Introduction

We multiply 2 Complex numbers z_1=a+bi, z_2=c+di with z=z_1z_2 by treating each of them as binomials and using the normal form of binomial multiplication, the FOIL process (First, Outer, Inner, Last)

z=z_1z_2=(a+bi)(c+di)=ac+adi+bci+bdi^{2}=(ac-bd)+(ad+bc)i

with i^{2}=-1 which makes bdi^{2}=-bd

ac-bd is the Real part of the product z=z_1z_2 or Re(z)=Re(z_1z_2)=ac-bd

ad+bc is the Imaginary part of the product z=z_1z_2 or Im(z)=Im(z_1z_2)=ad+bc multiplied by i

Example

z_1=3-2i and z_2=1+4i with z=z_1z_2

z=(3-2i)(1+4i)=3+12i-2i-8i^{2}=(3+8)+(12-2)i=11+10i

or straight into (ac-bd)+(ad+bc)i form

z=(3-2i)(1+4i)=(3-(8i^{2}))+(12-2)i=11+10i

The (ac-bd) form already takes into account that there will be an i^{2}=-1 term, which will make bd the opposite sign of the original product.

Example

z_1=1+i and z_2=2+i with z=z_1z_2

z=(1+i)(2+i)=(2-1)+(1+2)i=1+3i

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Complex Multiplication Laws

Commutative Law

z_1z_2=z_2z_1

Example

z_1=3-4i and z_2=5+2i then

z_1z_2=(3-4i)(5+2i)=((3)(5)-(-4i)(2i))+((3)(2)+(-4)(5))i=((5)(3)-(2i)(-4i))+((2)(3)+(5)(-4))i=(5+2i)(3-4i)=z_2z_1

Associative Law

z_1(z_2z_3)=(z_1z_2)z_3

Example

z_1=1-i, z_2=3+i and z_3=2+5i then

z_1(z_2z_3)=(1-i)((3+i)(2+5i))=(1-i)((6-5)+(15+2)i)=(1-i)(1+17i)=(1+17)+(17-1)i=18+16i

and

(z_1z_2)z_3=((1-i)(3+i))(2+5i)=((3+1)+(1-3)i)(2+5i)=(4-2i)(2+5i)=(8+10)+(20-4)i=18+16i

Multiplying 3 or more Complex numbers is the same process as 2 number multiplication, just with extra steps.

Distributive Law

z_1(z_2+z_3)=z_1z_2+z_1z_3

Example

z_1=2-3i, z_2=3+i and z_3=1-4i then

z_1(z_2+z_3)=(2-3i)((3+i)+(1-4i))=(2-3i)(4-3i)=(8-9)+(-6-12)i=-1-18i

and

z_1z_2+z_1z_3=(2-3i)(3+i)+(2-3i)(1-4i)=(6+3)+(2-9)i+(2-12)+(-8-3)i=(9-10)+(-7-11)i=-1-18i

Real and Imaginary Number Multiplication

Real Multiplication

Multiplying a Complex number by a Real number (x,0)=x+0i just magnifies or shrinks the components of the number by the magnitude of the Real number.

Example

z_1=(-2,0)=-2+0i and z_2=2-3i then

z_1z_2=(-2+0i)(2-3i)=(-4-0)+(6+0)i=-4+6i

Multiplying a Complex number by (-2,0)=-2+0i doubles the x and y components and changes their sign.

Example

z_1=(2,0)=2+0i and z_2=(3,0)=3+0i then

z_1z_2=(2+0i)(3+0i)=(6-0)+(0+0)i=6

Multiplying 2 Real numbers together gives back a Real number as a product.

Imaginary Multiplication

Justification of i^{2}=-1

Example

z_1=(0,1) and z_2=(0,1) then

z_1z_2=(0+i)(0+i)=(0-1)+(0+0)i=-1+0i=(-1,0)

Example

Ex. z_1=(0,1)=0+i and z_2=2-3i then

z_1z_2=(0+i)(2-3i)=(0+3)+(0+2)i=3+2i

Multiplying a Complex number by an Imaginary number (0,y)=0+iy magnifies or shrinks the components by the magnitude of the Imaginary number, switches the magnitudes of the components and changes the sign of the y component.

Multiplicative Identity and Multiplicative Inverse

Multiplicative Identity

(1,0) is the Multiplicative Identity of the Complex Numbers.

Example

z_1=(1,0)=1+0i and z_2=1+3i then

z_1z_2=(1+0i)(1+3i)=(1-0)+(3+0)i=1+3i

Multiplying a Complex number by (1,0)=1+0i gives back the Complex number as the product.

Multiplicative Inverse

The Multiplicative Inverse z^{-1} of a Complex number z=x+iy is

z^{-1}=\frac{x}{x^{2}+y^{2}}-\frac{y}{x^{2}+y^{2}}i

Example

z=1+\sqrt{3}i and x^{2}+y^{2}=(1)^{2}+(\sqrt{3})^{2}=1+3=4 then

zz^{-1}=(1+\sqrt{3})(\frac{1}{4}-\frac{\sqrt{3}}{4}i)=(\frac{1}{4}+\frac{3}{4})+(\frac{-\sqrt{3}}{4}i+\frac{\sqrt{3}}{4}i)=1+0i=(1,0)

This shows that zz^{-1}=(1,0) gives back the Multiplicative Identity as the product.

