SME
Grade 12 Math · Problem type
Powers by De Moivre’s theorem
Use this for “Find (1 + i)⁸ with De Moivre’s theorem.”
Diagram
z = 1 + i
z
=
√
2
(cos
45°
+
i
sin
45°)
Each
power
turns
45°
more
and
stretches
by
√
2
:
z
8
=
(
√
2
)
8
(cos
360°
+
i
sin
360°)
=
16
This is an example. Type your own values to start a new problem.
(
+
i)
=
p
+
q
i
r
Modulus of a + bi
worked out (not typed)
θ
Argument of a + bi
worked out (not typed) · °
R
Modulus of the power
worked out (not typed)
nθ
Argument of the power
worked out (not typed) · °
Formulas
r
=
√
a
2
+
b
2
1.4142
=
√
1
2
+
1
2
tan
θ
=
b
a
,
θ
in
the
quadrant
of
(
a
,
b
)
tan
45
=
1
1,
45
in
the
quadrant
of
(1,
1)
R
=
(
√
a
2
+
b
2
)
n
16
=
(
√
1
2
+
1
2
)⁸
nθ
=
n
×
θ
360 = 8 × 45
p
=
R
×
cos(
nθ
°)
16 = 16 × cos(360°)
q
=
R
×
sin(
nθ
°)
0 = 16 × sin(360°)
Assumptions
•
Write a + bi in polar form first: r = √(a² + b²), θ in the quadrant of (a, b).
•
De Moivre: (r(cos θ + i sin θ))ⁿ = rⁿ(cos nθ + i sin nθ).
•
Raise the modulus to the nth power and multiply the argument by n.
Step-by-step
WE KNOW
a = 1, b = 1, n = 8
FIND
modulus of a + bi (r), argument of a + bi (θ), modulus of the power (R), argument of the power (nθ), real part of the power (p), imaginary part of the power (q)
Step 1 · Find modulus of a + bi (r)
Use r = √(a² + b²)
The Pythagorean theorem: the distance from the pole to the point.
r
=
√
1
2
+
1
2
r
=
√
1 + 1
r
=
√
2
r
=
1.4142
Step 2 · Find argument of a + bi (θ)
Use tan θ = b ÷ a, θ in the quadrant of (a, b)
tan⁻¹ gives an angle from −90° to 90°; add 180° when the arrow to the point points left, 360° when it points down and right.
θ
=
tan⁻¹(
b
a
)
θ
=
tan⁻¹(
1
1
)
θ
=
45°
Step 3 · Find modulus of the power (R)
Use R = (√(a² + b²))ⁿ
Raise
the
modulus
r
=
√
a
2
+
b
2
to
the
nth
power,
with
no
rounded
r
.
R
=
(
√
1
2
+
1
2
)⁸
R
=
(
√
2
)
8
=
2
4
R
=
16
Step 4 · Find argument of the power (nθ)
Use nθ = n × θ
Multiply
the
argument
by
n
:
n
turns
of
θ
.
nθ
=
n
×
θ
°
nθ
=
8
×
45°
nθ
=
360°
Step 5 · Find real part of the power (p)
Use p = R × cos(nθ°)
Back
to
a
+
bi:
the
real
part
is
R
cos
nθ.
p
=
16
×
cos(360°)
p
=
16
Step 6 · Find imaginary part of the power (q)
Use q = R × sin(nθ°)
The
imaginary
part
is
R
sin
nθ.
q
=
16
×
sin(360°)
q
=
16
×
0
q
=
0
Check
Put the numbers back in. Both sides should match.
1.4142
≈
√
1
2
+
1
2
✓
tan(45°)
=
1
1
✓
16
=
(
√
1
2
+
1
2
)⁸
✓
360 = 8 × 45 ✓
16 = 16 × cos(360°) ✓
0 = 16 × sin(360°) ✓
Topic check
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