omple Numbers and Eponentials Definition and Basic Operations comple number is nothing more than a point in the plane. The sum and product of two comple numbers ( 1, 1 ) and ( 2, 2 ) is defined b ( 1, 1 )+( 2, 2 ) = ( 1 + 2, 1 + 2 ) ( 1, 1 )( 2, 2 ) = ( 1 2 1 2, 1 2 + 2 1 ) respectivel. It is conventional to use the notation + i (or in electrical engineering countr + j) to stand for the comple number (,). In other words, it is conventional to write in place of (,0) and i in place of (0,1). In this notation, the sum and product of two comple numbers z 1 = 1 +i 1 and z 2 = 2 +i 2 is given b z 1 +z 2 = ( 1 + 2 )+i( 1 + 2 ) The comple number i has the special propert z 1 z 2 = 1 2 1 2 +i( 1 2 + 2 1 ) i 2 = (0+1i)(0+1i) = (0 0 1 1)+i(0 1+1 0) = 1 For eample, if z = 1+2i and w = 3+4i, then z +w = (1+2i)+(3+4i) = 4+6i zw = (1+2i)(3+4i) = 3+4i+6i+8i 2 = 3+4i+6i 8 = 5+10i ddition and multiplication of comple numbers obe the familiar algebraic rules z 1 +z 2 = z 2 +z 1 z 1 z 2 = z 2 z 1 z 1 +(z 2 +z 3 ) = (z 1 +z 2 )+z 3 z 1 (z 2 z 3 ) = (z 1 z 2 )z 3 0+z 1 = z 1 1z 1 = z 1 z 1 (z 2 +z 3 ) = z 1 z 2 +z 1 z 3 (z 1 +z 2 )z 3 = z 1 z 3 +z 2 z 3 The negative of an comple number z = +i is defined b z = +( )i, and obes z +( z) = 0. Other Operations The comple conjugate of z is denoted z and is defined to be z = i. That is, to take the comple conjugate, one replaces ever i b i. Note that z z = (+i)( i) = 2 i +i + 2 = 2 + 2 is alwas a positive real number. In fact, it is the square of the distance from +i (recall that this is the point (,) in the plane) to 0 (which is the point (0,0)). The distance from z = +i to 0 is denoted z and is called the absolute value, or modulus, of z. It is given b z = 2 + 2 = z z c Joel Feldman. 2012. ll rights reserved. Januar 1, 2012 omple Numbers and Eponentials 1
Since z 1 z 2 = ( 1 +i 1 )( 2 +i 2 ) = ( 1 2 1 2 )+i( 1 2 + 2 1 ), z 1 z 2 = ( 1 2 1 2 ) 2 +( 1 2 + 2 1 ) 2 = 2 1 2 2 2 1 2 1 2 +1 22 2 +2 1 2 2 +2 1 2 2 1 + 2 2 2 1 = 21 22 +21 22 +21 22 +22 21 = ( 2 1 +2 1 )(2 2 +2 2 ) = z 1 z 2 for all comple numbers z 1,z 2. Since z 2 = z z, we have z ( ) z z = 1 for all comple numbers z 0. This sas that the multiplicative 2 inverse, denoted z 1 or 1 z, of an nonzero comple number z = +i is z 1 = z z 2 = i 2 + 2 = 2 + 2 2 + 2 i It is eas to divide a comple number b a real number. For eample 11+2i = 11 + 2 i In general, there is a trick for rewriting an ratio of comple numbers as a ratio with a real denominator. For eample, suppose that we want to find 1+2i 3 4i 3+4i. The trick is to multipl b 1 = 3 4i. The number 3 4i is the comple conjugate of 3+4i. Since (3+4i)(3 4i) = 9 12i+12i+16 = 1+2i 3+4i = 1+2i 3 4i 3+4i 3 4i = (1+2i)(3 4i) = 11+2i = 11 + 2 i ThenotationsRez andimz standfortherealandimaginarpartsofthecomplenumberz,respectivel. If z = +i (with and real) the are defined b Rez = Imz = Note that both Rez and Imz are real numbers. Just subbing in z = i gives Rez = 1 2 (z + z) Imz = 1 2i (z z) The omple Eponential Definition and Basic Properties. For an comple number z = +i the eponential e z, is defined b e +i = e cos +ie sin In particular, e i = cos+isin. This definition is not as msterious as it looks. We could also define e i b the subbing b i in the Talor series epansion e = n=0 n n!. The even terms in this epansion are e i = 1+i + (i)2 2! + (i)3 3! + (i)4 4! + (i)5 5! + (i)6 + 1+ (i)2 2! + (i)4 4! + (i)6 + = 1 2 2! + 4 4! 6 + = cos c Joel Feldman. 2012. ll rights reserved. Januar 1, 2012 omple Numbers and Eponentials 2
