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Unformatted text preview: Complex numbers, the exponential function, and factorization over C 1 Complex Numbers Recall that for every nonzero real number x , its square x 2 = x · x is always positive. Consequently, R does not contain the square roots of any negative number. This is a serious problem which rears its head all over the place. It is a nontrivial fact, however, that any positive number has two square roots in R , one positive and the other negative; the positive one is denoted √ x . One can show that for any x in R ,  x  = √ x · x. So if we can somehow have at hand a square root of − 1, we can find square roots of any real number. This motivates us to declare a new entity, denoted i , to satisfy i 2 = − 1 . One defines the set of complex numbers to be C = { x + iy  x, y ∈ R } and defines the basic arithmetical operations in C as follows: ( x + iy ) ± ( x ′ + iy ′ ) = ( x ± x ′ ) + i ( y ± y ′ ) , and ( x + iy )( x ′ + iy ′ ) = ( xx ′ − yy ′ ) + i ( xy ′ + x ′ y ) . 1 There is a natural onetoone function R → C , x → x + i. , compatible with the arithmetical operations on both sides. It is an easy exercise to check all the field axioms, except perhaps for the existence of multiplicative inverses for nonzero complex numbers. To this end one defines the complex conjugate of any z = x + iy in C to be z = x − iy. Clearly, R = { z ∈ C  z = z } . If z = x + iy , we have by definition, z z = x 2 + y 2 . In particular, z z is either 0 or a positive real number. Hence we can find a nonnegative square root of z z in R . Define the absolute value , sometimes called modulus or norm , by  z  = √ z z = √ x 2 + y 2 . If z = x + iy is not 0, we will put z − 1 = z z z = x x 2 + y 2 − i y x 2 + y 2 . It is a complex number satisfying z ( z − 1 ) = z z z z = 1 . Done. It is natural to think of complex numbers z = x + iy as being ordered pairs ( x, y ) of real numbers. So one can try to visualize C as a plane with two perpendicular coordinate directions, namely giving the x and y parts. Note in particular that 0 corresponds to the origin O = (0 , 0), 1 to (1 , 0) and i with (0 , 1). Geometrically, one can think of getting from − 1 to 1 (and back) by rotation about an angle π , and similarly, one gets from i to its square − 1 by rotating by half that angle, namely π/ 2, in the counterclockwise direction. 2 To get from the other square root of − 1, namely − i , one rotates by π/ 2 in the clockwise direction. (Going counterclockwise is considered to be in the positive direction in Math.) Addition of complex numbers has then a simple geometric interpretation: If z = x + iy , z ′ = x ′ + iy ′ are two complex numbers, represented by the points P = ( x, y ) and Q = ( x ′ , y ′ ) on the plane, then one can join the origin O to P and Q , and then draw a parallelogram with the line segments OP and OQ as a pair of adjacent sides. If R is the fourth vertex of this parallelogram, it corresponds to z + z ′ . This is called the....
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This note was uploaded on 12/02/2011 for the course MA 2a taught by Professor Makarov,n during the Fall '08 term at Caltech.
 Fall '08
 Makarov,N
 Factoring, Square Roots, Complex Numbers

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