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# Complex Numbers: Algebra of Complex Numbers

Module by: Louis Scharf. E-mail the author

## Note:

This module is part of the collection, A First Course in Electrical and Computer Engineering. The LaTeX source files for this collection were created using an optical character recognition technology, and because of this process there may be more errors than usual. Please contact us if you discover any errors.

The complex numbers form a mathematical “field” on which the usual operations of addition and multiplication are defined. Each of these operations has a simple geometric interpretation.

The complex numbers z 1 z 1 and z 2 z 2 are

z 1 + z 2 = ( x 1 + j y 1 ) + ( x 2 + j y 2 ) = ( x 1 + x 2 ) + j ( y 1 + y 2 ) . z 1 + z 2 = ( x 1 + j y 1 ) + ( x 2 + j y 2 ) = ( x 1 + x 2 ) + j ( y 1 + y 2 ) .
(1)

We say that the real parts add and the imaginary parts add. As illustrated in Figure 1, the complex number z1+z2z1+z2 is computed from a “parallelogram rule,” wherein z1+z2z1+z2 lies on the node of a parallelogram formed from z1 z1 and z 2 z 2 .

### Exercise 1

Let z1=r1ejθ1z1=r1ejθ1 and z2=r2ejθ2z2=r2ejθ2. Find a polar formula z3=z3=r3ejθ3r3ejθ3 for z3=z1+z2z3=z1+z2 that involves only the variables r1,r2,θ1r1,r2,θ1, and θ 2 θ 2 . The formula for r 3 r 3 is the “law of cosines.”

The product of z1 and z2 is

z 1 z 2 = ( x 1 + j y 1 ) ( x 2 + j y 2 ) = ( x 1 x 2 - y 1 y 2 ) + j ( y 1 x 2 + x 1 y 2 ) . z 1 z 2 = ( x 1 + j y 1 ) ( x 2 + j y 2 ) = ( x 1 x 2 - y 1 y 2 ) + j ( y 1 x 2 + x 1 y 2 ) .
(2)

If the polar representations for z 1 z 1 and z 2 z 2 are used, then the product may be written as 1

z 1 z 2 = r 1 e j θ 1 r 2 e j θ 2 = ( r 1 cos θ1+ j r 1 sin θ 1 ) ( r 2 cos θ 2 + j r 2 sin θ 2 ) = ( r 1 c o s θ 1 r 2 c o s θ 2 - r 1 s i n θ 1 r 2 s i n θ 2 ) + j ( r 1 s i n θ 1 r 2 c o s θ 2 + r 1 c o s θ 1 r 2 s i n θ 2 ) = r 1 r 2 cos ( θ 1 + θ 2 ) + j r 1 r 2 sin ( θ 1 + θ 2 ) = r 1 r 2 e j ( θ 1 + θ 2 ) . z 1 z 2 = r 1 e j θ 1 r 2 e j θ 2 = ( r 1 cos θ1+ j r 1 sin θ 1 ) ( r 2 cos θ 2 + j r 2 sin θ 2 ) = ( r 1 c o s θ 1 r 2 c o s θ 2 - r 1 s i n θ 1 r 2 s i n θ 2 ) + j ( r 1 s i n θ 1 r 2 c o s θ 2 + r 1 c o s θ 1 r 2 s i n θ 2 ) = r 1 r 2 cos ( θ 1 + θ 2 ) + j r 1 r 2 sin ( θ 1 + θ 2 ) = r 1 r 2 e j ( θ 1 + θ 2 ) .
(3)

We say that the magnitudes multiply and the angles add. As illustrated in Figure 2, the product z1z2z1z2 lies at the angle (θ1+θ2)(θ1+θ2).

Rotation. There is a special case of complex multiplication that will become very important in our study of phasors in the chapter on Phasors. When z 1 z 1 is the complex number z1=r1ejθ1z1=r1ejθ1 and z 2 z 2 is the complex number z2=ejθ2z2=ejθ2, then the product of z 1 z 1 and z 2 z 2 is

z1z2=z1ejθ2=r1ej(θ1+θ2).z1z2=z1ejθ2=r1ej(θ1+θ2).
(4)

As illustrated in Figure 3, z1z2z1z2 is just a rotation of z 1 z 1 through the angle θ 2 θ 2 .

### Exercise 2

Begin with the complex number z1=x+jy=rejθz1=x+jy=rejθ. Compute the complex number z2=jz1z2=jz1 in its Cartesian and polar forms. The complex number z 2 z 2 is sometimes called perp(z1)(z1). Explain why by writing perp(z1)(z1) as z1ejθ2z1ejθ2. What is θ 2 θ 2 ? Repeat this problem for z3=-jz1z3=-jz1.

Powers. If the complex number z 1 z 1 multiplies itself NN times, then the result is

(z1)N=r1NejNθ1.(z1)N=r1NejNθ1.
(5)

This result may be proved with a simple induction argument. Assume z1k=r1kejkθ1z1k=r1kejkθ1. (The assumption is true for k=1.k=1.) Then use the recursion z1k+1=z1kz1=r1k+1ej(k+1)θ1z1k+1=z1kz1=r1k+1ej(k+1)θ1. Iterate this recursion (or induction) until k+1=Nk+1=N. Can you see that, as n ranges from n=1,...,Nn=1,...,N, the angle of z§ranges from θ1 to 2θ1,...2θ1,..., to Nθ1Nθ1 and the radius ranges from r1 to r12,...r12,..., to r1Nr1N ? This result is explored more fully in Problem 1.19.

