complex analysis - METU | Department of Mechanical Engineering
Transkript
complex analysis - METU | Department of Mechanical Engineering
ME 210 Applied Mathematics for Mechanical Engineers 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. Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 1/39 ME 210 Applied Mathematics for Mechanical Engineers The general form of a complex number z, is z = x + i y or z = a + i b where i = -1 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 Using Euler’s identity eiθ = cosθ cos + i sinθ sin z = r [cosθ + i sinθ] = r ei θ = r θ r = |z| = x 2 + y 2 = mod(z) is called as the modulus (magnitude) of z y θ = arctan = arg(z) x is called as the argument (angle) of z Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 2/39 ME 210 Applied Mathematics for Mechanical Engineers 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. y z–plane z=x+iy z θ = tan−1 x y θ x z = r [ cos(θ ) + i sin(θ )] x z = r eiθ z=r Prof. Dr. Bülent E. Platin x = r cos(θ ) y y = r sin(θ ) z = r cos(θ ) + i r sin(θ ) r 0 x 2 + y2 r= Spring 2014 – Sections 02 & 03 θ 3/39 ME 210 Applied Mathematics for Mechanical Engineers Note the following: The modulus r, as the distance of the point z to the origin, is a non-negative quantity; i.e., r ≥ 0 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. Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 4/39 ME 210 Applied Mathematics for Mechanical Engineers Example: Determine the modulus and argument of the following complex numbers. z=-1+i ⇒ r = 1+ 1 = 2 1 3π θ = arctan = ≅ 2.356 rad (135o ) − 1 4 z=3-4i ⇒ r = 9 + 16 = 5 −4 o θ = arctan = −0.927 rad ( −53.1 ) 3 z = - 5 - 12 i ⇒ r = 25 + 144 = 13 − 12 o θ = arctan = −1.965 rad ( −112.6 ) −5 Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 5/39 ME 210 Applied Mathematics for Mechanical Engineers 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 Prof. Dr. Bülent E. Platin → x = 1 and y = 2 x = 4 and y = 1/8 Spring 2014 – Sections 02 & 03 6/39 ME 210 Applied Mathematics for Mechanical Engineers 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 , i6 = – 1 , i7 = – i , ..... modularity 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 Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 7/39 ME 210 Applied Mathematics for Mechanical Engineers Minus, Conjugate, Addition, and Subtraction in Polar Coordinates y y z1 + z2 z2 z=a+bi z1 x x _ -z=-a-bi z =a-ib z1 - z2 conjugate Prof. Dr. Bülent E. Platin - z2 Spring 2014 – Sections 02 & 03 8/39 ME 210 Applied Mathematics for Mechanical Engineers 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 z1 = r1 ( cos(θ1) + i sin(θ1) ) z 2 = r2 ( cos(θ 2 ) + i sin(θ 2 ) ) z1 z2 = r1 r2 ( cos(θ1 + θ 2 ) + i sin(θ1 + θ 2 ) ) Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 9/39 ME 210 Applied Mathematics for Mechanical Engineers Division z= y1 x 2 - x1 y 2 r1 z1 x1 x 2 + y1 y 2 = + i = 2 2 z 2 x 22 + y 22 x + y 2 2 r2 θ1 − θ2 r = r1 / r2 θ = θ1 – θ2 Easier to perform in polar form Not similar to any operation of vectors in plane z1 = r1 [ cos(θ1) + i sin(θ1)] z2 = r2 [ cos(θ 2 ) + i sin(θ 2 )] z1 r1 = z2 r2 Prof. Dr. Bülent E. Platin [cos(θ1 - θ2 ) + i sin(θ1 - θ2 )] Spring 2014 – Sections 02 & 03 , r2 ≠ 0 10/39 ME 210 Applied Mathematics for Mechanical Engineers 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: a+ib a + i b c - i d a c + b d b c - a d = = + i c 2 + d2 c 2 + d2 c+id c+id c-id c 2 + d2 ≠ 0 Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 11/39 ME 210 Applied Mathematics for Mechanical Engineers Note the following: z + z = 2x 2 → 2 z+z x = Re(z) = 2 2 z z = x + y =| z | z − z = i2y (z1 ± z 2 ) = z1 ± z2 z−z → y = Im(z) = i2 (z1z 2 ) = z1z2 z1 z1 = z 2 z2 Example: Solve the following (find z): z 2 + 2z = − 1 + i6 Let z = x + iy → z = x – iy then (x + iy)2 + 2(x – iy) = x2 – y2 + i2xy + 2x – i2y = –1 + i6 which gives the following two real equations x2 – y2 + 2x = –1 → y = ± (x+1) 2xy – 2y = 6 → y(x–1) = 3 whose solutions are found as x=2&y=3 and x = –2 & y = –1 Hence, the required solution for z is found as z1 = 2 + i3 and z2 = – 2 – i Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 12/39 ME 210 Applied Mathematics for Mechanical Engineers Integer powers of complex numbers (de Moivre’s formula) The idea of product of two complex numbers can be