+isinb)
cosb−isinb = e−a cosb−isinb = e−a e−ib = e−z
cosb−isinb
cos2 b+sin2 b
Exercise 3
(i) Show that log(z1 z2 ) = logz1 + logz2 as sets.
(ii) Show that Log(z1 z2 ) = Logz1 + Logz2 provided −π &lt; Argz1 + Argz2 &lt; π
(iii) Give an example of z1 , z2 with Log(z1 z2 ) 6= Logz1 + Logz2 .
Solution
(i) Let z1 = r1 eiθ1 and z2 = r2 eiθ2 , then z1 z2 = r1 r2 ei(θ1 +θ2 ) and hence log(z1 z2 ) =
logr1 r2 + i(θ1 + θ2 ) + i{2πk : k ∈ ZZ} = logr1 + iθ1 + logr2 + iθ2 + i{2πk : k ∈ ZZ} =
logz1 + logz2
(ii) We know from (i) that log(z1 z2 ) = logz1 + logz2 as sets, hence arg(z1 z2 ) = {θ1 +
θ2 + 2πk : k ∈ ZZ}, but −π &lt; Arg(z1 z2 ) &lt; π, so Arg(z1 z2 ) = Argz1 + Argz2 only
if −π &lt; Argz1 + Argz2 &lt; π. So, Log(z1 z2 ) = Logz1 + Logz2 only if −π &lt; Argz1 +
Argz2 &lt; π.
(iii) Let z1 = z2 = ei2π/3 . Then θ1 + θ2 = 4π/3 &gt; π. Logz1 + Logz2 = i4π/3, but
Log(z1 z2 ) = −i2π/3.
Exercise 4
Using the definition show that:
(i) Finite intersections and arbitrary unions of open sets are open.
(ii) Finite unions and arbitrary intersections of closed sets are closed.
Solution
(i) By the definition of an open set, for each point z in an open set U there exists an
ε &gt; 0 such that the open ball Bε (z) = {ξ : |ξ − z| &lt; ε} is contained in the set U .
S
S
If now α Uα is an arbitrary union, then z ∈ α Uα , then z ∈ Ui for some i, so
S
there is an ε &gt; 0 such that Bε (z) ⊂ Ui , and hence Bε (z) ⊂ α Uα .
If on the other hand, U1 ∩ &middot; &middot; &middot; ∩ Un is a finite intersection and z ∈ U1 ∩ &middot; &middot; &middot; ∩ Un
for Uj open, then there are εj such that Bεj (z) ⊂ Uj for every j = 1, . . . , n. Choose
ε = min{εj : 1 ≤ j ≤ n}, then Bε (z) ⊂ U1 ∩ &middot; &middot; &middot; ∩ Un , so finite intersections of open sets
are open.
(ii) Since we defined a closed set to be the complement of an open set, it suffices to note
S
T
that C\( α Uα ) = α (C\Uα ) and that C\(U1 ∩ . . . ∩ Un ) = (C\U1 ) ∪ . . . ∪ (C\Un )
and the problem is reduced to (i).
Exercise 5
Construct a branch of logz on the set C\{−iy : y ≥ 0}. Show that the branch you
constructed is indeed continuous.
Solution Since we know what logz is for polar coordinates, we should redefine the
set on which we want to define the branch:
C\{iy : y ≥ 0} = {reiθ : r &gt; 0, −π/2 &lt; θ &lt; 3π/2}
Now, we define our branch of logz to be:
logz =
r + i(Argz + 2π) if −π &lt; Argz &lt; −π/2;
r + iArgz
if −π/2 &lt; Argz ≤ π.
Now we need to show that this branch is continuous. The only points where continuity is an issue is at the line {reiθ : θ = &plusmn;π}. At the line the branch takes the value
r + iπ, and taking the limit as θ → −π the branch clearly tends towards r + iπ too.
Exercise 6
Sketch the set of points given by the condition:
(i) 1 &lt; |z| &lt; 2,
(ii) 1 &lt; |z + 2i| &lt; 2,
(iii) Re((1 + i)z) ≥ 4.
Solution Parts (i) &amp; (ii) are both annuli of inner radius 1 and outer radius 2,
centered at 0 and −2i respectively. If z = a + ib then Re((1 + i)z) = a + b, so part (iii)
is to the right of the diagonal line a + b = 4.
Exercise 7
Find all z for which the following identity holds:
∞
X zn
1
1
=
+
.
4z − z 2
4z n=0 4n+2
Solution
1
4z−z 2
=
1 1
z 4−z .
1
4−z
1
4z−z 2
Now,
1, or in other words |z| &lt; 4. Since
=
1
1
4 1−z/4
= 41 (
P∞
z n
n=0 ( 4 ) ),
whenever |z/4| &lt;
is not defined for z = 0 and the sum diverges
for |z| ≥ 4, we get that the identity holds for 0 &lt; |z| &lt; 4.
```