3F1-solutions-1 - Isaac Castillo Soto

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MSM3F1/4F1: Complex Variable Theory for Physicists
Solutions to Examples Sheet 1
Q1. Let z = x+ iy. Show that
a) ℜ(z) = z+z2 , iℑ(z) =
z−z
2 .
b) ¯̄z = z.
c) Find the image of the region 0 < Arg z < π/3 under the mapping
w = z3.
Solution.
a) We have z̄ = x − iy, so z + z = 2x = 2ℜ(z), and similarly,
z − z = 2iℑ(z), from which the other two statements follow.
b) Since z̄ = x− iy, we get ¯̄z = x+ iy = z.
c) The image is the region 0 < Argw < π.
Q2. Let z1 =
√
3 + i and z2 = 3 + i
√
3.
a) Write z1 and z2 in polar form.
b) Find z1z2 and z1/z2.
c) Solve the equation z2 =
√
3 + i.
Solution.
a)
z1 = 2e
iπ/6 z2 = 2
√
3eiπ/6
b)
z1z2 = 4
√
3eiπ/3 z1/z2 = 1/
√
3
c) From (a) we have
√
3 + i = 2eiπ/6, and we put z = |z|eiα. Then
z2 = |z|2e2iα. Identifying, we conclude that |z| =
√
2 and α =
π/12 + nπ.
1
2
Q3. Prove the equality |z1 + z2|2 + |z1 − z2|2 = 2(|z1|2 + |z2|2).
Solution. Let z1 = x1 + iy1 and z2 = x2 + iy2. Then,
|z1 + z2|2 + |z1 − z2|2 = (x1 + x2)2 + (y1 + y2)2︸ ︷︷ ︸
=|z1+z2|2
+(x1 − x2)2 + (y1 − y2)2︸ ︷︷ ︸
=|z1−z2|2
= (x21+2x1x2+x
2
2)+(y
2
1+2y1y2+y
2
2)+(x
2
1−2x1x2+x22)+(y21−2y1y2+y22)
= 2(x21 + x
2
2 + y
2
1 + y
2
2) = 2(x
2
1 + y
2
1︸ ︷︷ ︸
=|z1|2
+x22 + y
2
2︸ ︷︷ ︸
=|z2|2
) = 2(|z1|2 + |z2|2).
Q4. Solve the following equations
a) z3 = i
b) z2 = 3 + 4i
Solution.
a) Let z = reiθ. After representing i as i = ei(
π
2+2πn), n = 0, 1, 2, . . . ,
we can write down our equation z3 = i as(
reiθ
)3
= ei(
π
2+2πn)
that is
r3ei3θ = ei(
π
2+2πn).
Then we have r = 1 and θ =
π
6
+
2πn
3
, n = 0, 1, 2 (for greater n the
roots begin to coincide with the those we found, so, if represented
as z = x + iy be the same). Now, returning to the notation
z = x+ iy, we have
z1 = cos
π
6
+ i sin
π
6
=
√
3
2
+
i
2
,
z2 = cos
5π
6
+ i sin
5π
6
= −
√
3
2
+
i
2
,
z0 = cos
3π
2
+ i sin
3π
2
= −i.
b) We can solve z2 = 3 + 4i in the same (standard) way as the
one above or, alternatively, we can just use the representation
3
z = x+iy, so that our equation becomes z2 = x2−y2+2ixy = 3+4i
and hence we arrive at two equations
x2 − y2 = 3, 2xy = 4.
Then, from the second equation we have y = 2/x and the first one
becomes
x4 − 3x2 − 4 = 0,
giving us x2 =
3
2
± 5
2
. We remember that x is real, so we have to
choose the plus sign, and hence x1,2 = 4, i.e. x1 = 2 and x2 = −2.
These correspond to y1 = 2 and y2 = −2. Thus,
z1,2 = ±(2 + 2i).
Q5. What is the geometric meaning of |z − 2|+ |z + 7| = 22?
Solution. This equation describes an ellipse with foci at z = 2
and z = −7 and the large semi-axis 11. Here we had just to read
the question in geometric terms: ‘find the curve such that the sum of
distances from each of its points to two fixed point is the same’. As we
know (?) from analytic geometry, this is an ellipse.
Q6. Given that |z− z0| = R, R > 0 describes a circle with the centre
at z = z0 and radius R, show that∣∣∣∣z − z1z − z2
∣∣∣∣ = λ, 0 < λ ̸= 1,
also describes a circle. Find its centre and radius.
Solution. Once we get rid of the modulus, the rest is trivial:
|z − z1|2 = λ2|z − z2|2,
(z − z1)(z − z1) = λ2(z − z2)(z − z2),
(1) (1− λ2)zz − z(z1 − λ2z2)− z(z1 − λ2z2) = λ2|z2|2 − |z1|2.
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If we do the same with the standard equation of a circle |z − z0| = R,
we end up with
(2) zz − zz0 − zz0 + z0z0 = R2.
