1. Complex Numbers
Dave Saha 15 Jan 99

2. Complex Numbers by Dave Saha
Definition: A complex number is any number, z, which has a real part and an imaginary part:
z = a + bi, where a and b are real numbers and i =  – 1.
Examples of complex numbers:
1 + i, – i, – i, i, – i
The last example given is a complex number with real part zero; this is called a “pure imaginary” number.
The entire set of real numbers is embedded within the complex numbers: z = a + 0i
The entire set of imaginary numbers is embedded within the complex numbers: z = 0 + bi
Any point (x,y) on the Cartesian plane corresponds to a complex number z = x + yi
(the Cartesian plane then is referred to as the “complex plane”)
A complex number z = a + bi has another complex number associated with it, called its conjugate, z* = a – bi
(z and z* are conjugates of each other: z is the conjugate of z*)
11/15/2018
Complex Numbers by Dave Saha

3. Complex Numbers by Dave Saha
Operations with Complex Numbers
Addition
Just add real parts to real parts and imaginary parts to imaginary parts:
(a + bi) + (c + di) = (a + c) + (b + d)i
Examples:
(16 + i) + (7 – 11i) = (16 + 7) + (1 – 11)i = 23 – 10i
(–2 – 5i) + (–3 + 7i) = (–2 + (–3)) + (–5 + 7)i = –5 + 2i
Subtraction
Just subtract real parts from real parts and imaginary parts from imaginary parts:
(a + bi) – (c + di) = (a – c) + (b – d)i
Examples:
(16 + i) – (7 – 11i) = (16 – 7) + (1 – (– 11))i
= (16 – 7) + (1 + 11)i = i
(–2 – 5i) – (–3 + 7i) = (–2 – (–3)) + (–5 – 7)i
= (–2 + 3) + (–5 – 7)i = 1 – 12i
11/15/2018
Complex Numbers by Dave Saha

4. Complex Numbers by Dave Saha
Operations with Complex Numbers 2
Multiplication
For multiplication, we must use the distributive property:
(a + bi)(c + di) = (a + bi)(c) + (a + bi)(di)
= ac + bci + adi + bd(i2) = (ac – bd) + (ad + bc)i
Notice that we used the definition of i: it is the number which squared gives –1.
Examples:
(–2 – 5i)(–3 + 7i) = ((–2)(–3) – (–5)(7)) + ((–2)(7) + (–5)(–3))i
= (6 – (–35)) + (– )i = i
(7 + ½ i)(–2 + i) = ((7)(–2) – (½)(1)) + ((7)(1) + (½)(–2))i
= (–14 – ½) + (7 – 1)i = – i 2
Properties of i
i0 = 1, i1 = i, i2 = –1, i3 = –i, i4 = 1, i5 = i, i6 = –1, etc.
i–1 = –i, i–2 = –1, i–3 = i, i–4 = 1, i–5 = –i, i–6 = –1, etc.
11/15/2018
Complex Numbers by Dave Saha

5. Complex Numbers by Dave Saha
Operations with Complex Numbers 3
Division
For division, we must use the properties of complex conjugates (next slide):
a + bi (a + bi)(c – di) (ac + bd) + (– ad + bc)i
First we multiply the fraction (a + bi) / (c + di) by one. But we choose the form of one to
be the complex conjugate of the denominator: (c – di) / (c – di). Then just multiply the
numerator factors and the denominator factors. This makes the denominator pure real,
taking advantage of the fact that (c + di)(c - di) = c2 + d2. Examples:
16 + i (16 + i)(–3 – 7i) (–48 + 7) + (–112 – 3)i –41 – 115i
3 – 13i (3 – 13i)(– 5i) – 65 – 15i – 5(13 + 3i) –13 – 3i  = =
c + di (c + di)(c – di) c2 + d2  = = =
–3 + 7i (–3 + 7i)(–3 – 7i) = = = =
5i (5i)(– 5i)
11/15/2018
Complex Numbers by Dave Saha

