4.1 Complex numbers expressed in Cartesian form

4.1 Complex numbers expressed in
Cartesian form
4.1.1
Concept of complex numbers
4.1.2
Arithmetic operations of complex numbers
4.1.3
Complex conjugates and properties
Learning Outcomes
Include:
(a)
Extension of the number system from real numbers to complex numbers
(b)
Complex roots of quadratic equations
(c)
Four operations of complex numbers expressed in the form (x + iy)
(d)
Equating real parts and imaginary parts
(e)
Conjugate roots of a polynomial equation with real coefficients
4.1
complex numbers expressed in cartesian form
complex numbers
Content
4-1
4.1.1 Concept of complex numbers
Define
Complex numbers
A complex number is an expression of the form z = x + iy , where x
and y are real numbers, and i is a specific imaginary number,
such that i2 = −1 and i = + −1 .
† The real number x is called the real part, Re(z) of the complex
number, while the real number y is the imaginary part, Im(z).
•
Re ( z ) = Re ( x + iy ) = x
and Im ( z ) = Im ( x + iy ) = y
•
When the imaginary part y is 0, with only a real part, the complex
number is just the real number x. Correspondingly, when the real
part x is 0, with only an imaginary part, the complex number is a
pure imaginary number.
Examples
z = 2 + 3i is not a pure imaginary number, not a pure real number,
just a complex number.
z = 10 is a pure real number.
z = −6i is a pure imaginary number.


☺
A complex number is said to be a pure imaginary number if and only if (iff)
Re ( z ) = 0 and Im ( z ) ≠ 0 .
† The system of complex numbers was developed to cope with
algebraic equations like x 2 + 1 = 0 . Equations with non-real roots can
then be solved in terms of real and imaginary numbers.
•
The imaginary number i is introduced, where i = − 1 so that
i 2 = −1 .
Example
x2 + 1 = 0
⇒
x 2 = −1 ⇒
x = ± −1 = ±i

† Complex numbers can be added, subtracted, multiplied, and divided in
a similar way to real numbers.
† Complex numbers can also be written as an ordered pair of its real
and imaginary parts, i.e.,
z = x + yi
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⇔
z = ( x, y )
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† From the definition of i, it can be seen that:
i4 n = 1 ⎫
⎪
i4 n+1 = i ⎬
i4 n+3 = −i ⎪⎭
for any positive integer, n
Examples
z = 5 − 3i is an example of a complex number.
For the same complex number, Re ( z ) = 5 and Im ( z ) = −3

☺
A complex number of the form z = a + bi is said to be in the cartesian form or
algebraic form or standard form.
Modulus of complex numbers
The modulus (absolute magnitude) of a complex number
z = x + iy is defined as z =
x2 + y 2 .
† The absolute value has three important properties.
For all complex numbers z and w,
z = 0 if and only if (iff or ⇔) z = 0 + 0i = 0 .
z+w ≤ z + w
zw = z w
Equality of complex numbers
† Two complex numbers are equal if and only if they have the same real
and imaginary parts.
x1 + iy1 = x2 + i y 2
⇔
x1 = x2
y1 = y 2
&
4.1
complex numbers expressed in cartesian form
4-3
4.1.2
Arithmetic operations of complex
numbers
Carry out arithmetic operations
Addition and subtraction
† Complex numbers are added and subtracted by formally applying the
associative law of algebra, together with the equation i 2 = −1 ,
( x1 + iy 1 ) ± ( x2 + iy 2 ) = ( x1 ± x2 ) + i( y1 ± y 2 )
Examples
(3 + i) + (2 − 3i) = 5 − 2i
(3 + i) − (2 − 3i) = 1 + 4i

Multiplication
† Complex numbers are multiplied by formally applying the distributive
and commutative laws of algebra, together with the equation i 2 = −1 ,
( x 1 + iy 1 )( x 2 + i y 2 ) = x 1 x 2 + iy 1 x 2 + ix 1y 2 + i 2 y 1y 2
= ( x1x 2 − y1y 2 ) + i ( x2 y1 + x1y 2 )
Example
(3 + i)(2 − 3i) = 6 − 9i + 2i − 3i 2 = 9 − 7i

