Complex Numbers and Applications ME50/A1
1
Comp Comple lex x Numb Number erss
√
A complex number is an ordered pair (x, y) of real numbers x and y . For example, ( 2.1, 3.5), (π, 2), (0, (0, 0) are complex numbers. Let z = (x, y) be a complex number. The real part of z , denoted by Re z , is the real number x. The imaginary part of z , denoted by Im z , is the real number y.
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Re z = x Im z = y Two complex numbers z 1 = (a1, b1) and z 2 = (a2, b2) are equal, equal, written z 1 = z 2 or (a1, b1) = (a2, b2) if and only if a1 = a2 and b1 = b2. For example, if (x, 2) = (3, (3 , c) then x = 3 and c = 2. Since a complex number is denoted by an ordered pair (x, y) of real numbers x and y, then we may view the complex number (x, y ) as the point with abscissa x and ordinate y. The complex plane cons consis ists ts of all all the the poin points ts that that repr represe esent nt the the comp comple lex x numb number ers. s. For For exam exampl ple, e, let let us indi indica cate te the the foll follow owin ing g complex numbers in the complex plane: z 1 = ( 3, 2), 2), z 2 = (0, (0, 1), 1), z 3 = (4, (4, 2), 2), z 4 = (5, (5, 1)
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.. .... .......... .... .. ......... .... .... .. .. .......... .... .. ... .. .......... .... ... .. .. ............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................. .... ... ... .. .......... ... ... ... .. ......... ... ... ... ... ..
z 3
•
z 2
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............. ....
•
z 1
z 4
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The complex plane showing four complex numbers z 1, z 2, z 3, and z 4 1.1 Operations on Complex Numbers
Some binary operations on complex numbers are addition, multiplication, and division. They are defined as follows: Let z 1 = (a1, b1) and z 2 = (a2, b2). Then 1. Addition. z 1 + z 2 = (a1 + a2, b1 + b2). For example, (2, 3) + ( 1, 2) = (1, 1).
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− − 2. Multiplication. z 1z 2 = (a1a2−b1b2, a1b2 +a2b1). For example, (2, −3)(−1, 2) = ((2)( −1)−(−3)(2), (2)(2)+ (−3)(−1)) = (4, 7) (0, 0), then 3. Division. If z 2 = z 1 a1a2 + b1b2 −a1b2 + a2b1 = , z 2
For example,
a22 + b22
(2, 3) = ( 1, 2)
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−
a22 + b22
− − 8 , 5
1 5
Remark. The complex numbers of the form (x, 0) are actually the real numbers x in the following sense:
1. (a1, 0) + (a2, 0) = (a1 + a2, 0), which corresponds to the sum a1 + a2. 2. (a1, 0)(a2, 0) = (a1a2, 0), which corresponds to the product a1a2 (a1, 0) 3. = (a2, 0)
a1 a2 , 0
, which corresponds to the quotient
a1 a2 .
1.2 Scalar Multiple
A complex number z = (x, y) may be multiplied by a real number c and the result is cz = (cx,cy)
For example, if z = (2, 3), then 5z = (10, 15). The additive inverse or negative of a complex number z = (x, y), denoted by z , is defined by z = ( 1)z . For example, if z = (2, 3), then z = ( 1)z = ( 2, 3).
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Remark. We may define subtraction denoted by z 1 z 2 in terms of addition and negative as follows: z 1 z 2 = z 1 + ( 1)z 2. For example, (1, 2) (2, 2) = (1, 2) + ( 2, 2) = ( 1, 4).
