MATH 20142 Complex Analysis
Solution Sheet 1
1. (a) (3 + 4i)2 = 9 + 24i − 16 = −7 + 24i;
(b)
2 + 3i
(2 + 3i)(3 + 4i)
6
17
=
= − + i;
3 − 4i
25
25 25
(c)
1 − 5i
8 1
= − + i;
−1 + 3i
5 5
(d)
1−i
+ 2 − i = −i + 2 − i = 2 − 2i.
1+i
2. (a) Write z = a + ib; then a2 + 2abi − b2 = −5 + 12i, whence
a2 − b2 = −5,
2ab = 12.
The second equation yields b = 6/a (with a ̸= 0), whence a4 + 5a2 − 36 = 0. Solving this
bi-quadratic equation gives a = ±2, whence b = ±3. Hence there are two solutions: z = 2+ 3i
and z = −2 − 3i.
(b) We have (z +2)2 = −8+ 6i, whence it is more convenient to write z +2 = a + ib. Similarly,
to the previous solution, we have a = ±1, b = ±3, whence the solutions are z = −1 + 3i and
z = −3 − 3i.
3. (a) Writing z = x + iy, we obtain Re z = {(x, y) : x > 2}, i.e., a half-plane.
(b) Here we have the open strip {(x, y) : 1 < y < 2}.
(c) The condition |z| < 3 is equivalent to x2 + y 2 < 9, hence our set is an open disc of radius 3
centred at the origin.
(d) Write z = x + iy; we have |x − 1 + iy| < |x + 1 + iy|, which is equivalent to (x − 1)2 + y 2 <
(x + 1)2 + y 2 , i.e., after all cancellations, to x > 0. Hence we have an open half-plane.
4. We have
zw = rs(cos θ cos ϕ − sin θ sin ϕ + i(sin θ cos ϕ + cos θ sin ϕ))
= rs(cos(θ + ϕ) + i sin(θ + ϕ)),
whence arg(zw) = arg(z) + arg(w). Hence by induction, arg(z n ) = n arg(z). Put z = cos θ +
i sin θ. Since arg z = θ, we have arg z n = nθ. To complete the proof of De Moivre’s Theorem,
we observe that |z| = cos2 θ + sin2 θ = 1, whence |z n | = 1 as well. Hence z n = cos nθ + i sin nθ.
5. Apply De Moivre’s Theorem to the case n = 3:
(cos θ + i sin θ)3 = cos3 θ + 3 cos2 θ sin θi + 3 cos θi2 sin θ + i3 sin3 θ
= cos 3θ + i sin 3θ,
1
whence, in view of i2 = −1, i3 = −i, we obtain
cos 3θ = cos3 θ − 3 cos θ sin2 θ = 4 cos3 θ − 3 cos θ,
sin 3θ = 3 cos2 θ sin θ − sin3 θ = 3 sin θ − 4 sin2 θ.
Similarly,
cos 4θ = sin4 θ − 6 cos2 θ sin2 θ + cos4 θ,
sin 4θ = −4 cos θ sin3 θ + 4 cos3 θ sin θ.
6. We first show that any open disc Nδ (z0 ) is an open set. Choose z ∈ Nδ (z0 ); then we have
|z − z0 | < δ. Now, assume 0 < r < δ − |z − z0 | and consider z1 ∈ Nr (z). By the triangle
inequality,
|z1 − z0 | ≤ |z1 − z| + |z − z0 | < |z1 − z| + δ < r.
This proves z1 ∈ Nδ (z0 ), i.e., Nδ (z0 ) is open.
(a) Put X1 = {z : Im z > 0} = {(x, y) : y > 0}. Let us show that X1 is open. Suppose
z0 = (x0 , y0 ) ∈ X1 ; let δ = y0 /2 > 0. Then Nδ (z0 ) (the open disc centred at (x0 , y0 ) of radius
δ) lies in X1 as well, because |z − z0 | < δ, i.e.,
√
(x − x0 )2 + (y − y0 )2 < y0 /2,
whence |y − y0 | < y0 /2, and therefore, y > y0 /2 > 0, i.e., (x, y) ∈ X1 . Hence X1 is open.
(b) The set X2 = {z : Re z > 0, |z| < 2} is open as well. The best way to prove it is to use
the facts that the disc {z : |z| < 2} is open (see above) and so is {z : Re z > 0} (this is shown
exactly like in the previous example). The set X2 is the intersection of two open sets and
therefore, open. (Prove it!)
(c) The set X3 = {z√: |z 2 | ≤ 6} is closed.√Indeed, assume z0 to be a limit point of X3 , and
z0 ∈
/ X3 , i.e., |z0 | > 6. Put δ = 12 (|z0 | − 6) > 0; it follows from the triangle inequality that
the disc Nδ (z0 ) has an empty intersection with X3 , i.e., by definition, z0 cannot be a limit
point for X3 , a contradiction.
7. Fix ε > 0; by our assumption, there exists δ > 0 such that |f (z) − k| < ε/2 and
|g(z) − ℓ| < ε/2 whenever |z − z0 | < δ. Hence, by the triangle inequality,
|f (z) + g(z) − (k + ℓ)| ≤ |f (z) − k| + |g(z) − ℓ| < ε/2 + ε/2 = ε,
whence limz→z0 (f (z) + g(z)) = k + ℓ.
8. (a) We can use the fact that if for f (x + iy) = u(x, y) + iv(x, y) both functions u and v
are continuous at (x0 , y0 ), then f is continuous at z0 = x0 + iy0 .
We have Re z = u(x, y) = x, which is clearly continuous, and so is v(x, y) ≡ 0. Thus, f is
continuous as well.
√
(b) Here u(x, y) = x2 + y 2 + x2 − y 2 , v(x, y) = 2xy, both continuous everywhere in R2 .
Hence f is continuous everywhere in C.
(c) Here u(x, y) = x/(x2 + y 2 ), v(x, y) = −y/(x2 + y 2 ), both continuous in R2 \ {(0, 0)}.
Hence f is continuous in C \ {0}.
2
9. (a) The limit does not exist, since for z = x > 0 we have |z|/z = 1 and for z = x < 0 we
have |z|/z = −1.
(b) We have
whence limz→0
2
|z| z = |z|,
|z|2
z
= 0. (We may take ε = δ.)
(c)
z 2 − 3z + 2
(z − 1)(z − 2)
= lim
= lim (z − 2) = −1.
z→1
z→1
z→1
z−1
z−1
lim
3