Division
† Given a complex number ( a + ib ) which is to be divided by another
complex number ( c + id ) whose magnitude is non-zero, express the
division as a fraction, then multiply both numerator and denominator
by the complex conjugate of the denominator.
•
The complex conjugate of a complex number z = a + ib is the
complex number z * = a − ib or z .
•
This process is also known as the realization of the
denominator.
x 1 + iy 1
( x + iy 1 )( x 2 − iy 2 ) (x 1 x 2 + y 1y 2 ) + ( x 2 y 1 − x 1y 2 )i
= 1
=
2
2
x 2 + iy 2 ( x 2 + iy 2 )( x 2 − iy 2 )
x2 + y 2
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Example
(2 + i)(3 + 2i) = 6 + 4i + 3i + 2i 2 = 6 + 7i − 2
2+i
=
3 − 2i (3 − 2i)(3 + 2i)
13
32 + 22
=
1
13
( 4 + 7i )

☺
More on complex conjugates will be discussed in the next section.
Square root
† The square root of a complex number is also a complex number which
can be determined in the following way.
x + iy = a + ib , where a and b are real
⇒ x + iy = (a 2 − b 2 ) + (2ab )i
Equating both real and imaginary parts, x = a 2 − b 2 and y = 2ab .
Example
Let a + ib = 12 + 16i , where a, b ∈ \ .
Squaring both sides,
( a + ib )
2
=
(
12 + 16i
)
2
⇒ a2 − b2 + i ( 2ab ) = 12 + 16i
Equating both the real and imaginary parts,
a2 − b2 = 12
Substituting
and 2ab = 16 ⇒ b =
-
into
16 8
=
2a a
-
:
2
⎛8⎞
a2 − ⎜ ⎟ = 12 ⇒ a 4 − 64 = 12a2
⎝a⎠
a 4 − 12a2 − 64 = 0
⇒ a2 = 16
⇒ a = ±4
∴
or
⇒
(a
2
)(
⇒
)
− 16 a2 + 4 = 0
a2 = −4 (NA as a is defined always real)
and b = ±2
12 + 16i = 4 + 2i or −4 − 2i

4.1
complex numbers expressed in cartesian form
4-5
Worked Examples
Example 1
Given three complex numbers z1 = a + bi , z 2 = c + di and z 3 = e + fi ,
find
(a) z1 + z2 and z2 + z1 and compare the results.
(b)
( z1 + z2 ) + z3
and z1 + ( z2 + z3 ) and compare the results.
(c) the additive inverse of z1 .
(d) z1 − z2 .
(e) z1z2 and z2 z1 and compare the results.
(f)
( z1z2 ) z3
and z1 ( z2 z3 ) and compare the results.
(g) the multiplicative inverse of z1 .
(h) z1 ( z2 + z3 ) .
(i)
z1
, where z2 ≠ 0 .
z2
Solution:
(a) z1 + z 2 = (a + bi) + (c + di) = (a + c ) + (b + d )i
(ans)
z 2 + z1 = (c + di) + (a + bi) = (c + a ) + (d + b )i
= (a + c ) + (b + d )i = z 1 + z 2
(b)
(ans)
(z1 + z 2 ) + z 3 = [(a + bi) + (c + di)] + (e + fi)
= (a + c + e ) + (b + d + f )i (ans)
z1 + ( z 2 + z 3 ) = (a + bi) + [(c + di) + (e + fi)] = (a + bi) + [(c + e ) + (d + f )i]
= (a + c + e ) + (b + d + f )i = ( z1 + z2 ) + z3
(ans)
(c) The additive inverse of z1 = −a − bi . (ans)
(d) z1 − z 2 = (a + bi) − (c + di) = (a − c ) + (b − d )i
(ans)
(e) z1z 2 = (a + bi)(c + di) = (ac − bd ) + (ad + bc )i
(ans)
z 2 z1 = (c + di)(a + bi) = (ca − db ) + (da + cb )i
= (ac − bd ) + (ad + bc )i = z 1z 2
(f)
(ans)
( z1z2 ) z3 = [( a + bi )( c + di )] ( e + fi ) = [( ac − bd ) + ( ad + bc ) i ] ( e + fi )
= ⎡⎣( ac − bd ) e − ( ad + bc ) f ⎤⎦ + ⎡⎣( ac − bd ) f + ( ad + bc ) e ⎤⎦ i (ans)
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z1 ( z 2 z 3 ) = (a + bi)[(c + di)(e + fi)] = (a + bi)[(ce − df ) + (cf + de )i]
= [(ce − df )a − (cf + de )b ] + [(cf + de )a + (ce − df )b ]i
= [(ac − bd )e − (ad + bc )f ] + [(ac − bd )f + (ad + bc )e ]i
= ( z1z2 ) z3
(ans)
(g) The multiplicative inverse of z1 =
a − bi
a2 + b2
. (ans)
(h) z1 (z 2 + z 3 ) = (a + bi)[(c + di) + (e + fi)] = (a + bi )[(c + e ) + (d + f )i]
= [a(c + e ) − b(d + f )] + [a(d + f ) + b(c + e )]i
(i)
(ans)
z1
a + bi ⎛ a + bi ⎞⎛ c − di ⎞ (ac + bd ) + (bc − ad )i
=
=⎜
⎟⎜
⎟ =
z2
c + di ⎝ c + di ⎠⎝ c − di ⎠
c2 + d2
(ans)

