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MÄLARDALENS HÖGSKOLA
School of education, culture and communication
Erik Darpö
MAA150 Vektoralgebra
Autumn term 2014
This text is a translated excerpt from a compendium written by M. Herschend and
myself for a course at Uppsala University in 2007.
Equations with complex coefficients
We will study two types of equations: quadratic equations, which are of the form
az 2 + bz + c = 0, with a, b, c ∈ C, and binomial equations: z n = w, where w is a
fixed complex number and n a positive integer.
Quadratic equations
Using complex numbers all quadratic equations can be solved, including those with
complex coefficients.
We start by solving equations of the type
z2 = w
where w ∈ C is a given complex number. For example:
z 2 = 3 − 4i.
(1)
Setting z = x + iy gives z 2 = (x + iy)2 = x2 + 2xyi − y 2 . Equation (1) can thus be
written as
x2 − y 2 + 2xyi = 3 − 4i.
Identifying the real and imaginary parts of the two sides, we get the following system
of equations:
2
x − y 2 = 3,
(2)
2xy = −4.
Solving the second equation for y gives
2
y=− ,
x
(3)
which, inserted into the first equation produces the following equation:
4
=3
x2
4
2
⇔ x − 3x − 4 = 0
r
r
3
9
3
25
3 5
⇔ x2 = ±
+4= ±
= ± .
2
4
2
4
2 2
x2 −
Since x2 > 0, this means that
x2 =
3 5
8
+ = =4
2 2
2
1
and thus x = ±2. Using the equation (3), we can now find the values of y. If x = 2,
then
2
2
y = − = − = −1
x
2
whilst x = −2 gives
−2
y=−
= 1.
2
Hence, the solutions to Equation (1) is
z = x + iy = ±(2 − i).
An alternative way to solve the equation is to use the identity |z|2 = x2 + y 2 . If z
is a solution to (1) then
√
√
|z|2 = |z 2 | = |3 + 4i| = 9 + 16 = 25 = 5.
This means that in addition to the system (2) we get a “bonus equation” x2 +y 2 = 5.
Altogether:
 2
 x − y 2 = 3,
x2 + y 2 = 5,

2xy = −4.
Adding the two first of these equations to each other, we get 2x2 = 8. Hence x2 = 4
and x = ±2. As before, we use Equation (3) to find y, and get again the solutions
z = x + iy = ±(2 − i).
Using this method, any equation of the form z 2 = w can be solved. Combined with
the techinque of completing the square, it allows us to tackle general quadratics
with real or complex coefficients.
Example: Solve the equation z 2 + (2 − 4i)z + 2 − 16i = 0.
Solution: We start by completing the square:
z 2 + (2 − 4i)z + 2 − 16i = 0
2 2
2 − 4i
2 − 4i
−
+ 2 − 16i = 0
⇔ z+
2
2
⇔ (z + 1 − 2i)2 = (1 − 2i)2 − 2 + 16i = −3 − 4i − 2 + 16 = −5 + 12i,
that is, (z + 1 − 2i)2 = −5 + 12i. Setting w = z + 1 − 2i, we have
w2 = −5 + 12i.
This equation can be solved (for w) as in the previous example. We set w = x + iy
and obtain
x2 − y 2 + 2xyi = −5 + 12i
Identifying the real and imaginary parts in this equation yields the system
2
x − y 2 = −5,
2xy = 12.
As before, we utilise the fact that
x2 + y 2 = |w|2 = |w2 | = | − 5 + 12i| =
2
√
25 + 144 =
√
169 = 13
to get
 2
 x − y2
x2 + y 2

2xy
= −5,
= 13,
= 12.
Adding the first two equations gives 2x2 = −5 + 13 = 8, that is, x2 = 4 and thus
x = ±2.
From the equation 2xy = 12 follows that
y=
6
.
x
If x = 2 then y = 3, whilst x = −2 gives y = −3. Consequently,
w = x + yi = ±(2 + 3i).
Finally, we insert these solutions into the identity w = z + 1 − 2i, and solve for z.
This gives
z = −1 + 2i + w = −1 + 2i ± (2 + 3i),
and so
z1 = −1 + 2i + 2 + 3i = 1 + 5i,
z2 = −1 + 2i − (2 + 3i) = −3 − i.
are the solutions to the equation z 2 + (2 − 4i)z + 2 − 16i = 0.
The binomial equation
We shall now consider equations of the form
zn = w
where w ∈ C is a given complex number and n a positive integer.
Example: Solve the equation z 4 = −4.
Solution: The strategy here is to write both sides of the equation on polar form.
Since
z = r(cos θ + i sin θ),
where r = |z| and θ = arg(z), we have
z 4 = r4 (cos θ + i sin θ)n = r4 (cos 4θ + i sin 4θ).
The right hand side, on polar form, is
−4 = 4(cos π + i sin π).
The equation z 4 = −4 now becomes
r4 (cos 4θ + i sin 4θ) = 4(cos π + i sin π)
giving rise to the system
r4
4θ
= 4,
= π + k2π, k ∈ Z.
3
√
1
Since r > 0 it follows that r = 4 4 = 2. The argument θ of z is determined using
the second equation:
π
π
θ = θk = + k , k ∈ Z.
4
2
Hence we have the solutions
√
z = zk = 2(cos θk + i sin θk ).
Sine and cosine are periodic functions, with period 2π. Therefore, zk = zk+4 for all
k ∈ Z. This means that the equation has four distinct solutions
√ π
1
π √
1
= 2 √ + √ i = 1 + i,
z0 = 2 cos + i sin
4
4
2
2
√
√
3π
1
3π
1
z1 = 2 cos
+ i sin
= 2 − √ + √ i = −1 + i,
4
4
2
2
√
√
5π
1
5π
1
z2 = 2 cos
+ i sin
= 2 − √ − √ i = −1 − i,
4
4
2
2
√
√
7π
1
7π
1
z3 = 2 cos
+ i sin
= 2 √ − √ i = 1 − i.
4
4
2
2
The solutions may be viewed in the Argand plane as the corners of a square with
side 2, centered in the origin.
The method used above can be generalised to solve arbitrary equations on the form
zn = w
where w ∈ C is a fixed number different from 0, and n > 1.
First, write z on polar form:
z = r(cos θ + i sin θ)
with r = |z| and θ = arg(z). Then
z n = rn (cos nθ + i sin nθ).
Now write the right hand side on polar form:
w = |w|(cos α + i sin α).
n
The equation z = w becomes
rn (cos nθ + i sin nθ) = |w|(cos α + i sin α),
having the solutions
r
θ
1
= |w| n ,
= nπ + k 2π
n , k ∈ Z.
So the numbers z that satisfy the equation z n = w are
1
π
2π
π
2π
z = zk = |w| n cos
+k
+ i sin
+k
, k∈Z
n
n
n
n
Again, due to the periodicity of the trigonometic functions sin and cos, zn = z0 ,
zn+1 = z1 , etcetera. Hence we only need to count the solutions z0 , z1 , . . . , zn−1 ,
since every other solution zk is equal to one of these.
The solutions may also be expressed using the complex exponential function. The
numbers zk are then written as
z = |w| n ei( n +k
1
π
2π
n
) , k = 0, 1, . . . n − 1.
They constitute the corners of a regular n-gon in the Argand plane, centered in 0.
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