(1) Prove the following:
√
(a) Prove that |z| ≤ |Re (z)| + |Im(z)| ≤ 2 |z|.
Answer:
Observe that, for any two real numbers x and y, we have
2
(|x| − |y|) ≥ 0 =⇒ |x|2 + |y|2 ≥ 2|x| |y|.
Let z = x + iy be any complex number.
Now,
(|x| + |y|)2 = |x|2 + |y|2 + 2|x| |y| ≤ |x|2 + |y|2 + |x|2 + |y|2
= 2 |x|2 + |y|2 = 2 x2 + y 2 = 2|z|2 .
√
This gives that |Re (z)| + |Im(z)| = |x| + |y| ≤ 2 |z|.
|z| =
p
p
p
x2 + y 2 ≤ x2 + y 2 + 2|x| |y| = |x|2 + |y|2 + 2|x| |y|
q
= (|x| + |y|)2 = |x| + |y| .
That is, |z| ≤ |Re (z)| + |Im(z)|.
(b) |z1 + z2 | ≤ |z1 | + |z2 | and equality holds if and only if one is a nonnegative (real) scalar
multiple of the other.
Answer: Do yourself.
z 1 − z2 = 1. What exception
(c) If either |z1 | = 1 or |z2 | = 1, but not both, then prove that 1 − z1 z2 must be made for the validity of the above equality when |z1 | = |z2 | = 1?
Answer: Case I: |z1 | = 1 and |z2 | =
6 1
z1 − z2 2 |z1 |2 + |z2 |2 − 2Re (z1 z2 )
1 − z1 z2 = 1 + |z1 z2 |2 − 2Re (z1 z2 )
1 + |z2 |2 − 2Re (z1 z2 )
=
=1.
1 + |z2 |2 − 2Re (z1 z2 )
Observe that the denominator 1 + |z1 z2 |2 − 2Re (z1 z2 ) 6= 0 if |z1 | = 1 and |z2 | =
6 1.
Case II: |z2 | = 1 and |z1 | =
6 1. It can be worked out similarly as in the previous
case.
Case III: |z1 | = 1 and |z2 | = 1
z1 − z2 2 2 − 2Re (z1 z2 )
=
Then, = 1 if the denominator 2 − 2Re (z1 z2 ) 6= 0. That is,
1 − z1 z2 2 − 2Re (z1 z2 )
Re (z1 z2 ) 6= 1 if and only if z1 6= z2 . So, the exception is to be made for the validity of the
above equality in this case is z1 6= z2 .
(2) Show that the equation z 4 + z + 5 = 0 has no solution in the set {z ∈ C : | z |< 1}.
Answer: Suppose α is a solution. So |α| < 1 and α4 + α = −5. Then 5 = |α4 + α| ≤ 2.
(3) If z and w are in C such that Im(z) > 0 and Im(w) > 0, show that | z−w
z−w |< 1.
z−w 2
Answer: z−w̄ ≤ 1 ⇐⇒ (z − z̄)(w − w̄) < 0. Clearly (z − z̄)(w − w̄) < 0 if Im(z) > 0 and
Im(w) > 0.
(4) When does az + bz + c = 0 has exactly one solution?
Answer: Let z = x + iy, a = a1 + ia2 , b = b1 + ib2 and c = c1 + ic2 and put these values in the
given equation az + bz + c = 0. After simplification (please check it carefully!)we have,
(a1 x − a2 y) + i(a2 x + a1 y) + (b1 x + b2 y) + i(b2 x − b1 y) + c1 + ic2 = 0
After equating real and imaginary parts we get the following system of linear equations
(a1 + b1 )x + (b2 − a2 )y = c1
(a2 + b2 )x + (a1 − b1 )y = c2 .
1
Therefore given equation has exactly one solution if the above system of linear equations has
unique solution. In this case
(a1 + b1 )(a1 − b1 ) − (b2 − a2 )(a2 + b2 ) 6= 0.
In fact the given equation has exactly one solution if |a| =
6 |b|.
Pn−1 j
th
(5) If 1 = z0 , z1 , . . . , zn−1 are distinct n roots of unity, prove that Πn−1
j=0 z .
j=1 (z − zj ) =
n
n
Answer The points 1 = z0 , z1 , . . . , zn−1 are the roots of z − 1 = 0. So z − 1 = (z − 1)Πn−1
j=1 (z −
Pn−1 j
Pn−1 j
n−1
zj ) = (z − 1) j=0 z . Let f (z) = Πj=1 (z − zj ) and g(z) = j=0 z . Thus f (z) = g(z) for z 6= 1.
Since both f, g are continuous it follows that f (1) = g(1) as well.
(6) For each of the following subsets of C, determine whether it is open, closed or neither. Justify
your answers.
(a) S = {z ∈ C : Re(z) = 1 and Im(z) 6= 4}
Answer: The set S = {z = x + iy : x = 1} \ {1 + 4i} is neither open nor closed. S o = ∅ and
∂S = S 0 = S = {z = x + iy : x = 1}
(b) z ∈ C : Re(z) ∈ (−1, 2) ∪ (2, 52 )and Im(z) = 0
Answer: The S = z ∈ C : Re(z) ∈ (−1, 2) ∪ (2, 25 )and Im(z) = 0 is neither open nor
closed. S o = ∅ and ∂S = S 0 = S = {z = x + iy : x ∈ [−1, 52 ] and y = 0}.
(c) {z = x + iy ∈ C : xy > 1}
Answer: The set S = {z = x + iy ∈ C : xy > 1} is an open set, hence S = S o .
∂S = {z = x + iy ∈ C : xy = 1} and S = S 0 = {z = x + iy ∈ C : xy ≥ 1}.
(d) {z = x + iy ∈ C : x ∈ Q and y ∈ R \ Q}
Answer: The set S = {z = x + iy ∈ C : x ∈ Q and y ∈ R \ Q} is neither open nor closed.
S o = ∅ and ∂S = S 0 = S = C
(e) n1 + mi : n, m ∈ N
Answer: The set S = n1 + mi : n, m ∈ N is neither open nor closed.
S o = ∅, S 0 = { n1 + 0i} ∪ {0 + mi } ∪ {(0, 0)} and ∂S = S = S ∪ S 0 .
(f) z = reiθ ∈ C : 0 < r < 1 and θ ∈ ( π4 , π3 )
Answer: The set S = z = reiθ ∈ C : 0 < r < 1 and θ ∈ ( π4 , π3 ) is an open set, hence
S = S o . S = S 0 = z = reiθ ∈ C : 0 ≤ r ≤ 1 and θ ∈ [ π4 , π3 ] .
∂S = z = reiθ ∈ C : 0 ≤ r ≤ 1 and θ = π4 , π3 ∪ z = eiθ : θ ∈ [ π4 , π3 ] .
(g) r cos n1 + i sin n1 ∈ C : r > 0, n ∈ N ∪ {z ∈ C : Re(z) < 0}
Answer: The set r cos n1 + i sin n1 ∈ C : r > 0, n ∈ N ∪ {z ∈ C : Re(z) < 0} is neither open nor closed. S o = {z ∈ C : Re(z) < 0} .
∂S = r cos n1 + i sin n1 ∈ C : r > 0, n ∈ N ∪{ positive real axis }∪{ imaginary axis }.
S = S0 =
r cos n1 + i sin n1 ∈ C : r > 0, n ∈ N ∪ { positive real axis } ∪
{z ∈ C : Re(z) ≤ 0}