complex analysis - metucourses.me.metu.edu.tr/.../me210-14s-week13-complex_analysis.pdf · complex...
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ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 1/39
COMPLEX ANALYSIS
Complex Numbers
Complex numbers are useful in the fields such as:
� Solutions of some types of linear differential equations
� Electrical circuit analyses
� Inverse transformations
� Solutions of field problems
� etc.
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ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 2/39
The general form of a complex number z, is z = x + i y or z = a + i b
where
x = Re (z) is a real number and called as the real part of z .
y = Im (z) is a real number and called as the imaginary part of z .
Another representation of a complex number is its polar form which is obtained by
applying usual rectangular to polar coordinate transformation as
x = r cosθ & y = r sinθ
z = r [cosθ + i sinθ] = r ei θ = r θ
i = -1
r = |z| = = mod(z) is called as the modulus (magnitude) of z22 yx +
xy
θ = arctan = arg(z) is called as the argument (angle) of z
Using EulerUsing Euler’’s identity s identity eeiiθ = = coscosθ + i + i sinsinθ
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ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 3/39
y
x
y
z
x 0
r
θ
z–plane
The figure below denotes the geometric representation of a complex number in a
plane called complex z-plane , or Argand diagram , defined by a Cartesian
coordinate system whose abscissa is used to represent the real part of z, and
ordinate is used to represent the imaginary part of z.
[ ]
2 2
1
z = x + i y
r = x + y x = r cos( )
y = tan y = r sin( )
x
z = r cos( ) + i r sin( )
z = r cos( ) + i sin( )
θ
θ θ
θ θθ θ
−
z = r eiθ
z = r θ
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ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 4/39
Note the following :
Geometrically, the argument θ is the directed angle measured in radians from
positive x–axis in counterclockwise direction. For z = 0, this angle is undefined.
For a given z ≠ 0, this angle is determined only up to integer multiples of 2π.
The value of θ that lies in the interval –π < θ ≤ π is called as the principle value
of the argument of z and is denoted by Arg(z) , with capital A. Thus, – π < Arg(z) ≤ + π
It becomes extremely important to consider the quadrant of the z–plane in which
the point z lies when determining the value of the argument by using the above
equation.
The modulus r, as the distance of the point z to the origin, is a non-negative quantity;
i.e., r ≥ 0
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ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 5/39
Example:
Determine the modulus and argument of the following complex numbers.
z = - 1 + i ⇒
z = 3 - 4 i ⇒
z = - 5 - 12 i ⇒
211r =+= )(135rad2.3564
3π1
1arctanθ o≅=
−=
5169r =+= )53.1(rad0.92734
arctanθ o−−=
−=
1314425r =+= )112.6(rad1.9655
12arctanθ o−−=
−−=
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ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 6/39
There are two fundamental rules for the manipulation of complex numbers:
1. A complex number z = x + i y is zero iff both its real and imaginary parts are zero;
i.e., z = 0 iff x = 0 & y = 0
it follows that two complex numbers z1 = x1 + i y1 and z2 = x2 + i y2 are equal
iff both their real and imaginary parts are equal;
i.e., z1 = z2 iff x1 = x2 & y1 = y2
Example :
What are the x and y values that satisfy the equation
(x2y – 2) + i (x + 2xy – 5) = 0
x2y – 2 = 0 and x + 2xy – 5 = 0 → x = 1 and y = 2x = 4 and y = 1/8
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ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 7/39
2. Complex numbers obey the ordinary rules of algebra
(a + i b) ± (c + i d) = (a ± c) + i (b ± d)
(a + i b) (c + i d) = (a c – b d) + i (a d + b c)
(a + i b)2 = a2 – b2 + i (2 a b)
with the addition that
i2 = – 1 , i3 = – i , i4 = +1 , i5 = i
Addition and Subtraction
z = z1 ± z2 = (x1 ± x2) + i (y1 ± y2)