Conjugate Multiplication and the Modulus

Conjugate Multiplication gives the Square of the Modulus

When we multiply a Complex number z by its conjugate z^{*}, we obtain the square of the modulus

|z|=\sqrt{x^{2}+y^{2}}

of the Complex number z.

If z=x+iy then z^{*}=x-iy and

zz^{*}=(x+iy)(x-iy)=x^{2}-ixy+ixy-i^{2}y=x^{2}+y^{2}=|z|^{2}

Example

z=2-3i and |z|=|2-3i|=\sqrt{(2)^{2}+(-3)^{2}}=\sqrt{13}

The modulus of its conjugate z^{*}=2+3i is also equal to \sqrt{13}

|z^{*}|=|\sqrt{(2)^{2}+(3)^{2}}=\sqrt{13}

and |z||z^{*}|=(\sqrt{13})(\sqrt{13})=13

then zz^{*}=(2-3i)(2+3i)=4+9=13=|z|^{2}

The Conjugate of a Product

The conjugate of the product of 2 Complex numbers is equal to the product of the conjugates of 2 Complex numbers.

(z_1z_2)^{*}=z_1^{*}z_2^{*}

Example
z_1=1+3i and z_2=4-2i then

    \[z_1z_2=(1+3i)(4-2i)=(4+6)+(-2+12)i=10+10i\]

and (z_1z_2)^{*}=10-10iz_1^{*}=1-3i and z_2^{*}=4+2i thenz_1^{*}z_2^{*}=(1-3i)(4+2i)=(4+6)+(2-12)i=10-10i

The Modulus of a Product

We want to show that the identity |z_1z_2|^{2}=|z_1|^{2}|z_2|^{2} holds true

Example

z_1=x_1+iy_1=2+i and z_2=x_2+iy_2=3+2i then

|z_1z_2|^{2}=|(2+i)(3+2i)|^{2}=((2+i)(3+2i))((2-i)(3-2i))=((2+i)(2-i))((3+2i)(3-2i))=

(4+1)(9+4)=((x_1)^{2}+(y_1)^{2})((x_2)^{2}+(y_2)^{2})=|z_1|^{2}|z_2|^{2}

and then by taking the square root of both sides of the equation |z_1z_2|^{2}=|z_1|^{2}|z_2|^{2}

we can deduce that |z_1z_2|=|z_1||z_2|

which is an important and useful identity that shows that the modulus of a product is equal to the product of each modulus.

Equations with Complex Solutions

We may encounter equations that involve the square of an unknown Complex number z^{2} set equal to another known Complex number a+bi where we have to solve for the Complex number's Real and Imaginary parts

Example

z^{2}=7+24i with z=x+iy then

z^{2}=(x+iy)^{2}=x^{2}+2ixy-y^{2}=7+24i

we set the Real part of z^{2} equal to 7 and the Imaginary part of z^{2} equal to 24i and have two equations

x^{2}-y^{2}=7 and 2ixy=24i or xy=12

we can solve for x and y by finding the modulus of a+bi=7+24i

|z^{2}|=|7+24i|=\sqrt{(7)^{2}+(24)^{2}}=\sqrt{49+576}=\sqrt{625}=25

and by using the identity |z^{2}|=|z|^{2}=(\sqrt{x^{2}+y^{2}})^{2}=x^{2}+y^{2}

we can say x^{2}+y^{2}=25 and use this to find x and y

x^{2}-y^{2}=7\implies y^{2}=x^{2}-7 and

x^{2}+y^{2}=x^{2}+(x^{2}-7)=2x^{2}-7=25\implies 2x^{2}=32\implies

x^{2}=16 and x=\pm4

xy=12 and x=4\implies 4y=12 and y=3

then the first solution z_1=x_1+iy_1=4+3i

xy=12 and x=-4\implies -4y=12 and y=-3

then the second solution is z_2=x_2+iy_2=-4-3i

Checking the solutions

z_1^{2}=(4+3i)(4+3i)=(16-9)+(12+12)i=7+24i

z_2^{2}=(-4-3i)(-4-3i)=(16-9)+(12+12)i=7+24i

Visualization of Complex Number Multiplication

Introduction

Multiplication by i is a 90 degree rotation in the xy-plane.

Multiplying a positive Real number by i switches the number from the positive x-axis to the positive y-axis

Example

z_1=3+0i and z_2=0+i with z=z_1z_2

z=(3+0i)(0+i)=(0-0)+(3+0)i=0+3i

or just 3\times i=3i

and multiplying a negative Real number by i switches the number from the negative x-axis to the negative y-axis

Multiplying a positive Imaginary number by i switches the number from the positive y-axis to the negative x-axis

i Times
Example

z_1=0+2i and z_2=0+i

z=(0+2i)(0+i)=(0-2)+(0+0)i=-2+0i

and multiplying a negative Imaginary number by i switches the number from the negative y-axis to the positive x-axis

Multiplying a Complex number by i switches the x and y components and changes the sign of the switching y component

Example

i(3+4i)=3i+4i^{2}=-4+3i

90 Degree Rotation of z=3+4i

y=4 switches to x=-4 and x=3 switches to y=3

Example

i(3-4i)=3i-4i^{2}=4+3i

90 Degree Rotation of z=3-4i

 

The product of the multiplication of 2 Complex numbers is another Complex number

Example

z_1=2-i and z_2=3-2i then

z_1z_2=(2-i)(3-2i)=(6-2)+(-4-3)i=4-7i

Complex Multiplication

Multiplication is Angle Addition

Multiplication is Angle Addition

Triangle Method

Triangle Method
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