and the odd terms in this epansion are i+ (i)3 3! + (i)5 5! + = i ( 3 For an two comple numbers z 1 and z 2 3! + 5 e z1 e z2 = e 1 (cos 1 +isin 1 )e 2 (cos 2 +isin 2 ) = e 1+2 (cos 1 +isin 1 )(cos 2 +isin 2 ) 5! + ) = isin = e 1+2 {(cos 1 cos 2 sin 1 sin 2 )+i(cos 1 sin 2 +cos 2 sin 1 )} = e 1+2 {cos( 1 + 2 )+isin( 1 + 2 )} = e (1+2)+i(1+2) = e z1+z2 so that the familiar multiplication formula also applies to comple eponentials. For an comple number c = α+iβ and real number t e ct = e αt+iβt = e αt [cos(βt)+isin(βt)] so that the derivative with respect to t is also the familiar one. d dt ect = αe αt [cos(βt)+isin(βt)]+e αt [ βsin(βt)+iβcos(βt)] = (α+iβ)e αt [cos(βt)+isin(βt)] = ce ct Relationship with sin and cos. When is a real number e i = cos +isin e i = cos isin = e i are comple numbers of modulus one. Solving for cos and sin (b adding and subtracting the two equations) cos = 1 2 (ei +e i ) = Ree i sin = 1 2i (ei e i ) = Ime i These formulae make it eas derive trig identities. For eample coscosφ = 1 4 (ei +e i )(e iφ +e iφ ) = 1 4 (ei(+φ) +e i( φ) +e i( +φ) +e i(+φ) ) = 1 4 (ei(+φ) +e i(+φ) +e i( φ) +e i( +φ) ) ( ) cos( +φ)+cos( φ) = 1 2 and, using (a+b) 3 = a 3 +3a 2 b+3ab 2 +b 3, sin 3 ( = 1 8i e i e i) 3 ( = 1 8i e i3 3e i +3e i e i3) = 3 1 4 2i( e i e i) ( 1 1 4 2i e i3 e i3) = 3 4 sin 1 4 sin(3) c Joel Feldman. 2012. ll rights reserved. Januar 1, 2012 omple Numbers and Eponentials 3
and cos(2) = Ree i2 = Re ( e i) 2 = Re ( cos+isin ) 2 = Re ( cos 2 +2isincos sin 2 ) = cos 2 sin 2 Polar oordinates. Let z = + i be an comple number. Writing (,) in polar coordinates in the usual wa gives = rcos, = rsin and +i = re i +i = rcos +irsin = re i r In particular ( 1,0)= 1 i=(0,1) π π 2 1=(1,0) π 2 i=(0, 1) 1 = e i0 = e 2πi = e 2kπi for k = 0,±1,±2, 1 = e iπ = e 3πi = e (1+2k)πi for k = 0,±1,±2, i = e iπ/2 = e 5 2 πi = e (1 2 +2k)πi for k = 0,±1,±2, i = e iπ/2 = e 3 2 πi = e ( 1 2 +2k)πi for k = 0,±1,±2, The polar coordinate = tan 1 associated with the comple number z = +i is also called the argument of z. The polar coordinate representation makes it eas to find square roots, third roots and so on. Fi an positive integer n. The n th roots of unit are, b definition, all solutions z of Writing z = re i z n = 1 r n e ni = 1e 0i The polar coordinates (r,) and (r, ) represent the same point in the plane if and onl if r = r and = +2kπ for some integer k. So z n = 1 if and onl if r n = 1, i.e. r = 1, and n = 2kπ for some integer k. The n th roots of unit are all comple numbers e 2πi k n with k integer. There are precisel n distinct n th roots of unit because e 2πi k n = e 2πi k n if and onl if 2π k n 2πik n = 2πk k n is an integer multiple of 2π. That is, if and onl if k k is an integer multiple of n. The n distinct nth roots of unit are e 2πi2 6 e 2πi 1 6 1, e 2πi 1 n, e 2πi 2 n, e 2πi 3 n,, e 2πi n 1 n e 2πi3 6= 1 1=e 2πi0 6 e 2πi4 6 e 2πi 5 6 c Joel Feldman. 2012. ll rights reserved. Januar 1, 2012 omple Numbers and Eponentials 4
Sketching omple Numbers as Vectors lgebraic epressions involving comple numbers ma be evaluated geometricall b eploiting the following two observations. (ddition and subtraction) comple number is nothing more than a point in the plane. So we ma identif the comple number = a + ib with the vector whose tail is at the origin and whose head is at the point (a,b). Similarl, we ma identif the comple number = c+id with the vector whose tail is at the origin and whose head is at the point (c,d). Those two vectors form two sides of a parallelogram. The vector for the sum + = (a+c)+i(b+d) is that from the origin to the diagonall opposite corner of the parallelogram. The vector for the difference = (a c) +i(b d) has its tail at and its head at. + (Multiplication and Division) To multipl or divide two comple numbers, write them in their polar coordinate forms = re i, = ρe iϕ. So r and ρ are the lengths of and, respectivel, and and ϕ are the angles from the positive ais to and, respectivel. Then = rρe i(+ϕ). This vector has length equal to the product of the lengths of and. The angle from the positive ais to is the sum of the angles and ϕ. nd = r ρ ei( ϕ). This vector has length equal to the ratio of the lengths of and. The angle from the positive ais to is the difference of the angles and ϕ. φ +φ φ φ / c Joel Feldman. 2012. ll rights reserved. Januar 1, 2012 omple Numbers and Eponentials 5