� Easier to perform in Cartesian form
� Similar to addition and subtraction of vectors in a plane
Triangle Inequality:
Length of any one side of a triangle is less than or equal to the sum of the lengths
of other two sides. That is r ≤ r1 + r2
, i6 = – 1 , i7 = – i , ..... modularity
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ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 8/39
Minus, Conjugate, Addition, and Subtraction in Polar Coordinates
x
y
z = a + b i
_
z = a - i b- z = - a - b i
x
y
z1
z2z1 + z2
- z2
z1 - z2
conjugate
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ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 9/39
Multiplication
z = z1 z2 = (x1 x2 – y1 y2) + i (x1 y2 + y1 x2) = r1r2 θ1 + θ2
r = r1 r2
θ = θ1 + θ2
� Easier to perform in polar form
� Not similar to multiplication of vectors in plane
( )( )
( )
1 1 1 1
2 2 2 2
1 2 1 2 1 2 1 2
z = r cos( ) + i sin( )
z = r cos( ) + i sin( )
z z = r r cos( + ) + i sin( + )
θ θ
θ θ
θ θ θ θ
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ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 10/39
Division
1
2
1 2 1 2 1 2 1 22 2 2 22 2 2 2
z x x + y y y x - x yz = = + i
z x + y x + y
r = r1 / r2
θ = θ1 – θ2
� Easier to perform in polar form
� Not similar to any operation of vectors in plane
[ ][ ]
[ ]
1 1 1 1
2 2 2 2
1 11 2 1 2 2
2 2
z = r cos( ) + i sin( )
z = r cos( ) + i sin( )
z r = cos( - ) + i sin( - ) , r 0
z r
θ θ
θ θ
θ θ θ θ ≠
212
1 θθ rr −=
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ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 11/39
Complex conjugate
The complex conjugate of a complex number z = x + i y is defined as_
z = x - i y
Another practical way to obtain division of two complex numbers is to use the
conjugate of the denominator:
Example:
2 2 2 2
2 2
a + i b a + i b c - i d a c + b d b c - a d = = + i
c + i d c + i d c - i d c + d c + d
c + d 0
≠
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ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 12/39
Note the following:
Letz = x + iy → = x – iy
then(x + iy)2 + 2(x – iy) = x2 – y2 + i2xy + 2x – i2y = –1 + i6
which gives the following two real equationsx2 – y2 + 2x = –1 → y = ± (x+1)
2xy – 2y = 6 → y(x–1) = 3whose solutions are found as
x = 2 & y = 3 and x = –2 & y = –1Hence, the required solution for z is found as
z1 = 2 + i3 and z2 = – 2 – i
2 z + z
Re(z)x2x = z + z ==→i2
z zIm(z)yi2y = z z
−==→−
222 |z| = y+ x=z z 2121 zz)z(z ±=± 2121 zz)z(z =2
1
2
1
zz
zz =
Example :
Solve the following (find z): i6 1 z2 z2 +−=+
z
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ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 13/39
Integer powers of complex numbers (de Moivre’s formu la)
The idea of product of two complex numbers can be extended to the product of n complex numbers
If all the complex numbers multiplied are the same, the above expression gives an important result
zn = rn einθ = rn [cosθ + i sinθ]n = rn [cos(nθ) + i sin(nθ)]
where n is either an integer or a rational number (that is, n=p/q where p and q areintegers) with the condition that z ≠ 0 for n = –1.
For r = 1, the expression reduces the form
(cosθ + i sinθ)n = cos(nθ) + i sin(nθ)
called as the de Moivre’s formula named after Abraham de Moivre (1667-1754)
∑
=…=…= =∏∏
=
+++
=
n
1kk
n21
θin
1k
k)θ...θi(θ
n21n21
n
1k
k erer rr z zzz
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ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 14/39
Example: Find z5 for z = 1 + i 3
Solution 1 : Taking the power of a binomial
( )55 2 3 4 5z = 1 + i 3 = 1 + 5 ( i 3) + 10 ( i 3) + 10 ( i 3) + 5 ( i 3) + ( i 3)
= 1 + i 5 3 - 30 - i 30 3 + 45 + i 9 3
= 16 - i 16 3
Solution 2 : Using de Moivre’s formula
( )5 55
5 i 5 3
5i 5 5 3
1 3z = 1 + i 3 = 2 + i = 2 cos + i sin = 2 e
2 2 3 3
5 5 = 2 cos + i sin = 2 e
3 3
1 3 = 32 - i = 16 - i 16 3
2 2
π
π
π π
π π
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ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 15/39
Let
z = wn where w = R eiφ = ?
