Complex Homework I - Solutions
1. Write in the form a + bi:
(a) (2 + i) − (3 + i);
(b)
2 1
3
+ i +
+i ;
3 4
2
(c) (1 + 4i)(2 + 4i);
Solution.
(a) (2 + i) − (3 + i) = −1
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(b) 32 + 14 i + 32 + i = 13
6 + 4i
(c) (1 + 4i)(2 + 4i) = −14 + 12i
(d) (2 − 3i)(2 + 3i) = 13
2. Write in the form a + bi:
(a)
2+i
;
3+i
(b)
1 + 4i
;
2 + 8i
(c)
2 − 3i
.
3 + 2i
Solution.
(a)
2+i
3+i
(b)
1+4i
2+8i
(c)
2−3i
3+2i
2+i 3−i
3+i 3−i
=
7
10
=
1 1+4i
2 1+4i
=
1
2
=
−i(3+2i)
3+2i
=
+
1
10 i
= −i
3. Write in polar form:
(a) 1 −
√
3i;
(b) − 5 + 5i;
(c) πi.
Solution.
√
(a) 1 − 3i = 2ei5π/3
√
(b) −5 + 5i = 5 2ei3π/4
(c) πi = πeiπ/2
4. Write in the form a + bi (using De Moivre’s Theorem):
√
(a) ( 3 − i)7 ;
(b) (1 + i)9 .
Solution.
√
√
(a) ( 3 − i)7 = (2ei11π/6 )7 = 27 ei77π/6 = 128ei5π/6 = −64 3 + 64i
√
√ 9
√
(b) (1 + i)9 = ( 2eiπ/4 )9 = 2 ei9π/4 = 16 2eiπ/4 = 16 + 16i
1
(d) (2 − 3i)(2 + 3i).
5. Write in the form a + bi:
(a) e−πi/4 ;
(b) e1+πi ;
(c) e3+i .
Solution. We use Euler’s Formula throughout.
√
√
(a) e−πi/4 = cos − π4 + i sin − π4 = 22 − i 22
(b) e1+πi = eeπi = e(cos (π) + i sin (π)) = −e
(c) e3+i = e3 ei = e3 (cos (1) + i sin (1))
6. Find all complex numbers z such that z 4 = −1. Write the answer in both polar and cartesian
coordinates. How many different solutions are there?
Solution. We have −1 = cos(π) + i sin(π) in polar coordinates, with r = 1, θ = π. Therefore,
using de Moivre’s Theorem, the fourth roots of −1 are of the form
π 2πk
π 2πk
+
+ i sin
+
, where k = 0, 1, 2, 3.
zk = cos
4
4
4
4
Explicitly, we have
π π z0 = cos
+ i sin
4
4
3π
3π
z1 = cos
+ i sin
4
4
5π
5π
z2 = cos
+ i sin
4
4
7π
7π
+ i sin
z3 = cos
4
4
√
=
=
=
=
2
√2
2
−
2
√
2
−
2
√
2
2
√
2
√2
2
+i
2
√
2
−i
2
√
2
−i
.
2
+i
This gives four distinct solutions, as expected.
7. Find all complex numbers z such that z 5 = −2 − 2i. (You can leave your answer in polar form.)
How many different solutions are there?
Solution. As before, we apply de Moivre’s Theorem. We have −2 − 2i = 23/2 (cos(5π/4) +
i sin(5π/4)) in polar coordinates, with r = 23/2 and θ = 5π/4. Therefore, the solutions are all of
the form
π 2πk
π 2πk
3/10
zk = 2
cos
+
+ i sin
+
, where k = 0, 1, 2, 3, 4.
4
5
4
5
2
Explicitly, we have
π π z0 = 23/10 cos
+ i sin
4
4 13π
13π
3/10
cos
+ i sin
z1 = 2
20
20
21π
21π
3/10
cos
+ i sin
z2 = 2
20
20
29π
29π
3/10
z3 = 2
cos
+ i sin
20
20
37π
37π
3/10
+ i sin
z4 = 2
cos
20
20
We get five solutions.
8. Solve the equation z 2 +
√
32iz − 6i = 0.
Solution. We apply the quadratic formula:
√
√
√
√
− 32i ± −32 + 24i
z=
= − 8i ± −8 + 6i.
2
√
To evaluate −8 + 6i, we may proceed as follows. We have | − 8 + 6i| = 10, so we may write
−8 + 6i = 10(cos(θ) + i sin(θ)) for some angle θ with π/2 ≤ θ ≤ π. Using de Moivre’s Theorem,
the square roots are
√
θ
θ
+ kπ + i sin
+ kπ
, where k = 0, 1.
zk = 10 cos
2
2
We can be more precise about the values cos(θ/2), sin(θ/2), etc., using double angle formulas.