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1.3
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Conjugate
The conjugate of a complex number z = (x, y), denoted by the symbol z , is the complex number (x, y). For example, if z = (5, 2), then z = (5, 2). Note that if we plot z and z on the complex plane, then
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these two points are reflections of each other with respect to the x axis. .. ... ......... .. .... . ......... ... .... ... .. ... .... .... ... .... ....... .. .... ... .. ... .... .... ... ... .... . . . . ................................................................................................................................................................................................. ............. ... ... ... ... .... ... ... ... ... ... .. ... ... ... .... ... . ... ... ... ... .. .... ... ..
z
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•
z
A complex number z and its conjugate z
The following properties of conjugate are easy to verify. 1. z = z
3. z 1 + z 2 = z 1 + z 2
2. (cz ) = cz
4. z 1z 2 = z 1z 2
5.
z 1 z 2
=
z 1 z 2
1.4 Forms of a complex number
The norm of a complex number z = (x, y), denoted by z is the real number z = x2 + y2. For example, if z = (4, 3), then z = 42 + ( 3)2 = 5. Observe that z is numerically equal to the distance from the point z in the complex plane to the origin (0, 0). A complex number with norm 1 is called a unit complex number. Some examples of unit complex √ 2 √ 2 3 4 numbers are 5 , 5 , (1, 0), 2 , 2 , (0, 1).
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Of particular importance to us are the unit complex numbers (1, 0) and (0, 1). Observe that (1, 0)(x, y) = (x, y) for all complex numbers (x, y). For this reason, we simply denote (1, 0) by 1. Let us denote the unit complex number (0, 1) by i. Then i2 = (0, 1)(0, 1) = ( 1, 0) = ( 1)(1, 0) = (1, 0) = 1. Therefore, i2 = 1. The reason why 1 and i are important unit complex numbers is because every complex num ber can be written as a linear combination of them. To see this, let z = (x, y) be any complex number. Then
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z = (x, y) = (x, 0) + (0, y) = x(1, 0) + y(0, 1) = x 1+y i = x + iy
·
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The form x + iy of the complex number (x, y) is called the Cartesian form. For example, the Cartesian form of (2, 3) is 2 3i. Operations on complex numbers become more convenient if we write complex numbers in Cartesian form and remember that i2 = 1. For example, to multiply (2, 3) and ( 1, 2), we do it this way:
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(2, 3)( 1, 2) = (2 3i)( 1 + 2i) = 2 + 4i + 3i 6i2 = 2 + 7i 6( 1) = 2 + 7i + 6 = 4 + 7i
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− − −
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Let us observe that zz = z 2. This helps in performing division because z 1 z 1z 2 z 1z 2 = = z 2 z 2z 2 z 2 2 As an example, (2, 3) (2, 3)( 1, 2) = ( 1, 2) ( 1, 2) 2 (2 3i)( 1 2i) = ( 1)2 + (2)2 2 4i + 3i + 6i2 = 5 8 i = 5 8 1 = i 5 5 Let z be a non-zero complex number and let P be the point denoting z . Join the origin to P and let θ be the angle measured from the positive x-axis to the line OP . Then it is easy to see that x = z cos θ and y = z sin θ.
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− − − − − −− − −− −−
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. .... ......... ... .... ... ... ... ... ... .... ... .......... ........ ... ... ........ ... ........ ..... ........ . . . . ... ... . . . . ........ .... .... ........ . . . . . . ... . . . .. ...... .... . . . . ... . . ... . ... ...... . . . . .... . . ... . .. ... ... .... ... ................ .. . .............. .. ....................................................................................................................................................................................................................................................
P
•
θ
O
y
x
We can therefore write z = z (cos θ + i sin θ)
............. ....
This is called the polar form of the complex number z . Example. Write the complex number z = 3 4i in polar form.
− √ SOLUTION. First, we get the norm z = 32 + 42 = 5. Next, we get the argument θ. x θ y
............................................................................................................................................................ .... .. .. .... .. ... .... ... ... .... .... ....... ... ........ ... .... .. .... .... .... .... ... .... .. .....
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tan θ = =
y x
−4 3
4 3 0.9273 radians
θ = tan−1 =
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Therefore, z = 5(cos( 0.9273) + i sin( 0.9273)) .