thenthen
z = Rn (cos nφ + i sin nφ) = Rn einφ
Hence
r (cos θ + i sin θ) = Rn (cos nφ + i sin nφ) or r eiθ = Rn einφ
De Moivre’s formula may also be employed to evaluate the nth root of a complex
number z = r (cos θ + i sin θ)
n 1/n nr = R or R = r = r
Therefore
1/n 1/n θ + 2 k θ + 2 k z = r cos + i sin , k = 0, 1, ..., n-1
n nπ π
and cos θ = cos nφ & sin θ = sin nφ
1n , 1, 0, k,n2kπθ
or2kπθn −…=+=φ+=φ
w =w =
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ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 16/39
Example: Find 3 8
Let z = 8 + i 0
1/3 1/3 0 + 2 k 0 + 2 k z = 8 cos + i sin , k = 0, 1, 2
3 3π π
1/31z = 8 cos(0) + i sin(0) = 2
1/32
0 + 2 0 + 2 1 3z = 8 cos + i sin = 2 - + i = - 1 + i 3
3 3 2 2π π
1/33
0 + 4 0 + 4 1 3z = 8 cos + i sin = 2 - - i = - 1 - i 3
3 3 2 2π π
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ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 17/39
Note that all solutions have a common modulus of 2, but their arguments differ
from each other, and they are located on a circle of radius of 2 about the origin of
the z–plane, equally spaced around it with an incremental angle of 2π/3 as
illustrated
(81/3)2
Im
Re
(81/3)1
i2
0 2
2π/3
z–plane
(81/3)3
4π/3
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ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 18/39
Example: Find i
Let i /2z = 0 + i = 1 e π
1/2 1/2 /2 + 2 k /2 + 2 k z = 1 cos + i sin , k = 0, 1
2 2π π π π
i /41
2 2z = 1 cos + i sin = e = + i
4 4 2 2ππ π
i 5 /42
5 5 2 2z = 1 cos + i sin = e = - - i
4 4 2 2ππ π
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ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 19/39
Complex Functions
Let z and w are two complex variables defined as z = x + i y and w = u + i v
where x, y, u, and v are real variables.
If, for each value of z in some portion of the complex z–plane, one or more value(s)
of w are defined, then w is said to be a complex function of z.
w = f(z) = u + i v = f(x + i y) = u(x,y) + i v(x,y)
This complex functional relationship between z and w may be regarded as a
complex mapping or complex transformation of points P within a region in
the z–plane (called as the Domain ) to corresponding image point(s) Q within a
region in the w–plane (called as the Range ).
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ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 20/39
w = f(z) mapping
y
x
domain
P
z–plane v
u
range
Q
w–plane
Complex Functions as Complex Mapping
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ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 21/39
Example:
Find the ranges, R, in the complex w–plane of the following complex functions
corresponding to their domains, D, in z–plane. Plot D and R regions.
Use the standard notation, z = x + i y = r eiθ & w = u + i v
(a) w = f(z) = i z , D: Re(z) ≥ 0
w = i z = i (x + i y) = – y + i x
u(x,y) = – y & v(x,y) = x
Re(z) = x ≥ 0 => v(x,y) ≥ 0
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ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 22/39
(b) w = f(z) = 3 z – π , D: – π ≤ Re(z) ≤ π
w = 3 z – π = (3x – π) + i 3 y
u(x,y) = 3 x – π & v(x,y) = 3 y
–π ≤ x ≤ π => – 4 π ≤ u(x,y) ≤ 2 π
(c) w = f(z) = z2 , D: |z| ≤ 1 and 0 ≤ Arg(z) ≤ π /4
w = z2 = (r ei θ)2 = r2 ei 2θ
|w| = r2 = |z|2 & Arg(w) = 2 Arg(z)
|z| ≤ 1 => |w| ≤ 1
0 ≤ Arg(z) ≤ π /4 => 0 ≤ Arg(w) ≤ π /2
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ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 23/39
In dealing with complex functions, it is possible to distinguish the following two cases.
