SME
Grade 12 Math · Problem type
The nth roots of a complex number
Use this for “Find the cube roots of 8i” or “the fourth roots of −16.”
Diagram
The
3
roots
of
z
=
8i:
modulus
∛8
=
2,
arguments
90° + 360°k
3
=
30°,
150°,
270°
Roots: 1.73 + i, −1.73 + i, −2i; the corners of a regular 3-gon
This is an example. Type your own values to start a new problem.
w
=
z
=
a
+
b
i
w₀
=
p
+
q
i
Each root w has wⁿ = z. The first, w₀, has modulus ρ = ⁿ√|z| and argument θ₀ = arg z ÷ n.
|z|
Modulus of z
worked out (not typed)
arg z
Argument of z
worked out (not typed) · °
ρ
Modulus of each root
worked out (not typed)
θ₀
Argument of the first root
worked out (not typed) · °
Formulas
|z|
=
√
a
2
+
b
2
8
=
√
0
2
+
8
2
tan
arg
z
=
b
a
,
arg
z
in
the
quadrant
of
(
a
,
b
)
tan
90
=
8
0,
90
in
the
quadrant
of
(0,
8)
ρ
=
|z|^(
1
n
)
2
=
8
1
3
θ₀
=
arg z
n
30
=
90
3
p
=
ρ
×
cos(
θ₀
°)
√
3
=
2
×
cos(30°)
q
=
ρ
×
sin(
θ₀
°)
1 = 2 × sin(30°)
Assumptions
•
Write z = a + bi in polar form first: modulus √(a² + b²), argument in the quadrant of (a, b).
•
Each nth root has modulus the nth root of |z|; the first has argument arg z ÷ n.
•
The n roots are 360° ÷ n apart on one circle, the corners of a regular polygon.
Step-by-step
WE KNOW
a = 0, b = 8, n = 3
FIND
modulus of z (|z|), argument of z (arg z), modulus of each root (ρ), argument of the first root (θ₀), real part of the first root (p), imaginary part of the first root (q)
Step 1 · Find modulus of z (|z|)
Use |z| = √(a² + b²)
The modulus of z: its distance from 0, by the Pythagorean theorem.
|z|
=
√
0
2
+
8
2
|z|
=
√
64
|z| = 8
Step 2 · Find argument of z (arg z)
Use tan arg z = b ÷ a, arg z in the quadrant of (a, b)
The point is on the vertical axis: 90° when it is above 0, 270° when below.
arg z = 90°
Step 3 · Find modulus of each root (ρ)
Use ρ = |z|1/n
ρⁿ
must
equal
|z|,
so
ρ
is
the
nth
root
of
|z|.
ρ
=
∛|z|
ρ
=
∛8
ρ
=
2
Step 4 · Find argument of the first root (θ₀)
Use θ₀ = (arg z) ÷ n
n
times
the
root’s
argument
must
be
arg
z,
so
share
it
among
n
equal
turns.
θ₀
=
arg
z°
÷
n
θ₀
=
90°
÷
3
θ₀
=
30°
→
the
others:
150°,
270°
(each
120°
on)
Step 5 · Find real part of the first root (p)
Use p = ρ × cos(θ₀°)
Back
to
a
+
bi:
the
real
part
of
the
first
root
is
ρ
cos
θ₀
.
p
=
2
×
cos(30°)
p
=
2
×
√
3
2
p
=
√
3
≈
1.7321
Step 6 · Find imaginary part of the first root (q)
Use q = ρ × sin(θ₀°)
And
its
imaginary
part
is
ρ
sin
θ₀
.
q
=
2
×
sin(30°)
q
=
2
×
0.5
q
=
1
Check
Put the numbers back in. Both sides should match.
8
=
√
0
2
+
8
2
✓
0 = 8 × cos(90°) ✓
2 = ∛8 ✓
30
=
90
3
✓
√
3
=
2
×
cos(30°)
✓
1 = 2 × sin(30°) ✓
Topic check
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