SME
Grade 11 Math · Problem type
Divide complex numbers
Use this for “Divide (3 + 4i) ÷ (1 + 2i).”
Diagram
|z|
=
√
1
2
+
2
2
=
√
5
≈
2.24
arg z = 63.43°.
The conjugate z̄ = 1 − 2i: z reflected across the real axis.
This is an example. Type your own values to start a new problem.
+
i
+
i
=
11
5
p
+
(
−2
5
q
)
i
N
Bottom times its conjugate, c² + d²
worked out (not typed)
Formulas
N
=
c
2
+
d
2
5
=
1
2
+
2
2
p
=
a
×
c
+
b
×
d
N
11
5
=
3 × 1 + 4 × 2
5
q
=
b
×
c
−
a
×
d
N
−
2
5
=
4 × 1 − 3 × 2
5
Assumptions
•
Multiply the top and the bottom by the conjugate of the bottom, c − di.
•
The bottom becomes (c + di)(c − di) = c² + d², a real number.
•
Check: the answer times c + di gives back a + bi.
Step-by-step
WE KNOW
a = 3, b = 4, c = 1, d = 2
FIND
bottom times its conjugate, c² + d² (N), real part of the answer (p), imaginary part of the answer (q)
Step 1 · Find bottom times its conjugate, c² + d² (N)
Use N = c² + d²
The
new
bottom:
(
c
+
di)(
c
−
di)
=
c
2
−
d
2
i
2
=
c
2
+
d
2
,
a
real
number.
N
=
1
2
+
2
2
N
=
1
+
4
N
=
5
Step 2 · Find real part of the answer (p)
Use p = (a × c + b × d) ÷ N
The
real
part
of
(
a
+
bi)(
c
−
di)
is
ac
+
bd,
since
−bd
×
i
2
=
bd.
p
=
3 × 1 + 4 × 2
5
p
=
3 + 8
5
p
=
11
5
p
=
11
5
Step 3 · Find imaginary part of the answer (q)
Use q = (b × c − a × d) ÷ N
The
imaginary
part:
bc
from
bi
×
c
and
−ad
from
a
×
(−di).
q
=
4 × 1 − 3 × 2
5
q
=
4 − 6
5
q
=
−
2
5
3 + 4i
1 + 2i
=
11
5
−
2
5
i
q
=
−
2
5
Check
Put the numbers back in. Both sides should match.
5
=
1
2
+
2
2
✓
11
5
=
3 × 1 + 4 × 2
5
✓
−
2
5
=
4 × 1 − 3 × 2
5
✓
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