� Complex valued functions of a real variable
� Complex valued functions of a complex variable
Complex valued functions of a real variable
In this case, to each value of a real variable, t (a ≤ t ≤ b), one or more complex
value(s) of z are assigned, which can be shown as z(t) = x(t) + i y(t) = f(t)
Some examples for this type of complex function are:
z = ei t (0 ≤ t ≤ 2 π) => z = cos(t) + i sin(t) => x(t) = cos(t) & y(t) = sin(t)
z = t + i t2 (for all t) => x(t) = t & y(t) = t2
z = (1 – i ) (1 + i 2 ) (t ≥ 0) => x(t) = (1 + 2 t) & y(t) = tt t t t
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ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 24/39
Example:
For the slider-crank mechanism shown in the figure, it is desired to represent the
position of the point Q as a (complex valued) function of the horizontal position t
(a real variable) of the slider P as z(t) = x(t) + i y(t) (h – r ≤ t ≤ h + r)
y
x
P
Q
θ
r h
β
t
This form can conveniently be used to represent the parametric equations of planar
curves in complex z–plane.
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ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 25/39
Note that as P moves on a straight line, Q moves on a circle.
For every position of P denoted by t, it is possible to find two positions for Q denoted by z.
Therefore, the relationship between the positions of P and Q can be considered as
a mapping of points lying on a straight line to points lying on a curve (circle).
By using either θ or β angle, the coordinates x and y of Q can be related to the
position t of P.
y
x
P
Q
θ
r h
β
t
![Page 26: COMPLEX ANALYSIS - METUcourses.me.metu.edu.tr/.../ME210-14S-Week13-Complex_Analysis.pdf · COMPLEX ANALYSIS Complex Numbers Complex numbers are useful in the fields such as: Solutions](https://reader034.vdocuments.mx/reader034/viewer/2022042508/5f56af80af0e2f32646fbfa3/html5/thumbnails/26.jpg)
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 26/39
y
x
P
Q
θ
r h
β
t
x(t) = r cos(θ) = t – h cos(β)
y(t) = r sin(θ) = h sin(β)
Since x2 + y2 = r2 => [t – h cos(β)]2 + [h sin(β)]2 = r2
2 2 2t + h - rcos(β) =
2 h t
22 2 2t + h - rsin( ) = 1 -
2 h tβ
±
![Page 27: COMPLEX ANALYSIS - METUcourses.me.metu.edu.tr/.../ME210-14S-Week13-Complex_Analysis.pdf · COMPLEX ANALYSIS Complex Numbers Complex numbers are useful in the fields such as: Solutions](https://reader034.vdocuments.mx/reader034/viewer/2022042508/5f56af80af0e2f32646fbfa3/html5/thumbnails/27.jpg)
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 27/39
The parametric representation of the position of Q in the complex z–plane becomes
( )
−+−±
+−=
−+−±
−+−=
t2
rhtth4i
t2
rht
ht2
rht1hi
t2
rhtt)t(z
222222222
2222222
y
x
P
Q
θ
r h
β
t
![Page 28: COMPLEX ANALYSIS - METUcourses.me.metu.edu.tr/.../ME210-14S-Week13-Complex_Analysis.pdf · COMPLEX ANALYSIS Complex Numbers Complex numbers are useful in the fields such as: Solutions](https://reader034.vdocuments.mx/reader034/viewer/2022042508/5f56af80af0e2f32646fbfa3/html5/thumbnails/28.jpg)
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 28/39
For r = 1 and h = 4, x(t) and y(t) are shown for 3 ≤ t ≤ 5 in the Figure.