extended to the product of n complex numbers n ∏z k = z1z 2 … zn = r1r2 … rne i(θ1 +θ2 + ... +θn ) k =1 = n i ∑ θk rk e k =1 k =1 n ∏ 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 are integers) 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) Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 13/39 ME 210 Applied Mathematics for Mechanical Engineers Example: Find z5 for z = 1 + i 3 Solution 1: Taking the power of a binomial ( 5 z = 1+i 3 ) 5 = 1 + 5 ( i 3) + 10 ( i 3)2 + 10 ( i 3)3 + 5 ( i 3)4 + ( i 3)5 = 1 + i 5 3 - 30 - i 30 3 + 45 + i 9 3 = 16 - i 16 3 Solution 2: Using de Moivre’s formula 5 ( z = 1+i 3 ) 5 = 2 5 1 3 + i = 2 2 2 i 5π 5π 5 = 2 cos + i sin =2 e 3 3 5 1 3 = 32 - i = 16 - i 16 2 2 Prof. Dr. Bülent E. Platin 5 i π π cos 3 + i sin 3 = 2 e π 3 5 5π 3 3 Spring 2014 – Sections 02 & 03 14/39 ME 210 Applied Mathematics for Mechanical Engineers De Moivre’s formula may also be employed to evaluate the nth root of a complex number z = r (cos θ + i sin θ) Let z = wn where w = R eiφ = ? then 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 Therefore r = Rn R = r1/n = n r or nφ = θ + 2kπ or 1/n w= z Prof. Dr. Bülent E. Platin =r 1/n and cos θ = cos nφ & sin θ = sin nφ , k = 0, 1, ..., n-1 θ + 2kπ φ= , k = 0, 1, … , n − 1 n θ + 2 k π θ+ 2k π + i sin cos n n Spring 2014 – Sections 02 & 03 15/39 ME 210 Applied Mathematics for Mechanical Engineers Example: Find Let 3 8 z=8+i0 0 + 2 k π 0 + 2 k π z1/3 = 81/3 cos + i sin 3 3 , k = 0, 1, 2 z1 = 81/3 cos(0) + i sin(0) = 2 0 + 2 π 0 + 2 π z2 = 81/3 cos + i sin 3 3 1/3 z3 = 8 Prof. Dr. Bülent E. Platin = 2 0 + 4 π 0 + 4 π cos + i sin =2 3 3 Spring 2014 – Sections 02 & 03 1 3 + i = - 1 + i 3 2 2 1 3 -i = - 1 - i 3 2 2 16/39 ME 210 Applied Mathematics for Mechanical Engineers 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 i2 z–plane 2π/3 (81/3)1 4π/3 0 2 Re (81/3)3 Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 17/39 ME 210 Applied Mathematics for Mechanical Engineers Example: Find Let i z = 0 + i = 1 ei π /2 π /2 + 2 k π π /2 + 2 k π z1/2 = 11/2 cos + i sin 2 2 , k = 0, 1 2 2 π π z1 = 1 cos + i sin = ei π /4 = +i 2 2 4 4 2 2 5 π 5 π i 5π /4 z2 = 1 cos + i sin = e = i 4 2 2 4 Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 18/39 ME 210 Applied Mathematics for Mechanical Engineers 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). Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 19/39 ME 210 Applied Mathematics for Mechanical Engineers Complex Functions as Complex Mapping y v z–plane P w = f(z) mapping Q domain range x Prof. Dr. Bülent E. Platin w–plane Spring 2014 – Sections 02 & 03 u 20/39 ME 210 Applied Mathematics for Mechanical Engineers 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 Prof. Dr. Bülent E. Platin => v(x,y) ≥ 0 Spring 2014 – Sections 02 & 03 21/39 ME 210 Applied Mathematics for Mechanical Engineers (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 & –π ≤ x ≤ π – 4 π ≤ u(x,y) ≤ 2 π => v(x,y) = 3 y (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 Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 22/39 ME 210 Applied Mathematics for Mechanical Engineers 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 = t (1 – i t ) (1 + i 2 t ) (t ≥ 0) => x(t) = t (1 + 2 t) & y(t) = t Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 23/39 ME 210 Applied Mathematics for Mechanical Engineers This form can conveniently be used to represent the parametric equations of planar curves in complex z–plane. 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) Q h r θ P y β x t Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 24/39 ME 210 Applied Mathematics for Mechanical Engineers Q h r θ P y β x t 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. Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 25/39 ME 210 Applied Mathematics for Mechanical Engineers Q h r θ P y β x 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 t2 + h2 - r 2 cos(β) = 2ht Prof. Dr. Bülent E. Platin t 2 + h2 - r 2 sin( β ) = ± 1 - 2 h t Spring 2014 – Sections 02 & 03 2 26/39 ME 210 Applied Mathematics for Mechanical Engineers Q h r θ P y β x t The parametric representation of the position of Q in the complex z–plane becomes 2 2 2 2 t +h −r t +h −r z ( t ) = t − ± i h 1 − 2t 2ht 2 2 2 ( t 2 − h 2 + r 2 4h 2 t 2 − t 2 + h 2 − r 2 = ± i 2t 2t Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 ) 2 27/39 ME 210 Applied Mathematics for Mechanical Engineers For r = 1 and h = 4, x(t) and y(t) are shown for 3 ≤ t ≤ 5 in the Figure. 