Example 2
⎛ 2−i ⎞
⎛ 4 + 2i ⎞
Given that ⎜
⎟z = ⎜
⎟ , find z in the form x + iy.
⎝ 1 + 3i ⎠
⎝ 3+i ⎠
Solution:
⎛ 2−i ⎞
⎛ 4 + 2i ⎞
⎜
⎟z = ⎜
⎟
⎝ 1 + 3i ⎠
⎝ 3+i ⎠
⎛ 4 + 2i ⎞ ⎛ 1 + 3i ⎞ 4 + 12i + 2i + 6i 2 14i − 2
⇒z=⎜
=
⎟⋅⎜
⎟=
7−i
6 − 3i + 2i − i 2
⎝ 3+i ⎠ ⎝ 2−i ⎠
Rationalising the denominator,
14i − 2 14i − 2 7 + i 98i + 14i 2 − 14 − 2i − 28 + 96i
=
⋅
=
=
7−i
7−i 7+i
50
72 + 1
Thus, z =
2
25
( −7 + 24i )
(ans)

4.1
complex numbers expressed in cartesian form
4-7
4.1.3 Complex conjugates and properties
Define
Complex conjugate
The complex conjugate of a complex number z = x + iy is
the complex number z * = x − iy or z .
† Properties of complex conjugates for complex numbers z1 and z2:
z1 + z2 = z1 + z2
z1z2 = z1.z2
z1
⎛ z1 ⎞
⎜z ⎟ =
z
⎝ 2⎠
2
z =z
z = z iff z is real
z = z
z
2
= zz
z −1 = z z
−2
if z is non-zero
† Algebraic properties of conjugates include
(z * )* = z
z + z * = 2 Re(z )
z − z * = i[2 Im(z )]
zz * = Re(z )2 + Im(z )2
Example
The conjugate z of the complex number z = 2 + 3i is
z = 2 − 3i .