-1.00
-0.75
-0.50
-0.25
0.00
0.25
0.50
0.75
1.00
3.0 3.5 4.0 4.5 5.0Real variable t
y(t)
x(t)
Position of Q as a function of Slider Position
![Page 29: COMPLEX ANALYSIS - METUcourses.me.metu.edu.tr/.../ME210-14S-Week13-Complex_Analysis.pdf · COMPLEX ANALYSIS Complex Numbers Complex numbers are useful in the fields such as: Solutions](https://reader034.vdocuments.mx/reader034/viewer/2022042508/5f56af80af0e2f32646fbfa3/html5/thumbnails/29.jpg)
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 29/39
Complex valued functions of a complex variable
In this case, to each value of a complex variable z in a domain D of the z-plane,
one or more complex values w are assigned, which can be shown as w = f(z)
For a given complex variable z, a complex value w may be obtained regardless
of the function, f, itself being a real function or a complex function.
Some examples for this type of complex function are:
w = i z → i (x + i y) = – y + i x → u = – y & v = x
w = z2 → (x + i y)2 = (x2 – y2) + i 2 x y → u = x2 – y2 & v = 2 x y
w = z1/4 → (r eiθ)1/4 = R eiΦ → R = r1/4 & Φ = (θ + 2 k π) / 4
w = ez → ex + iy = ex (cos y + i sin y) → u = ex cos y & v = ex sin y
![Page 30: COMPLEX ANALYSIS - METUcourses.me.metu.edu.tr/.../ME210-14S-Week13-Complex_Analysis.pdf · COMPLEX ANALYSIS Complex Numbers Complex numbers are useful in the fields such as: Solutions](https://reader034.vdocuments.mx/reader034/viewer/2022042508/5f56af80af0e2f32646fbfa3/html5/thumbnails/30.jpg)
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 30/39
Note that once w = f(z) is known, it is a straightforward algebra to obtain u & v (or R
& f) in terms of x & y (or r & q).
However, if u(x,y) & v(x,y) are given in turn, to express w as a function of z is not a
trivial problem at all.
� In fact, it may even not have a solution.
� If there is a solution, then a suitable manipulation must be carried out in order
to come up with the correct w = f(z) expression.
Example :
Given w = u(x,y) + i v(x,y) = (x2 + x – y2 + 1) + i y (2x + 1) express w as a function
of z = x + i y
Rearranging gives
w = x2 + x – y2 + 1 + i 2 x y + i y
= x2 + i 2 x y – y2 + x + i y + 1
= (x + i y)2 + (x + i y) + 1
= z2 + z + 1
![Page 31: COMPLEX ANALYSIS - METUcourses.me.metu.edu.tr/.../ME210-14S-Week13-Complex_Analysis.pdf · COMPLEX ANALYSIS Complex Numbers Complex numbers are useful in the fields such as: Solutions](https://reader034.vdocuments.mx/reader034/viewer/2022042508/5f56af80af0e2f32646fbfa3/html5/thumbnails/31.jpg)
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 31/39
Some elementary functions of z
Exponential function: Using the basic definition of exponential function for real
variables, one gets
( )1
n 2 3 4z z z z z
f z = e = = 1 + z + + + +…n! 2! 3! 4!
∞
=∑k
u(x,y) = ex cos(y) & v(x,y) = ex sin(y) Mod(ez) = |ez| = ex & Arg(ez) = y
Note that ( )1
n 2 3 4- z (-z) z z z
f z = e = = 1 - z + - + - …n! 2! 3! 4!
∞
=∑k
( )1
n 2 3 4i z (i z) z i z z
f z = e = = 1 + i z - - + - …n! 2! 3! 4!
∞
=∑k
( )1
n 2 3 4- i z (- i z) z i z z
f z = e = = 1 - i z - + + - …n! 2! 3! 4!