6. Complex Numbers by Dave Saha
Properties of Conjugates
You have encountered conjugates before! The special factoring A2 - B2 = (A + B)(A – B)
from algebra involved conjugate binomials. There are other uses for conjugates in algebra.
COMPLEX CONJUGATES
Every complex number z = x + yi has a corresponding conjugate z* = x – yi
Complex conjugates correspond to points on the complex plane which are “mirror images” of each other (as shown)
The sum of conjugates is the same as the conjugate of the sum: z* + w* = (z + w)*
The product of conjugates is the same as the conjugate of the product: (z*)(w*) = (zw)*
The power of a conjugate is the same as the conjugate of the power: (z*)n = (zn)* . {
(x, y) { .
(x, –y)
11/15/2018
Complex Numbers by Dave Saha

7. Complex Numbers by Dave Saha
Polar Form of Complex Numbers
POLAR COORDINATES
On the complex plane, the ray of the positive x-axis is called the pole.
Each point (x, y) on the plane corresponds to an angle  measured counterclockwise from the pole, and a distance out from the origin r along the ray determined by the angle. .
(x, y)
(r, ) r 
the pole
The relations among x, y, r, and  are:
x = r cos 
y = r sin 
r =  x2 + y2
 = tan –1 ( )
For the complex number z = x + yi, r is called the magnitude of z and written |z|.
The angle  is called the argument of z, also written  = arg(z). The argument of a complex number is not unique; if  = arg(z), then   2m is also an argument of z.
The relation z = x + yi becomes z = r(cos  + i sin ).
Around and around we go.
11/15/2018
Complex Numbers by Dave Saha

8. Complex Numbers by Dave Saha
CiS Notation
PRODUCTS and QUOTIENTS
We now have a very wonderful result:
z1z2 = r1(cos 1 + i sin 1) r2(cos 2 + i sin 2)
= r1r2(cos 1 cos 2 + i2 sin 1 sin 2 + i (cos 1 sin 2 + sin 1 cos 2))
= r1r2(cos 1 cos 2 – sin 1 sin 2 + i (cos 1 sin 2 + sin 1 cos 2))
= r1r2(cos [1 + 2] + i sin [1 + 2] )
In words, to multiply two complex numbers, simply multiply their magnitudes and add their arguments.
For division, such as z1/z2 = r1(cos 1 + i sin 1)/ r2(cos 2 + i sin 2), we need to take a look at the denominator:
r2(cos 2 + i sin 2) r2(cos 2 + i sin 2) r2(cos 2 – i sin 2)
r2(cos2 2 – i2 sin2 2) 1
r2(cos 2 – i sin 2) =
cos 2 – i sin 2 = =
r2–1(cos 2 – i sin 2)
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Complex Numbers by Dave Saha

9. Complex Numbers by Dave Saha
CiS Notation 2
Because of the form of the reciprocal of a complex number we have just investigated,
division of complex numbers has a simple formula, just like multiplication:
z1/z2 = r1(cos 1 + i sin 1)/ r2(cos 2 + i sin 2)
= r1 r2–1(cos [1 – 2] + i sin [1 – 2] )
NOTATION
Because of the algebraic form of the argument, cos  + i sin , a complex number
z = r(cos  + i sin ) may be written as z = r(cis ).
This shortened form makes the formulas for multiplication and division much nicer:
Multiplication of complex numbers: z1z2 = r1r2 cis [1 + 2]
Division of complex numbers: z1/z2 = r1 r2–1 cis [1 – 2]
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Complex Numbers by Dave Saha

10. Complex Numbers by Dave Saha
Practice Problems
1. Add 3i and –5.
2. Add 3 – 5i and its conjugate.
3. Subtract 3 – 5i from its conjugate.
4. (4x – yi) + (x + 3yi) =
5. (– 11 – 2i) – (– 9 – 4i) =
6. (4 + 3i)(4 – 3i) =
7. (4 + 3i)(3 + 4i) =
8. (6 + 3i)  (1 – i) =
10. (1 – i)(4 + 3i)
11. Show that 2 i2
is a square root of i. + =
4 + 3i – 3i =
(2 + i)(1 + 2i)
( ) +
12. Two complex numbers are equal
if their real parts are equal and their
imaginary parts are equal.
Solve for a and b so that
( a + bi)(2 – 3i) = (2 + 3i) + (a + bi).
11/15/2018
Complex Numbers by Dave Saha