Complex Conjugate. Corresponding to every complex number z=z= x+jy=rejθx+jy=rejθ is the complex conjugate

z*=x-jy=re-jθ.z*=x-jy=re-jθ.
(6)

The complex number zz and its complex conjugate are illustrated in Figure 4. The recipe for finding complex conjugates is to “change jto-jjto-j. This changes the sign of the imaginary part of the complex number.

Magnitude Squared. The product of z and its complex conjugate is called the magnitude squared of z and is denoted by |z|2|z|2 :

|z|2=z*z=(x-jy)(x+jy)=x2+y2=re-jθrejθ=r2.|z|2=z*z=(x-jy)(x+jy)=x2+y2=re-jθrejθ=r2.
(7)

Note that |z|=r|z|=r is the radius, or magnitude, that we defined in "Geometry of Complex Numbers".

### Exercise 3

Write z * z * as z*=zwz*=zw. Find ww in its Cartesian and polar forms.

### Exercise 4

Prove that angle (z2z1*)=θ2-θ1(z2z1*)=θ2-θ1.

### Exercise 5

Show that the real and imaginary parts of z=x+jyz=x+jy may be written as

Re [ z ] = 1 2 ( z + z * ) Re [ z ] = 1 2 ( z + z * )
(8)
Im [ z ] = 2 j ¯ ( z - z * ) . Im [ z ] = 2 j ¯ ( z - z * ) .
(9)

Commutativity, Associativity, and Distributivity. The complex numbers commute, associate, and distribute under addition and multiplication as follows:

z 1 + z 2 = z 2 + z 1 z 1 z 2 = z 2 z 1 z 1 + z 2 = z 2 + z 1 z 1 z 2 = z 2 z 1
(10)
( z 1 + z 2 ) + z 3 = z 1 + ( z 2 + z 3 ) z 1 ( z 2 z 3 ) = ( z 1 z 2 ) z 3 z 1 ( z 2 + z 3 ) = z 1 z 2 + z 1 z 3 . ( z 1 + z 2 ) + z 3 = z 1 + ( z 2 + z 3 ) z 1 ( z 2 z 3 ) = ( z 1 z 2 ) z 3 z 1 ( z 2 + z 3 ) = z 1 z 2 + z 1 z 3 .
(11)

Identities and Inverses. In the field of complex numbers, the complex number 0+j00+j0 (denoted by 0) plays the role of an additive identity, and the complex number 1+j01+j0 (denoted by 1) plays the role of a multiplicative identity:

z + 0 = z = 0 + z z 1 = z = 1 z . z + 0 = z = 0 + z z 1 = z = 1 z .
(12)

In this field, the complex number -z=-x+j(-y)-z=-x+j(-y) is the additive inverse of z, and the complex number z-1=xx2+y2+j(-yx2+y2)z-1=xx2+y2+j(-yx2+y2) is the multiplicative inverse:

z + ( - z ) = 0 z z - 1 = 1 . z + ( - z ) = 0 z z - 1 = 1 .
(13)

### Exercise 6

Show that the additive inverse of z=rejθz=rejθ may be written as rej(θ+π).rej(θ+π).

### Exercise 7

Show that the multiplicative inverse of z z may be written as

z - 1 = 1 z * z z * = 1 x 2 + y 2 ( x - jy ) . z - 1 = 1 z * z z * = 1 x 2 + y 2 ( x - jy ) .
(14)
Show that z*zz*z is real. Show that z-1z-1 may also be written as
z - 1 = r - 1 e - j θ . z - 1 = r - 1 e - j θ .
(15)
Plot zz and z-1z-1 for a representative zz.

### Exercise 8

Prove (j)-1=-j(j)-1=-j.

### Exercise 9

Find z-1z-1 when z=1+j1z=1+j1.

### Exercise 10

Prove (z-1)*=(z*)-1=r-1ejθ=1z*zz(z-1)*=(z*)-1=r-1ejθ=1z*zz. Plot zz and (z-1)*(z-1)* for a representative zz.

### Exercise 11

Find all of the complex numbers zz with the property that jz=-z*jz=-z*. Illustrate these complex numbers on the complex plane.

Demo 1.2 (MATLAB). Create and run the following script file (name it Complex Numbers)2

clear, clg
j=sqrt(-1)
z1=1+j*.5,z2=2+j*1.5
z3=z1+z2,z4=z1*z2
z5=conj(z1),z6=j*z2
avis([-4 4 -4 4]),axis('square'),plot(z1,'0')
hold on
plot(z2,'0'),plot(z3,'+'),plot(z4,'*'),
plot(z2,'0'),plot(z3,'+'),plot(z4,'*'),
plot(z5,'x'),plot(z6,'x')

With the help of Appendix 1, you should be able to annotate each line of this program. View your graphics display to verify the rules for add, multiply, conjugate, and perp. See Figure 5.

### Exercise 12

Prove that z 0 = 1 z 0 =1.

### Exercise 13

(MATLAB) Choose z 1 = 1.05 e j 2 π / 16 z 1 =1.05 e j 2 π / 16 and z 2 = 0.95 e j 2 π / 16 z 2 =0.95 e j 2 π / 16 . Write a MATLAB program to compute and plot z 1 n z 1 n and z 2 n z 2 n for n = 1 , 2 , ... , 32 . n=1,2,...,32. You should observe a figure like Figure 6.

## Footnotes

1. We have used the trigonometric identities cos(θ1+θ2)=cosθ1cos(θ1+θ2)=cosθ1 cos θ2-θ2- sin θ1 sin θ2 and sin(θ1+θ2)=sinθ1sin(θ1+θ2)=sinθ1 cos θ2+cosθ1θ2+cosθ1 sin θ2 to derive this result.
2. If you are using PC-MATLAB, you will need to name your file cmplxnos.m.

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