∞
=∑k
![Page 32: COMPLEX ANALYSIS - METUcourses.me.metu.edu.tr/.../ME210-14S-Week13-Complex_Analysis.pdf · COMPLEX ANALYSIS Complex Numbers Complex numbers are useful in the fields such as: Solutions](https://reader034.vdocuments.mx/reader034/viewer/2022042508/5f56af80af0e2f32646fbfa3/html5/thumbnails/32.jpg)
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 32/39
Hyperbolic functions:
cosh(z) = (ez + e–z)/2 = cosh(x) cos(y) + i sinh(x) sin(y)
sinh(z) = (ez – e–z)/2 = sinh(x) cos(y) + i cosh(x) sin(y)
If z = 0 + iy => cosh(iy) = cos(y) sinh(iy) = i sin(y)
Familiar laws for the hyperbolic functions:
cosh2(z) – sinh2(z) = 1
cosh(z1 + z2) = cosh(z1) cosh(z2) + sinh(z1) sinh(z2)
sinh(z1 + z2) = sinh(z1) cosh(z2) + cosh(z1) sinh(z2)
cosh(2 z) = cosh2(z) + sinh2(z) = 1 + 2 sinh2(z) = 2 cosh2(z) – 1
sinh(2 z) = 2 sinh(z) cosh(z)
![Page 33: COMPLEX ANALYSIS - METUcourses.me.metu.edu.tr/.../ME210-14S-Week13-Complex_Analysis.pdf · COMPLEX ANALYSIS Complex Numbers Complex numbers are useful in the fields such as: Solutions](https://reader034.vdocuments.mx/reader034/viewer/2022042508/5f56af80af0e2f32646fbfa3/html5/thumbnails/33.jpg)
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 33/39
Trigonometric functions:
cos(z) = (eiz + e–iz)/2 = cos(x) cosh(y) – i sin(x) sinh(y)
sin(z) = (eiz – e–iz)/2i = sin(x) cosh(y) + i cos(x) sinh(y)
If z = 0 + iy => cos(iy) = cosh(y) sin(iy) = i sinh(y)
Familiar laws for the trigonometric functions:
cos2(z) + sin2(z) = 1
cos(z1 + z2) = cos(z1) cos(z2) – sin(z1) sin(z2)
sin(z1 + z2) = sin(z1) cos(z2) + cos(z1) sin(z2)
cos(2 z) = cos2(z) – sin2(z) = 1 – 2 sin2(z) = 2 cos2(z) – 1
sin(2 z) = 2 sin(z) cos(z)
![Page 34: COMPLEX ANALYSIS - METUcourses.me.metu.edu.tr/.../ME210-14S-Week13-Complex_Analysis.pdf · COMPLEX ANALYSIS Complex Numbers Complex numbers are useful in the fields such as: Solutions](https://reader034.vdocuments.mx/reader034/viewer/2022042508/5f56af80af0e2f32646fbfa3/html5/thumbnails/34.jpg)
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 34/39
Logarithmic function:
The logarithm of z = r eiθ that is defined implicitly as the function w = ln(z) which
satisfies the equation z = ew or r eiθ = eu + i v = eu eiv
Hence, eu = r or u = ln(r), and v = θ
Thus, w = u + i v = ln(r) + i θ = ln |z| + i arg(z)
If the principal argument of z is denoted by Arg(z), then this equation can
be rewritten as ln(z) = ln |z| + i [Arg(z) + 2 k π] k = 0, ±1, ±2, …
This indicates that complex logarithmic function is infinitely multi–valued.
For k = 0, the part (branch) of the logarithmic function is called as the principal
value .
![Page 35: COMPLEX ANALYSIS - METUcourses.me.metu.edu.tr/.../ME210-14S-Week13-Complex_Analysis.pdf · COMPLEX ANALYSIS Complex Numbers Complex numbers are useful in the fields such as: Solutions](https://reader034.vdocuments.mx/reader034/viewer/2022042508/5f56af80af0e2f32646fbfa3/html5/thumbnails/35.jpg)
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 35/39
Familiar laws for the logarithms of real quantities all hold for the logarithms of
complex quantities in the following sense:
ln(z1z2) = ln(z1) + ln (z2)
ln(z1/z2) = ln(z1) – ln (z2)
ln(zk) = k ln(z) k = 0, ±1, ±2, …
Example:
Compute w(z) = ln(1 – i)
ln(z) = ln |z| + i [Arg(z) + 2 k π] k = 0, ±1, ±2, …
( ) L2,1,0,k,2kπ4π
i2lni)ln(1 ±±=
+−+=−
( ) ( ) ( ) L,4
9πi2ln,
47π
i2ln,4π
i2lni)ln(1 −+−=−
–iπ/4
ln( )2
i7π/4
i15π/4
–i9π/4
i23π/4
–i17π/4
w-plane
Re(w)
Im(w)
![Page 36: COMPLEX ANALYSIS - METUcourses.me.metu.edu.tr/.../ME210-14S-Week13-Complex_Analysis.pdf · COMPLEX ANALYSIS Complex Numbers Complex numbers are useful in the fields such as: Solutions](https://reader034.vdocuments.mx/reader034/viewer/2022042508/5f56af80af0e2f32646fbfa3/html5/thumbnails/36.jpg)
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 36/39
General Powers of z
w(z) = zc , c is any number, real or complex
( )0c ln(r) + i + 2 k c ln(z)z = e = e , k = 0, 1, 2, ...c θ π ± ±
If c = n:( ) ( )0 0 0
ln rn ln(r) + i i n i n n ln(z)
i 2 n k
z = e = e = e e = r e
e = 1
nn nθ θ θ
π
Because for all k’s
If c = m/n: ( )
( ) ( )
0i m/n + 2 k / /
/0 0
z = r e
m m = r cos + 2 k + i sin + 2 k
n n
k = 0, 1, 2, ...
