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Unformatted text preview: 11 Complex numbers Read: Boas Ch. 2 Represent an arb. complex number z ∈ C in one of two ways: z = x + iy ; x, y ∈ R “rectangular” or ”Cartesian” form” z = re iθ ; r, θ ∈ R “polar” form . (1) Here i is √ 1, engineers call it j (ychh! The height of bad taste.). If z 1 = z 2 , both real and imaginary parts are equal, x 1 = x 2 and y 1 = y 2 . This implies of course that “0” ∈ C ” is the complex number with both real and imaginary parts 0. Defs.: The complex conjugate of z = x + iy is defined to be z * = x iy (Boas sometimes calls the same thing z ). Note zz * = x 2 + y 2 , a real number ≥ 0. The modulus or magnitude of z is then defined to be  z  = √ zz * = p x 2 + y 2 , has obvious analogies to the distance function in Euclidean space. Some useful relations you can easily work out are Re z = z + z * 2 ; Im z = z z * 2 i ; ( z 1 z 2 ) * = z * 1 z * 2 . (2) You should practice simplifying complex numbers which are not given explicitly in the form x + iy . For example, √ 2 + i 1 i = √ 2 + i 1 i 1 + i 1 + i ¶ = ( √ 2 + i )(1 + i ) 2 = √ 2 1 2 + i √ 2 + 1 2 (3) Polar form: Just as all real numbers can be represented as points on a line, complex numbers can be represented and manipulated as points in a 2D space spanned by the real and imaginary parts x and y . Euler theorem Recall the expansion of the exponential function e x = ∑ n x n /n !, which converges for arbitrary size of x . Let’s define a complex exponential e z in the same way, and choose in particular z = iθ . Then e iθ = ∞ X n =0 ( iθ ) n n ! = ∞ X n =0 ( 1) n ( θ ) 2 n (2 n )!  {z } + i ∞ X n =0 ( 1) n ( θ ) 2 n +1 (2 n + 1)!  {z } even terms, cos θ odd terms, sin θ = cos θ + i sin θ, (4) 1 Figure 1: Representation of a complex number z and its conjugate z * . where I used i 2 = 1, i 3 = i , i 4 = 1, etc. and identified the series for sin and cos of a real variable! Then going back and forth between rectangular and polar form is as easy as z = x + iy = r [ x r + i y r ] = r [cos θ + i sin θ ] = re iθ (5) Note cos θ = Re e iθ = e iθ + e iθ 2 ; sin θ = Im e iθ = e iθ e iθ 2 i . (6) Now we can see an interesting relationship between these functions and their hy perbolic analogs, cosh θ = e θ + e θ 2 ; sinh θ = e θ e θ 2 . (7) 11.1 Roots and powers of complex variable Powers: z = a + ib = re iθ ; z n = r n e inθ (8) Ex.: (1 + i ) 3 = ( √ 2 e iπ/ 4 ) 4 = 4 e iπ = 4 Roots require a bit more care since they are fractional powers: n √ z = n √ re iθ = r 1 /n e i ( θ +2 mπ ) /n m = 0 , ± 1 , ± 2 , ... (9) This seems a little paradoxical. Adding 2 mπ to the argument of the exponential doesn’t change z . Nevertheless it has different n th roots according to how you...
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This note was uploaded on 04/17/2008 for the course PHZ 3113 taught by Professor Hirshfeld during the Fall '07 term at University of Florida.
 Fall '07
 HIRSHFELD

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