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The Shorthand Notation. If the polar form of the complex number z is z = r(cos θ + i sin θ) then for simplicity, we shall write
z = r cis θ
This is known as the shorthand notation for a complex number. For example, if z = 2 cis π6 , then π 6
z = 2(cos + i sin
π 6)
=2
√ 3 2
+
i 12
=
√ 3 + i
Multiplication and Division of Complex Numbers in Polar Form.
We will need here the following trigonometric identities:
• cos(θ1 + θ2) = cos θ1 cos θ2 − sin θ1 sin θ2 • sin(θ1 + θ2) = sin θ1 cos θ2 + cos θ1 sin θ2 Let z 1 = r1 cis θ1 and z 2 = r2 cis θ2. Then z 1z 2 = r1(cos θ1 + i sin θ1)r2(cos θ2 + i sin θ2) = r1r2 (cos θ1 cos θ2 sin θ1 sin θ2 + i(cos θ1 sin θ2 + sin θ1 cos θ2)) = r1r2 (cos(θ1 + θ2) + i sin(θ1 + θ2)) = r1r2 cis (θ1 + θ2)
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Example. Let z 1 = 1 + i and z 2 = 1
− i. Find z 1z 2 using multiplication in polar form.
SOLUTION. In polar form we have z 1 = 1 + i =
√
2 cis
z 2 = 1 =
−i √
2 cis
π 4 π 4
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z 1z 2 =
√ √
2 2 cis
π 4
π 4
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= 2 cis 0 = 2(cos 0 + i sin 0) = 2(1 + i0) =2
It is an easy exercise to prove that if z = r cis θ, then 1 1 = cis ( θ) z r
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For example, if z = 1 + i, then z =
√
2 cis
π 4
1 1 π cis = z 4 2 1 = (cos( π/4) + i sin( π/4)) 2 1 = 2/2 i 2/2 2 1 i = 2 2
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√ √ − √ √ √ − −
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Now suppose that z 1 = r1 cis θ1 and z 2 = r2 cis θ2. Then,
z 1 1 cis ( θ2) = (r1 cis θ1) z 2 r2 r1 = cis (θ1 θ2) r2 r1 = (cos(θ1 θ2) + i sin(θ1 r2
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− θ2))
De Moivre’s Theorem. Let z = r cis θ be a complex number and n a positive integer. Then
z n = rn cis (nθ) Exercise. Derive De Moivre’s theorem. Example. Evaluate (1 + i)9.
SOLUTION. We first write 1 + i in polar form (shorthand). We have 1 + i = 9 9π (1 + i)9 = 2 cis 4 = 16 2 cis (2π + π/4) π = 16 2 cis 4 2 2 = 16 2 +i 2 2
√ √ √
√ √
= 16 + 16i
√
√ 2 cis π . Therefore, 4
Roots of a Complex Number . Consider a complex number z = r cis θ. Since cosine and sine are periodic functions with period 2π , then z = r cis (θ + 2kπ) where k is any integer.
Let n be a positive integer and consider the complex number wk = theorem, we have wkn = r cis (θ + 2kπ) = r cis θ = z
√ r cis n
Therefore, wk is an nth root of z . To summarize, the n n th roots of z = r cis θ are wk
√ = r n
cos( θ+2πk n )+
i sin( θ+2kπ) n
Example. Find all the cube roots of 8.
, k = 0, 1, 2, . . . , n
SOLUTION. 8 = 8 cis 0. The cube roots of 8 are wk =
√ 3
8 cis
0 + 2kπ 3
,
k = 0, 1, 2
θ+2kπ n
−1
. By De Moivre’s
w0 = 2 cis 0 =2 2π w1 = 2 cis 3 1 3 =2 +i 2 2
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=
√
√ −1 + i 3
w2 = 2 cis =2
4π 3 1 3 i 2 2
− − √ = −1 − i 3
√
Example. Find the cube roots of 8.
