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by: AVDHUT VYAS

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# MATHS 1ST WEEK NOTE ME-5332

AVDHUT VYAS
UTA
GPA 2.0

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IT INCLUDES BASIC INFORMATION OF COMPLEX NUMBER.
COURSE
engineering analysis
PROF.
C.luo
TYPE
Class Notes
PAGES
4
WORDS
CONCEPTS
MATHS CALCULATION
KARMA
25 ?

## Popular in Aerospace Engineering

This 4 page Class Notes was uploaded by AVDHUT VYAS on Tuesday March 15, 2016. The Class Notes belongs to ME-5332 at University of Texas at Arlington taught by C.luo in Spring 2016. Since its upload, it has received 24 views. For similar materials see engineering analysis in Aerospace Engineering at University of Texas at Arlington.

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Date Created: 03/15/16
Complex Variables Why is it necessary to learn the complex variable theory ? † Useful for certain types of integrals (the residue theorem). Z ∞ 4 Z 2π x cos θ (Examples:) 6 dx; 2 dθ 0 1+x 0 1+sin θ 1 † Useful for 2-D ﬂuid mechanics/solid mechanics problems . Δu = Δv = 0; w = u+iv: 2 † Real variable theories are subsets of the complex variable theory (the fundamental theorem of algebra ). With a complex number in the form of z = a+ib; p where i = ¡1, divisions and multiplications between two complex numbers are computed as z1 2 = (a 1ib )1a +i2 ) 2 = (a a ¡b b )+i(a b +a b ); (1) 1 2 1 2 1 2 2 1 and z (a +ib )(a ¡ib ) 1 = 1 1 2 2 z2 (a2+ib2)(2 ¡i2 ) (2) = a1 2+b 2 2+i a2 2¡a 2 2: a2+b 2 a2+b 2 The complex conjugate of z = a+ib is deﬁned as ﬂ= a¡ib: It follows z1+z 2 = z1+z ;2 z1 2 = z1 2; (3) ‡ · z1 = 1: z2 2 Note that 2 zﬂ= jzj : 1 3 The real part and the imaginary part of any analytic functions (e.g. f(z) = z ) satisfy the Laplace equation. For instance, z = (x+iy) = x ¡3xy +i(3x y¡y );3 so u = x ¡3xy ; v = 3x y¡y : Verify that both u and v satisfy Δu = 0; Δv = 0: 2An algebraic equation of the n-th order has n zeros. Complex Plane With the introduction of the imaginary number, i, it is possible to associate z = a+ib with a point (a;b) in a 2-D plane. y z=a+i b r b q x a p 2 2 jzj = p a +b (4) = zﬂ; and argz = θ: It is noted that z = a+ib µ ¶ p = a +b 2 p a +i p b a +b a +b2 (5) = jzj(cos θ +isin θ) = re ; where Euler’s formula iθ 3 e = cos θ +isin θ; was used. On the complex plane, it is possible to visualize the operations among complex numbers as shown in the table below: z1§z 2 addition of vectors i(1 +2 ) z1 2 r1 2e (7) z1 1ei(1 ¡θ2) z2 2 Using Euler’s formulas, iθ e = cos θ +isin θ; (8) e¡iθ = cos θ ¡isin θ; trigonometric functions can be written as cos θ = e +e¡iθ; iθ2¡iθ (9) sin θ = e ¡e : 2i 3 2 3 4 eiθ = 1+(iθ)+ (i) + (iθ)+ (iθ::: ‡ θ θ4 ·! ‡ 4!θ3 θ5 · = 1¡ 2!+ 4!¡::: +i θ ¡ 3!+ 5!¡::: (6) = cos θ +isin θ: It is seen from the above that sine and cosine functions are the even and odd parts of f(θ) = e . Similarly, the hyperbolic sine and cosine functions can be deﬁned as (the even and odd parts of f(θ) = e ) cosh θ = e+e¡ θ; θ 2¡ θ (10) sinh θ = e¡e : 2 The following similarity can be immediately veriﬁed: (sin θ) = cos θ (sinh θ) = cosh θ 0 0 (cos θ) = ¡sin θ (cosh θ) = sinh θ cos θ +sin θ = 1 cosh θ ¡sinh θ = 1 Example 1 Evaluate sin i. ii ¡ii sin i = e ¡e e1¡e = 2i (11) = i(e¡ ): 2 e Note that this is a pure imaginary number. Example 2 Evaluate Z ∞ ¡x I · e sin xdx: 0 R ∞ ¡x ix 4 I = ℑ R0 e e dx = ℑ ∞e(¡1+i)xdx h0 i ∞ = ℑ 1 e(¡1+i)x ¡ ¡1+i ¢ 0 = ℑ 1 (0¡1) 5 (12) ¡ ¡¡1 ¡1¡i ¢ = ℑ ¡ ¡1+i ¢1¡i = ℑ 1+ i1 1 2 2 = 2: 6 Example 3 Solve sin z = 3 . Let z = x+iy, then, sin z = sin(x+iy) (13) 1 ‡ i(x+iy) ¡i(x+iy) = e ¡e (14) 2i 1 ¡ ¢ = cos x(e¡y¡e )+isin x(e ¡y+e ) ; (15) 2i so ¡y y ¡y y cos x(e ¡e )+isin x(e +e ) = 6i; (16) or by comparing the imaginary part and real part of both sides, cos x(e¡y¡e ) = 0; (17) sin x(e¡y+e ) = 6: (18) 4 5(f) is th(¡1+i)xn¡x ixart of f¡x ix As x ! ∞, e = e e ! 0 since e! 0 and e = cos x+isin x is bounded (but indeﬁnite). This equation does not make sense if z is a real number. If ey¡e = 0 in eq.(17), it follows = 1 or y = 0 which implies that sin x = 3 from eq.(18) which is impossible, so cos x = 0 must be satisﬁed from eq.(17). Therefore, π x = 2 §n π; (19) and π ¡y y sin( §nπ)(e +e ) = 6: (20) 2 For the right hand side to be positive, π sin( §nπ); 2 must be positive which implies that π x = §2n π: (21) 2 From eq.(20), ¡y y e +e = 6; (22) or (e ) ¡6e +1 = 0; (23) which can be solved as p y = ln(3§2 2); (24) so ﬁnally ¡π ¢ p z = §2n π +i ln(3§2 2): (25) 2

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