Grade 12 Math · Problem type

Multiplying in polar form

Use this for “Multiply 2(cos 30° + i sin 30°) by 3(cos 60° + i sin 60°).”

Diagram

−6−4−2246−4i−2i2i4i8iReIm60°90°w = 1.5 + 2.6izw = 6i30°z = 1.73 + i
z = 2(cos 30° + i sin 30°) ≈ 1.73 + i.
z × w = (1.73 + i)(1.5 + 2.6i)
z × w = 6i
The moduli multiply, 2 × 3 = 6, and the arguments add, 30° + 60° = 90°.
This is an example. Type your own values to start a new problem.
r₁
(cos
θ₁
°
+
i
sin
θ₁
°)
×
r₂
(cos
θ₂
°
+
i
sin
θ₂
°)
=
r
(cos
θ
°
+
i
sin
θ
°)
p
Real part of zw
worked out (not typed)
q
Imaginary part of zw
worked out (not typed)

Formulas

r
=
r₁
×
r₂
6 = 2 × 3
θ
=
θ₁
+
θ₂
90 = 30 + 60
p
=
r
×
cos(
θ
°)
0 = 6 × cos(90°)
q
=
r
×
sin(
θ
°)
6 = 6 × sin(90°)

Assumptions

•
Multiply the moduli and add the arguments.
•
So multiplying by w stretches z by |w| and turns it by w’s argument.
•
Subtract 360° from an argument past 360° if you want it on one turn.

Step-by-step

WE KNOW
r₁ = 2, θ₁ = 30°, r₂ = 3, θ₂ = 60°
FIND
modulus of zw (r), argument of zw (θ), real part of zw (p), imaginary part of zw (q)
Step 1 · Find modulus of zw (r)
Use r = r₁ × r₂
Moduli multiply.
r
=
2
×
3
r
=
6
Step 2 · Find argument of zw (θ)
Use θ = θ₁ + θ₂
Arguments add: the turns add up.
θ
=
θ₁
°
+
θ₂
°
θ
=
30°
+
60°
θ
=
90°
Step 3 · Find real part of zw (p)
Use p = r × cos(θ°)
Back
to
a
+
bi:
the
real
part
of
zw
is
r
cos
θ
.
p
=
6
×
cos(90°)
p
=
6
×
0
p
=
0
Step 4 · Find imaginary part of zw (q)
Use q = r × sin(θ°)
The
imaginary
part
of
zw
is
r
sin
θ
.
q
=
6
×
sin(90°)
q
=
6
Check
Put the numbers back in. Both sides should match.
6 = 2 × 3 ✓
90 = 30 + 60 ✓
0 = 6 × cos(90°) ✓
6 = 6 × sin(90°) ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.

Refresh

Related lessons

© 2026 SME. All rights reserved.