Alternatively, let α + βi be a square root of −8 + 6i; this implies
α2 − β 2 + 2αβi = −8 + 6i.
If we take α = 1, β = 3, we obtain the√solution 1 + 3i, and the
√ other solution is given by −1 − 3i.
Hence, our solutions are z = 1 + 3i − 8i and z = −1 − 3i − 8i.
9. Recall from class that for any two complex numbers z1 , z2 ∈ C, we have the triangle inequality:
|z1 + z2 | ≤ |z1 | + |z2 |
(a) Give an example when this inequality is strict; that is, when |z1 + z2 | < |z1 | + |z2 |.
(b) When can equality occur?
(c) Using the triangle inequality and a judicious choice of z1 and z2 , prove the reverse triangle
inequality:
|z1 − z2 | ≥ |z1 | − |z2 |
Solution.
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(a) If we let z1 be any nonzero complex number, and z2 = −z1 , we have
0 = |z1 + z2 | < |z1 | + |z2 | = 2|z1 |.
(b) This is most easily seen geometrically. If we consider the number z1 and z2 as vectors, then
the vector z1 + z2 is the third side of a triangle formed by z1 , z2 , and z1 + z2 . The only time
we can have the lengths of two sides equalling the length of the third is when the triangle
is degenerate, that is, when the angle between the vectors z1 and z2 is 0 or π. Since π will
give a vector pointing in the opposite direction, we conclude that equality occurs if and only if
z2 = αz1 , with α > 0 a real number.
(c) Set w1 = z1 − z2 and w2 = z2 . Applying the triangle inequality to w1 and w2 , we get
|z1 | = |w1 + w2 | ≤ |w1 | + |w2 | = |z1 − z2 | + |z2 |.
Subtracting |z2 | from both sides gives the result.
10. Write the function f (z) in the form u + iv:
(a) z + iz 2 ;
(b) 1/z 2 ;
(c) z/z.
Solution.
(a) z + iz 2 = (x + iy) + i(x + iy)2 = (x − 2xy) + i(y + x2 − y 2 )
(b)
z2
1
x2 − y 2
−2xy
=
=
+i 4
z2
|z|4
x4 + 2x2 y 2 + y 4
x + 2x2 y 2 + y 4
(c)
z
z2
x2 − y 2
−2xy
= 2 = 2
+i 2
z
|z|
x + y2
x + y2
11. Is the function z/z continuous at 0? Why or why not? Is the function z/z analytic where it is
defined? Why or why not?
Solution. The function z/z is not continuous at 0; approaching 0 along the real axis gives a value
of 1, while approaching 0 along the imaginary axis gives a value of −1 (we saw something similar
in class). To check whether or not it is analytic, one only needs to check that the Cauchy-Riemann
equations are satisfied. We shall proceed in a different way. If the function z/z was analytic, then
so would its product with the function z; that is, the function
z=
z
·z
z
would be analytic. We have seen in class that the function z is not analytic, so z/z cannot be
analytic either.
12. Compute the derivatives of the following analytic functions:
(a)
z2
iz + 3
;
− (2 + i)z + (4 − 3i)
2
(b) ez ;
Solution.
4
(c)
ez
1
.
+ e−z
(a)
d
dz
iz + 3
2
z − (2 + i)z + (4 − 3i)
=
=
i(z 2 − (2 + i)z + (4 − 3i)) − (iz + 3)(2z − (2 + i))
(z 2 − (2 + i)z + (4 − 3i))2
9 + 7i − 6z − iz 2
(z 2 − (2 + i)z + (4 − 3i))2
(b)
d z2 2
e
= 2zez
dz
(c)
d
dz
1
ez + e−z
= −
ez − e−z
(ez + e−z )2
13. Let f (z) be a complex function. Is it possible for both f (z) and f (z) to be analytic? (Hint: if
they are both analytic, they both satisfy the Cauchy-Riemann equations.)
Solution. Write f (z) = u(x, y) + iv(x, y), so that f (z) = u(x, y) − iv(x, y). If both f (z) and
f (z) were analytic, they would both satisfy the Cauchy-Riemann equations. The Cauchy-Riemann
equations for f (z) are
∂u
∂v
∂u
∂v
=
,
=− ,
∂x
∂y
∂y
∂x
and the Cauchy-Riemann equations for f (z) are
∂u
∂v
=− ,
∂x
∂y
∂u
∂v
=
.
∂y
∂x
These two sets of equations in tandem imply
∂v
∂v
=
=0
∂x
∂y
∂u
∂u
=
= 0,
∂x
∂y
and therefore both u(x, y) and v(x, y) must be constant. Therefore the only functions for which
f (z) and f (z) are both analytic are the constant functions.
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