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Complex Number

Properties of Multiplication of Complex Numbers

Properties of Multiplication Operation

Just like arithmetic operations on real numbers, the multiplication operation on complex numbers also has several important properties. Let z1,z2,z_1, z_2, and z3z_3 be any complex numbers.

Commutative Property

The commutative property means that the order in the multiplication of two complex numbers does not affect the result.

z1×z2=z2×z1z_1 \times z_2 = z_2 \times z_1

Example:

Let z1=1+2iz_1 = 1+2i and z2=3−iz_2 = 3-i.

z1×z2=(1+2i)(3−i)=1(3−i)+2i(3−i)=3−i+6i−2i2=3+5i−2(−1)=5+5iz_1 \times z_2 = (1+2i)(3-i) = 1(3-i) + 2i(3-i) = 3-i+6i-2i^2 = 3+5i-2(-1) = 5+5i
z2×z1=(3−i)(1+2i)=3(1+2i)−i(1+2i)=3+6i−i−2i2=3+5i−2(−1)=5+5iz_2 \times z_1 = (3-i)(1+2i) = 3(1+2i) - i(1+2i) = 3+6i-i-2i^2 = 3+5i-2(-1) = 5+5i

The results are proven to be the same.

Associative Property

The associative property states that when multiplying three or more complex numbers, the grouping of the multiplication does not change the result.

(z1×z2)×z3=z1×(z2×z3)(z_1 \times z_2) \times z_3 = z_1 \times (z_2 \times z_3)

Example:

Let z1=iz_1 = i, z2=2z_2 = 2, and z3=3−iz_3 = 3-i.

(z1×z2)×z3=(i×2)×(3−i)=2i(3−i)=6i−2i2=6i−2(−1)=2+6i(z_1 \times z_2) \times z_3 = (i \times 2) \times (3-i) = 2i(3-i) = 6i - 2i^2 = 6i - 2(-1) = 2+6i
z1×(z2×z3)=i×(2×(3−i))=i×(6−2i)=6i−2i2=6i−2(−1)=2+6iz_1 \times (z_2 \times z_3) = i \times (2 \times (3-i)) = i \times (6-2i) = 6i - 2i^2 = 6i - 2(-1) = 2+6i

The results are proven to be the same.

Multiplicative Identity

The complex number 1=1+0i1 = 1+0i is the identity element for multiplication. This means that any complex number multiplied by 1 results in the complex number itself.

z×1=z=1×zz \times 1 = z = 1 \times z

Example:

Let z=4−7iz = 4-7i.

z×1=(4−7i)(1+0i)=4(1)−(−7)(0)+i(4(0)+(−7)(1))=4−7i=zz \times 1 = (4-7i)(1+0i) = 4(1) - (-7)(0) + i(4(0) + (-7)(1)) = 4 - 7i = z
1×z=(1+0i)(4−7i)=1(4)−(0)(−7)+i(1(−7)+0(4))=4−7i=z1 \times z = (1+0i)(4-7i) = 1(4) - (0)(-7) + i(1(-7) + 0(4)) = 4 - 7i = z

Distributive Property of Multiplication over Addition

This property connects the operations of multiplication and addition of complex numbers.

z1×(z2+z3)=(z1×z2)+(z1×z3)z_1 \times (z_2 + z_3) = (z_1 \times z_2) + (z_1 \times z_3)

Example:

Let z1=2z_1=2, z2=1+iz_2 = 1+i, z3=3−2iz_3 = 3-2i.

z1×(z2+z3)=2×((1+i)+(3−2i))=2×(1+3+i−2i)=2×(4−i)=8−2iz_1 \times (z_2 + z_3) = 2 \times ((1+i) + (3-2i)) = 2 \times (1+3 + i-2i) = 2 \times (4-i) = 8-2i
(z1×z2)+(z1×z3)=(2(1+i))+(2(3−2i))=(2+2i)+(6−4i)=(2+6)+(2i−4i)=8−2i(z_1 \times z_2) + (z_1 \times z_3) = (2(1+i)) + (2(3-2i)) = (2+2i) + (6-4i) = (2+6) + (2i-4i) = 8-2i

The results are proven to be the same.

Example Proof Using Properties

We can prove several algebraic identities using these properties. Let's prove that (z1+z2)2=z12+2z1z2+z22(z_1 + z_2)^2 = z_1^2 + 2z_1z_2 + z_2^2 for any z1,z2z_1, z_2.

(z1+z2)2=(z1+z2)(z1+z2)(Definition of square)(z_1 + z_2)^2 = (z_1 + z_2)(z_1 + z_2) \quad \text{(Definition of square)}
=z1(z1+z2)+z2(z1+z2)(Distributive Property)= z_1(z_1 + z_2) + z_2(z_1 + z_2) \quad \text{(Distributive Property)}
=(z1z1+z1z2)+(z2z1+z2z2)(Distributive Property)= (z_1z_1 + z_1z_2) + (z_2z_1 + z_2z_2) \quad \text{(Distributive Property)}
=z12+z1z2+z2z1+z22(Definition of square)= z_1^2 + z_1z_2 + z_2z_1 + z_2^2 \quad \text{(Definition of square)}
=z12+z1z2+z1z2+z22(Commutative Property: z2z1=z1z2)= z_1^2 + z_1z_2 + z_1z_2 + z_2^2 \quad \text{(Commutative Property: } z_2z_1 = z_1z_2)
=z12+2z1z2+z22(Combining like terms)= z_1^2 + 2z_1z_2 + z_2^2 \quad \text{(Combining like terms)}

Multiplicative Inverse

Every non-zero complex number z=x+iy≠0z = x + iy \neq 0 has a multiplicative inverse, denoted as z−1z^{-1} or 1/z1/z, such that z×z−1=1z \times z^{-1} = 1.

Let z−1=u+ivz^{-1} = u + iv. Then:

z×z−1=1z \times z^{-1} = 1
(x+iy)(u+iv)=1+0i(x+iy)(u+iv) = 1 + 0i
(xu−yv)+i(xv+uy)=1+0i(xu - yv) + i(xv + uy) = 1 + 0i

Based on the equality of two complex numbers, we obtain the system of equations:

  1. xu−yv=1xu - yv = 1
  2. xv+uy=0xv + uy = 0

By solving this system of equations (for example, by multiplying equation 1 by xx, equation 2 by yy, then adding them, and using the substitution method), we will get:

u=xx2+y2u = \frac{x}{x^2+y^2}
v=−yx2+y2v = -\frac{y}{x^2+y^2}

So, the multiplicative inverse of z=x+iyz = x+iy is:

z−1=xx2+y2−iyx2+y2z^{-1} = \frac{x}{x^2+y^2} - i\frac{y}{x^2+y^2}

Note that x2+y2=∣z∣2x^2+y^2 = |z|^2 and x−iy=zˉx-iy = \bar{z}. Thus the inverse formula can also be written as:

z−1=zˉ∣z∣2z^{-1} = \frac{\bar{z}}{|z|^2}
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