So, all we need is to get (1) in the form of (2). This is similar to
‘completing the square’ that we know from school and looks as follows:
zz − z z1 − λ
2z2
1− λ2︸ ︷︷ ︸
=z0
−z z1 − λ
2z2
1− λ2︸ ︷︷ ︸
=z0
+
z1 − λ2z2
1− λ2
z1 − λ2z2
1− λ2︸ ︷︷ ︸
=z0z0
=
z1 − λ2z2
1− λ2
z1 − λ2z2
1− λ2︸ ︷︷ ︸
=z0z0
+
λ2|z2|2 − |z1|2
1− λ2
,
(z − z0)(z − z0) =
λ2|z1 − z2|2
(1− λ2)2︸ ︷︷ ︸
=R2
,
|z − z0|2 = R2,
|z − z0| = R,
where
z0 =
z1 − λ2z2
1− λ2
, R =
λ|z1 − z2|
|1− λ2|
.
Q7. Given that z1, z2 and z3 are three different complex numbers,
what condition do they have to satisfy (expressed in terms of z1, z2 and
z3, not their modulus or argument) to be on the same straight line?
Solution. The necessary and sufficient condition is that
z1 − z2
z3 − z2
is
real.
Q8. Find the real and imaginary part of z−2 + z−2 if z = x+ iy ̸= 0.
Solution. Following the definitions, we have
z−2 + z−2 =
z2 + z2
z2z̄2
=
(x− iy)2 + (x+ iy)2
(x2 + y2)2
= 2
x2 − y2
(x2 + y2)2
.
5
Q9. Prove that z1Im (z2z3)+z2Im (z3z1)+z3Im (z1z2) = 0 if z1, z2, z3 ∈
C.
Solution. Using Im z =
z − z
2i
, we get
z1Im (z2z3) + z2Im (z3z1) + z3Im (z1z2)
= z1
z2z3 − z2z3
2i
+ z2
z3z1 − z3z1
2i
+ z3
z1z2 − z1z2
2i
= 0.
Q10. If |z1| 6 1 and |z2| 6 1, prove that |z1 + z2| 6 |1 + z1z2|.
Solution.
|1 + z1z2|2 − |z1 + z2|2 = (1 + z1z2)(1 + z1z2)− (z1 + z2)(z1 + z2)
= 1 + |z1|2|z2|2 − |z1|2 − |z2|2 = (1− |z1|2)(1− |z2|2) > 0.
The questions below are to be considered after the corresponding ma-
terial is covered in the lectures.
Q11.
a) Determine at which points of the complex plane the function
f(z) = z2 and the function f(z) = |z|2 are analytic.
b) Show that f(z) = 3y + 2ix is not analytic at any point of the
complex plane.
c) Let f(z) be an analytic function in C with ℜ(f(z)) = 0. Prove
that f(z) = ia, where a is a real constant.
Solution.
a) For f(z) = z2, we have f(z) = u(x, y) + iv(x, y), with u(x, y) =
x2 − y2 and v(x, y) = 2xy. Thus, for the partial derivatives, ux =
2x, uy = −2y, vx = 2y, vy = 2x. Clearly, the Cauchy-Riemann
conditions
ux = vy and uy = −vx,
hold in all of the complex plane and since the partial derivatives
are continuous, we conclude that f(z) = z2 is an analytic function
in all of the complex plane.
For f(z) = |z|2 = x2+ y2, we have f(z) = u(x, y)+ iv(x, y) with
u(x, y) = x2+ y2 and v(x, y) = 0. The partial derivatives of u and
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v are ux = 2x, uy = 2y, vx = 0, vy = 0. The Cauchy-Riemann
conditions
ux = vy and uy = −vx
give
2x = 0 and 2y = 0,
i.e. these conditions are only satisfied at x = 0 and y = 0. We
conclude that f(z) = |z|2 is a differentiable function only at the
origin.
b) The second equation in the Cauchy-Riemann conditions gives 3 =
−2, which is clearly impossible. Thus, f(z) is not analytic at any
point of the complex plane.
c) Since ℜ(f(z)) = 0, we have f(z) = iv(x, y). From the Cauchy-
Riemann conditions we obtain vx = 0 and vy = 0, and from this
we conclude that v(x, y) = a, where a is a real constant. Thus,
f(z) = ia.
Q12. Let u(x, y) = 2xy. Show that u is harmonic in C and determine
its harmonic conjugate function v.
Solution. Direct calculations give uxx + uyy = 0 − 0 = 0, thus u is
harmonic. We wish to find an analytic function f(z) = 2xy + iv(x, y).
The Cauchy-Riemann conditions give the equations
2y = vy and 2x = −vx.
Integrating the first equation with respect to y gives v(x, y) = y2+g(x).
Using this expression in the second equation 2x = −g′, i.e. g = −x2+c.
Therefore, v(x, y) = y2 − x2 + c is the harmonic conjugate of 2xy.
Last LATEXed on: 16:10:2019. Last edited on or after: Thursday, Oct 10, 2019 at 19:12 YDS, Oct 19