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SOLUTION. 8 =
−
−8 + 0i. r =
−
( 8)2 + 12 = 8, θ = π wk =
√ 3
8 cis
π + 2kπ , 3
π 3 w1 = 2 cis π 5π w2 = 2 cis 3 w0 = 2 cis
k = 0, 1, 2
It is easy to see from the formula for the nth roots that these n roots are arranged evenly on the circumference of a circle with center at the origin and radius r. From a previous example, we show below the three cube roots of 8, and the three cube rots of 8 in the complex plane w1 w0
√ n
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....... .... .... .... ... .. .... .... .... . .... .. . .... .. .. .... ........ .... . ..... . .... ..... .... . ... . . . . . . .... ... .... ... . ... ... . . . . . ... ... .. .. .. . ... ... .. . ... .. .. ... .. ... ... .. ... ... ... ... ....... ... ............... ........ ........ ........ ........ ........ ........ .... ... . . ... . . .. ... ... . . . . ... ... .. ... ... ... . . . . ... .. ... ... ... ... ... ... ... ... ... ... . ... . . . . ... .... ... ... .... .... . .... ...... ... .... .... .. . .... .... .... .. .. .... . . . . . . . . . .. .... .... ....... .... .... .
.... ....... .... .... .... .... .... .... .... .... .... .... .... .... .. . .... ... .. ....... ..... .... ... .... . . .. ... . .. .. . . ... . . ... ... ... .. .. .. .. .. . . .. . .. . .. .. . .. . .. ... ... . .. . . . .. .. ... ... . ... .. ........ ........ ........ ........ ........ ........ ........ ............... . .. .. .. .. ... ... .. ... . ... ... . . ... ... ... ... ... ... ... ... ... ... ... ... ... . ... . . . ... ... ... ... .... ... ... .... ... ....... .... . . .. .... ... ... . . . .... . . .... .... . .. .. ... .... .... ....... .... .... .... .... ..
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r=2
r=2 • w 0
w1 •
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w2 Cube roots of 8
2
w2 Cube roots of 8
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Exponential Form
In elementary calculus, the following series are well-known and each series converges for all real values of x: x
e =
∞ xn
− − n=0
(1)
n!
∞
n
∞
n
x2n cos x = ( 1) (2n)! n=0 x2n+1 sin x = ( 1) (2n + 1)! n=0
(2) (3)
If we assume that equation (1) is valid even for complex exponents, then for a real number θ, eiθ =
∞ (iθ)n
n=0
n!
iθ i2θ2 i3θ3 i4θ4 i5θ5 i6θ6 i7θ7 = 1+ + + + + + + 1! 2! 3! 4! 5! 6! 7! 2 4 6 3 5 θ θ θ θ θ θ7 = 1 + + +i θ + + 2! 4! 6! 3! 5! 7! ∞ ∞ 2n 2n+1 n θ n θ = ( 1) +i ( 1) (2n)! (2n + 1)! n=0 n=0
− −
−
= cos θ + i sin θ = cis θ
···
··· − −
−
···
Therefore, if z is a complex number with r = z , and argument θ, then in shorthand form z = r cis θ and in exponential form.
z = reiθ
Let z 1 = r1eiθ1 and z 2 = r2eiθ2 . Then obviously z 1z 2 = r1 r2ei(θ1+θ2) z 1 r1 i(θ1 −θ2) = e , z 2 r2
provided r2 = 0
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Logarithm of a Complex Number
Let z be a complex number with exponential form z = reiθ . We define the logarithm of z , denoted by ln z , as follows: ln z = ln r + iθ
Actually, ln is multiple-valued because z has infinitely many arguments. We can write z = r cis (θ + 2kπ) = rei(θ+2kπ). Hence, ln z = ln r + i(θ + 2kπ) are all the logarithms of z . If we restrict θ to be in the interval ( π, π] then ln r + iθ is called the principal value of the logarithm of z , denoted by Log z . π As an example, if z = 3i, then z = 3 cis i 3π 2 and so Log z = ln 3 2. π As a second example, Log (3i) = ln 3 + i 2 . Verify the following:
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Log ( 1) = iπ Log (1 + i) = 12 ln 2 + i π4 Log (1 i) = 12 ln 2 i π4 Log ( 1 + i) = 12 ln 2 + i 3π 4 1 3π Log ( 1 i) = 2 ln 2 i 4
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Complex Variables
Let w be a function of a complex variable z , i.e., w = f (z ). We define derivative as follows: dw f (z + ∆z ) = lim dz ∆z →0 ∆z
− f (z )
if this limit exists. Other notations for derivative are f (z ) or w . Let us observe that ∆z is a complex number and is viewed as a point in the complex plane. The notion of ∆z 0 is more complicated since the complex number ∆z may approach 0 following an arbitrary path in the complex plane.
