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Complex Number

Multiplication of Complex Numbers

Multiplying Two Complex Numbers

Multiplying two complex numbers is similar to multiplying two binomial algebraic expressions. We can use the distributive property of multiplication over addition.

Let's see how to multiply z1=x1+iy1z_1 = x_1 + iy_1 by z2=x2+iy2z_2 = x_2 + iy_2.

z1×z2=(x1+iy1)(x2+iy2)z_1 \times z_2 = (x_1 + iy_1)(x_2 + iy_2)
=x1(x2+iy2)+iy1(x2+iy2)= x_1(x_2 + iy_2) + iy_1(x_2 + iy_2)
=(x1x2+ix1y2)+(ix2y1+i2y1y2)= (x_1x_2 + ix_1y_2) + (ix_2y_1 + i^2y_1y_2)

Remember that i2=1i^2 = -1, so we can substitute:

=(x1x2+ix1y2)+(ix2y1+(1)y1y2)= (x_1x_2 + ix_1y_2) + (ix_2y_1 + (-1)y_1y_2)
=x1x2+ix1y2+ix2y1y1y2= x_1x_2 + ix_1y_2 + ix_2y_1 - y_1y_2

Now, let's group the real and imaginary parts:

=(x1x2y1y2)+(ix1y2+ix2y1)= (x_1x_2 - y_1y_2) + (ix_1y_2 + ix_2y_1)
=(x1x2y1y2)+i(x1y2+x2y1)= (x_1x_2 - y_1y_2) + i(x_1y_2 + x_2y_1)

So, the general formula for complex number multiplication is:

z1×z2=(x1x2y1y2)+i(x1y2+x2y1)z_1 \times z_2 = (x_1x_2 - y_1y_2) + i(x_1y_2 + x_2y_1)

Calculation Example

Let z1=2+iz_1 = 2+i and z2=12iz_2 = 1-2i. Find z1×z2z_1 \times z_2.

Solution:

Using the distributive property:

z1×z2=(2+i)(12i)z_1 \times z_2 = (2+i)(1-2i)
=2(12i)+i(12i)= 2(1-2i) + i(1-2i)
=(24i)+(i2i2)= (2 - 4i) + (i - 2i^2)
=(24i)+(i2(1))(since i2=1)= (2 - 4i) + (i - 2(-1)) \quad \text{(since } i^2 = -1)
=(24i)+(i+2)= (2 - 4i) + (i + 2)
=(2+2)+(4i+i)= (2+2) + (-4i + i)
=43i= 4 - 3i

Or using the general formula with x1=2,y1=1,x2=1,y2=2x_1 = 2, y_1 = 1, x_2 = 1, y_2 = -2:

z1×z2=(x1x2y1y2)+i(x1y2+x2y1)z_1 \times z_2 = (x_1x_2 - y_1y_2) + i(x_1y_2 + x_2y_1)
=(2)(1)(1)(2)+i((2)(2)+(1)(1))= (2)(1) - (1)(-2) + i((2)(-2) + (1)(1))
=(2(2))+i(4+1)= (2 - (-2)) + i(-4 + 1)
=(2+2)+i(3)= (2+2) + i(-3)
=43i= 4 - 3i

The result is the same!

Visualization of Complex Number Multiplication
Visualization of z1=2+iz_1 = 2+i (blue), z2=12iz_2 = 1-2i (green), and their product z1×z2=43iz_1 \times z_2 = 4-3i (red).

Exercise

Let z1=1+iz_1 = 1+i and z2=122iz_2 = \frac{1}{2} - 2i. Find z1×z2z_1 \times z_2.

Answer Key

Using the distributive property:

z1×z2=(1+i)(122i)z_1 \times z_2 = (1+i)(\frac{1}{2} - 2i)
=1(122i)+i(122i)= 1(\frac{1}{2} - 2i) + i(\frac{1}{2} - 2i)
=(122i)+(12i2i2)= (\frac{1}{2} - 2i) + (\frac{1}{2}i - 2i^2)
=(122i)+(12i2(1))= (\frac{1}{2} - 2i) + (\frac{1}{2}i - 2(-1))
=(122i)+(12i+2)= (\frac{1}{2} - 2i) + (\frac{1}{2}i + 2)
=(12+2)+(2i+12i)= (\frac{1}{2} + 2) + (-2i + \frac{1}{2}i)
=(12+42)+(42i+12i)= (\frac{1}{2} + \frac{4}{2}) + (-\frac{4}{2}i + \frac{1}{2}i)
=5232i= \frac{5}{2} - \frac{3}{2}i

Using the general formula with x1=1,y1=1,x2=12,y2=2x_1 = 1, y_1 = 1, x_2 = \frac{1}{2}, y_2 = -2:

z1×z2=(x1x2y1y2)+i(x1y2+x2y1)z_1 \times z_2 = (x_1x_2 - y_1y_2) + i(x_1y_2 + x_2y_1)
=((1)(12)(1)(2))+i((1)(2)+(12)(1))= ((1)(\frac{1}{2}) - (1)(-2)) + i((1)(-2) + (\frac{1}{2})(1))
=(12(2))+i(2+12)= (\frac{1}{2} - (-2)) + i(-2 + \frac{1}{2})
=(12+2)+i(42+12)= (\frac{1}{2} + 2) + i(-\frac{4}{2} + \frac{1}{2})
=(12+42)+i(32)= (\frac{1}{2} + \frac{4}{2}) + i(-\frac{3}{2})
=5232i= \frac{5}{2} - \frac{3}{2}i