SME
Grade 11 Math · Algebra 2
Complex numbers and quadratic equations with complex solutions
Use this for “Multiply (2 + 3i)(1 − 4i).”
Diagram
z × w = (2 + 3i)(1 − 4i) = 14 − 5i.
The
moduli
multiply,
√
13
×
√
17
=
14.87,
and
the
arguments
add,
56.31°
+
284.04°
=
340.35°.
This is an example. Type your own values to start a new problem.
(
a
+
b
i)(
c
+
(
d
)
i)
=
p
+
(
q
)
i
Formulas
p
=
a
×
c
−
b
×
d
14 = 2 × 1 − 3 × (−4)
q
=
a
×
d
+
b
×
c
−5 = 2 × (−4) + 3 × 1
Assumptions
•
i² = −1, so the i × i term moves to the real part with its sign changed.
•
Multiply every term by every term, as with two binomials.
•
On the plane the lengths multiply and the angles add.
Step-by-step
WE KNOW
a = 2, b = 3, c = 1, d = −4
FIND
real part of the product (p), imaginary part of the product (q)
Step 1 · Find real part of the product (p)
Use p = a × c − b × d
The
real
part:
a
×
c
,
and
bi
×
di
=
bd
×
i
2
,
which
is
−bd.
p
=
2
×
1
−
3
×
(−4)
(2 + 3i)(1 − 4i) = 2 − 8i + 3i − 12i²
= 2 − 5i + 12
p
=
2
+
12
p
=
14
Step 2 · Find imaginary part of the product (q)
Use q = a × d + b × c
The
imaginary
part:
a
×
di
and
bi
×
c
,
the
two
outer
and
inner
terms.
q
=
2
×
(−4)
+
3
×
1
q
=
−8
+
3
(2 + 3i)(1 − 4i) = 14 − 5i
q
=
−5
Check
Put the numbers back in. Both sides should match.
14 = 2 × 1 − 3 × (−4) ✓
−5 = 2 × (−4) + 3 × 1 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.