1.00 0.75 y(t) 0.50 x(t) 0.25 0.00 -0.25 -0.50 -0.75 -1.00 3.0 3.5 4.0 Real variable t 4.5 5.0 Position of Q as a function of Slider Position Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 28/39 ME 210 Applied Mathematics for Mechanical Engineers 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=iz → → 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 Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 29/39 ME 210 Applied Mathematics for Mechanical Engineers 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 Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 30/39 ME 210 Applied Mathematics for Mechanical Engineers Some elementary functions of z Exponential function: Using the basic definition of exponential function for real variables, one gets ∞ zn z2 z3 z4 f (z) = e = ∑ =1+z+ + + +… 2! 3! 4! k =1 n! z u(x,y) = ex cos(y) & v(x,y) = ex sin(y) Note that f ( z ) = e- z f ( z ) = ei z f (z) = e Prof. Dr. Bülent E. Platin -iz Mod(ez) = |ez| = ex & Arg(ez) = y ∞ (-z)n z 2 z3 z4 =∑ =1-z+ + -… 2! 3! 4! k =1 n! ∞ (i z)n z 2 i z3 z4 =∑ =1+iz+ -… 2! 3! 4! k =1 n! ∞ (- i z)n z2 i z3 z4 =∑ =1-iz+ + -… n! 2! 3! 4! k =1 Spring 2014 – Sections 02 & 03 31/39 ME 210 Applied Mathematics for Mechanical Engineers 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) Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 32/39 ME 210 Applied Mathematics for Mechanical Engineers 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) Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 33/39 ME 210 Applied Mathematics for Mechanical Engineers Logarithmic function: The logarithm of z = r eiθ that is defined implicitly as the function w = ln(z) which satisfies the equation z = ew Hence, eu = r or u = ln(r), or r eiθ = eu + i v = eu eiv 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. Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 34/39 ME 210 Applied Mathematics for Mechanical Engineers Familiar laws for the logarithms of real quantities all hold for the logarithms of complex quantities in the following sense: Im(w) i23π/4 ln(z1z2) = ln(z1) + ln (z2) w-plane ln(z1/z2) = ln(z1) – ln (z2) ln(zk) = k ln(z) i15π/4 k = 0, ±1, ±2, … Example: i7π/4 Compute w(z) = ln(1 – i) ln( 2 ) ln(z) = ln |z| + i [Arg(z) + 2 k π] k = 0, ±1, ±2, … –iπ/4 Re(w) ( ) π ln(1− i) = ln 2 + i − + 2kπ , k = 0, ± 1, ± 2,L 4 ( ) ( ) ( ) –i9π/4 π 7π 9π ln(1− i) = ln 2 − i , ln 2 + i , ln 2 − i ,L 4 4 4 –i17π/4 Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 35/39 ME 210 Applied Mathematics for Mechanical Engineers General Powers of z w(z) = zc , c is any number, real or complex zc = ec ln(z) = e If c = n: n z =e n ln(z) Because If c = m/n: c ln(r) + i (θ0 + 2 k π ) =e n ln(r) + i (θ0 ) ei 2 n k π = 1 zm / n = r m / n e =e , k = 0, ± 1, ± 2, ... ( ) ei n θ = r n ei n θ ln r n 0 0 for all k’s i m/n (θ0 + 2 k π ) m m = r m / n cos θ + 2 k π + i sin θ + 2 k π (0 ) ) n ( 0 n k = 0, 1, 2, ... , n-1 Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 36/39 ME 210 Applied Mathematics for Mechanical Engineers If c is an irrational number, not expressible in the form m/n zc = r c e i c (θ0 + 2 k π ) = r c cos ( c (θ0 + 2 k π ) ) + i sin ( c (θ0 + 2 k π ) ) k = 0, ± 1, ± 2, ... If c is complex, i.e., c = a + i b zc = e =e =e (a + i b) ln(r) + i (θ0 + 2 k π ) a ln(r) - b (θ0 + 2 k π ) a ln(r) - b (θ0 + 2 k π ) e i b ln(r) +4 a (θ0 + 2 k π ) cos (b ln(r) + a (θ0 + 2 k π ) ) + i sin (b ln(r) + a (θ0 + 2 k π ) ) k = 0, ± 1, ± 2, ... Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 37/39 ME 210 Applied Mathematics for Mechanical Engineers Example: Find all possible values of (1)i → r = 1 , θo = 0 , c = i (1)i = ei ln(1) = ei[ln(1)+i(0+2kπ )] = e −2kπ , k = 0, ± 1, ± 2,... Example: → Find all possible values of (i)i π i[ln(1)+i( + 2kπ )] 2 (i)i = ei ln(i) = e r = 1 , θo = π/2 , c = i = e −( 4k +1)π / 2 , k = 0, ± 1, ± 2,... 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 Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 38/39 ME 210 Applied Mathematics for Mechanical Engineers END OF WEEK 13 Prof. Dr. Bülent E. Platin Spring 2014 – Sections 02 & 03 39/39