m n m n
m n
θ π
θ π θ π
, n, n--11
![Page 37: COMPLEX ANALYSIS - METUcourses.me.metu.edu.tr/.../ME210-14S-Week13-Complex_Analysis.pdf · COMPLEX ANALYSIS Complex Numbers Complex numbers are useful in the fields such as: Solutions](https://reader034.vdocuments.mx/reader034/viewer/2022042508/5f56af80af0e2f32646fbfa3/html5/thumbnails/37.jpg)
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 37/39
If c is an irrational number, not expressible in the form m/n
( )
( )( ) ( )( )0i c + 2 k
0 0
z = r e
= r cos c + 2 k + i sin c + 2 k
k = 0, 1, 2, ...
c c
c
θ π
θ π θ π
± ±
If c is complex, i.e., c = a + i b
( )
( ) ( )
( ) ( )( )( )( )
0
00
0
(a + i b) ln(r) + i + 2 k
i b ln(r) 4 a + 2 k a ln(r) - b + 2 k
0a ln(r) - b + 2 k
0
z = e
= e e
cos b ln(r) + a + 2 k = e
+ i sin b ln(r) + a + 2 k
k =
c θ π
θ πθ π
θ πθ π
θ π
0, 1, 2, ... ± ±
+
![Page 38: COMPLEX ANALYSIS - METUcourses.me.metu.edu.tr/.../ME210-14S-Week13-Complex_Analysis.pdf · COMPLEX ANALYSIS Complex Numbers Complex numbers are useful in the fields such as: Solutions](https://reader034.vdocuments.mx/reader034/viewer/2022042508/5f56af80af0e2f32646fbfa3/html5/thumbnails/38.jpg)
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 38/39
Example:
Find all possible values of (1)i
Summary: zc is
single valued if c is an integer
n-valued if c = 1/n
n-valued if c = m/n
infinite valued if c is real and irrational
infinite valued if Im(c) ≠ 0
→→ r = 1 , r = 1 , θθoo = 0 , c = i = 0 , c = i
Example:
Find all possible values of (i)i →→ r = 1 , r = 1 , θθoo = = ππ/2/2 , c = i , c = i
...2,1,0,k,eee(1) 2kπ)]2ki(0i[ln(1)ln(1)ii ±±==== −π++
...2,1,0,k,eee(i) 21)4k)]2k2
i(i[ln(1)ln(i)ii ±±==== π+−π+π+ /(
![Page 39: COMPLEX ANALYSIS - METUcourses.me.metu.edu.tr/.../ME210-14S-Week13-Complex_Analysis.pdf · COMPLEX ANALYSIS Complex Numbers Complex numbers are useful in the fields such as: Solutions](https://reader034.vdocuments.mx/reader034/viewer/2022042508/5f56af80af0e2f32646fbfa3/html5/thumbnails/39.jpg)
ME 210 Applied Mathematics for Mechanical Enginee rs
Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 39/39
END OF WEEKEND OF WEEKEND OF WEEKEND OF WEEKEND OF WEEKEND OF WEEKEND OF WEEKEND OF WEEK 1313131313131313