Multiplication of Complex Numbers

Now that we can add complex numbers, let's look at how we multiply them.

Multiplying Complex Numbers

Let z1=a+biz_1 = a + bi and z2=c+diz_2 = c + di. Then the product is given by:

(a+bi)(c+di)(a + bi)(c + di)

=a(c+di)+bi(c+di)= a(c + di) + bi(c + di)

=ac+adi+bci+bdi2= ac + adi+ bci + bdi^2

=ac+bdi2+adi+bci= ac + bdi^2 + adi + bci

=ac+bd(1)+i(ad+bc)= ac + bd \cdot(-1) + i(ad + bc) \quad \quad where (i2=1) (i^2 = -1)

=(acbd)+(ad+bc)i= (ac - bd) + (ad + bc)i

Example

Multiply (3+2i)(14i)(3 + 2i)(1 - 4i).

(3+2i)(1+4i)(3 + 2i)(1 + 4i)

=3(1+4i)+b(1+4i)= 3(1 + 4i) + b(1 + 4i)

=3+12i+i+4i2= 3 + 12i+ i + 4i^2

=3+4i2+13i= 3 + 4i^2 + 13i

=3+4(1)+13i= 3 + 4\cdot(-1) + 13i

=34+13i= 3 - 4 + 13i

=1+13i= -1 + 13i

Geometric Interpretation

Multiplying complex numbers has a geometric meaning too: it rotates and scales a number on the complex plane. We'll explore this in more detail later.

Summary

To multiply two complex numbers (a+bi)(c+di)(a + bi)(c + di), expand the product like you would binomials and simplify using i2=1i^2 = -1.

Simplify the following

(58i)(5+9i)(-5 - 8i) \cdot (-5 + 9i)

Simplify the following

(10+i)(77i)(10 + i) \cdot (7 - 7i)

Simplify the following

(6+6i)(152i)(6 + 6i) \cdot (-15 - 2i)

Simplify the following

(3+6i)(6+5i)(3 + 6i) \cdot (-6 + 5i)

Simplify the following

(314i)(12+7i)(-3 - 14i) \cdot (12 + 7i)


End of Lesson

 Previous
Coverting Square Roots to Iota
Next
Properties of Complex Numbers I