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Complex roots of quadratic equations
† Any quadratic equations of the form ax 2 + bx + c = 0 where
b 2 < 4ac , has a pair of conjugate solutions given by
−
4ac − b 2
4ac − b 2
b
b
+i
−i
and −
2a
2a
2a
2a
† It can be said that for any polynomial equation with real coefficients,
any non-real roots occur in conjugate pairs.
Example
For x 2 − 10 x + 34 = 0,
x=
− ( −10) ± ( −10) 2 − 4(1)(34 )
− b ± b 2 − 4ac
=
2a
2
10 ± 100 − 136 10 ± − 36 10 ± 36i 2
=
=
2
2
2
10 ± 6i
=
2
=
∴ x = 5 + 3i or 5 – 3i

Worked Examples

Example 1
Given the complex numbers z1 = a + bi and z 2 = c + di , find
(a) z1 + z2 and z1 + z2 , and compare the results.
(b) z1 − z2 and z1 − z2 , and compare the results.
(c) z1z2 and z1.z2 , and compare the results.
z
⎛z ⎞
(d) ⎜ 1 ⎟ and 1 , and compare the results.
z
z2
⎝ 2⎠
Solution:
(a) z1 + z 2 = (a + bi) + (c + di) = (a + c ) + (b + d )i
z1 + z 2 = (a + c ) − (b + d )i
(ans)
z1 + z 2 = (a − bi) + (c − di) = (a + c ) − (b + d )i = z 1 + z 2
4.1
(ans)
complex numbers expressed in cartesian form
4-9
(b) z1 − z 2 = (a + bi) − (c + di) = (a − c ) + (b − d )i
z1 − z 2 = (a − c ) − (b − d )i (ans)
z1 − z 2 = (a − bi) − (c − di) = (a − c ) − (b − d )i = z 1 − z 2
(ans)
(c) z1z 2 = (a + bi)(c + di) = (ac − bd ) + (ad + bc )i
z1z 2 = (ac − bd ) − (ad + bc )i
(ans)
z1.z 2 = (a − bi)(c − di) = (ac − bd ) − (ad + bc )i = z 1z 2
(d)
(ans)
z1 a + bi ⎛ a + bi ⎞⎛ c − di ⎞ (ac + bd ) + (bc − ad )i
=
=⎜
⎟=
⎟⎜
z2
c + di ⎝ c + di ⎠⎝ c − di ⎠
c 2 +d 2
⎛ z1
⎜⎜
⎝ z2
⎞ (ac + bd ) − (bc − ad )i (ac + bd ) + (ad − bc )i
⎟⎟ =
=
c 2 +d 2
c2 + d2
⎠
z1
=
z2
(ans)
a − bi ⎛ a − bi ⎞⎛ c + di ⎞ (ac + bd ) + (ad − bc )i
=⎜
⎟⎜
⎟=
c − di ⎝ c − di ⎠⎝ c + di ⎠
c 2 +d 2
⎛z ⎞
=⎜ 1⎟
⎝ z2 ⎠
(ans)

Example 2
If z = 2 − i is a factor of a polynomial f(z) with real coefficients, find a
quadratic factor of f(z).
Solution:
( z + 2 − i ) = ( z − (−2 + i))
Since complex roots of any polynomial equation with real coefficients will
occur in conjugate pairs, the other factor must be ( z − (−2 − i) ) .
Thus the quadratic factor of f(z)
= (z + 2 − i)(z + 2 + i) =
(( z + 2) + 1) = (z
2
2
+ 4z + 5
)
(ans)

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Worked Problems
Example 1
Express (4 – i)2 in the form a + ib . Hence or otherwise find the roots of
the equation (z + i) = −15 + 8i .
2
Solution:
(4 − i)2
= 16 − 8i + i 2 = 15 − 8i
(z + i)2
= −15 + 8i = −(15 − 8i)
(ans)
⇒ z + i = ± − 1(15 − 8i) = ±i(4 − i)
∴ z = 1 + 4i or –1 – 4i (ans)

Example 2
Given that (2 + i) + λ (2 + i) + μ = 0 , find the real numbers λ and μ .
2
Solution:
(2 + i)2 + λ (2 + i) + μ = 0
4 + 2i + i 2 + 2λ + λi + μ = 0
4 − 1 + 2λ + μ + i(2 + λ ) = 0
Equating real and imaginary parts,
∴
Re:
3 + 2λ + μ = 0
Im:
2+λ =0
λ = −2, μ = 1 (ans)


4.1
complex numbers expressed in cartesian form
4-11