→
The so-called Cauchy-Riemann equations gives us a set of necessary and sufficient conditions for a function f (z ) to be differentiable at z . If we express f (z ) in the form u(x, y) + iv(x, y), then the CauchyRiemann equations are ∂u ∂v = ∂x ∂y ∂u ∂v = ∂y ∂x
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Furthermore, if the Cauchy-Riemann equations are satisfied, then f (z ) = ux iuy f (z ) = vy + ivx
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Example. Verify that f (z ) = z 2 satisfies the Cauchy-Riemann equations and find f (z ). SOLUTION. Let z = x + iy. Then f (x + iy) = z 2 = (x + iy)2 = x2 y2 + i2xy. Therefore
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u(x, y) = x2 y2 v(x, y) = 2xy
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We have the following partial derivatives: ux = 2x uy = 2y vx = 2y vy = 2x
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Therefore, ux = 2x = vy and uy = 2y = vx. Therefore, the Cauchy-Riemann equations are satisfied. This means that the function f (z ) is differentiable. The derivative is given by:
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f (z ) = z 2 = ux ivy = 2x i( 2y) = 2x + i2y = 2(x + iy) = 2z
− − −
Notice that if we imagine z to be a real variable, and we apply the usual derivative formula, we would get 2z as the derivative of z 2 . It can be shown that if f (z ) satisfies the Cauchy-Riemann equations, then the usual derivative rules for real-valued functions of a real variable can be applied to find the derivative of f (z ).
1 As an example, let f (z ) = . Then z 1 z z = zz x iy = 2 x + y2 x = 2 x + y2
f (z ) =
−
− i x2 +y y2
Therefore,
x x2 + y2 y v= x2 + y2 By taking partial derivatives, one can verify that the Cauchy-Riemann equations are satisfied provided z = 0. Therefore, for every complex number z = 0, we have 1 f z ) = z 2 This is easily obtained by writing f (z ) = z −1 and then differentiating as f (z ) = 1z −2 = z −2. u=
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5
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Locus Problems
Sometimes, we are interested to find all points (complex numbers) z satisfying a given mathematical condition.
Example. Find the locus of all points z satisfying the equation
SOLUTION.
−
z i = 1. z + 1
z − i = z + 1 x + iy − i = x + iy + 1 x + i(y − 1) = (x + 1) + iy x2 + (y − 1)2 = (x + 1)2 + y2 x2 + (y − 1)2 = (x + 1)2 + y2 x2 + y2 − 2y + 1 = x2 + 2x + 1 + y 2 −2y = 2x y = −x
We identify this Cartesian equation as the straight line passing through the origin and having a slope of 1. However, in the original complex equation given, z + 1 appears in the denominator. Therefore, z cannot take the value 1. We need to check if this point is on the line or not. The complex number z = 1 = 1 + 0i is represented by the point ( 1, 0). This point is not on the line y = x so there is nothing to remove from the line.
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Exercises. Describe the loci of the following equations by writing them in Cartesian form:
1. z
− 4 + 3i = 2
−
z 1 2. = z + 2
√ 3
−
z + 2i 3. =1 z 3i
4. arg
z + i z 1
−
=
− π4