Mathematical Methods for Physics and Engineering - Matematica.NET

24.4 SOME ELEMENTARY FUNCTIONSreal-variable counterpart it is called the exponential function; also like its realcounterpart it is equal to its own derivative.The multiplication of two exponential functions results in a further exponentialfunction, in accordance with the corresponding result for real variables.◮Show that exp z 1 exp z 2 =exp(z 1 + z 2 ).From the series expansion (24.15) of exp z 1 and a similar expansion for exp z 2 ,itisclearthat the coefficient of z1 rzs 2 in the corresponding series expansion of exp z 1 exp z 2 is simply1/(r!s!).But, from (24.15) we also have∞∑ (z 1 + z 2 ) nexp(z 1 + z 2 )=.n!n=0In order to find the coefficient of z1 rzs 2 in this expansion, we clearly have to consider theterm in which n = r + s, namely(z 1 + z 2 ) r+s 1 (=r+s)C 0 z r+s1+ ···+ r+s C s z r(r + s)! (r + s)!1z2 s + ···+ r+s C r+s z r+s2 .The coefficient of z1 rzs 2 in this is given byr+s 1 (r + s)! 1C s =(r + s)! s!r! (r + s)! = 1r!s! .Thus, since the corresponding coefficients on the two sides are equal, and all the seriesinvolved are absolutely convergent for all z, we can conclude that exp z 1 exp z 2 =exp(z 1 +z 2 ). ◭As an extension of (24.15) we may also define the complex exponent of a realnumber a>0 by the equationa z = exp(z ln a), (24.16)where ln a is the natural logarithm of a. The particular case a = e and the factthat ln e = 1 enable us to write exp z interchangeably with e z .Ifz is real then thedefinition agrees with the familiar one.The result for z = iy,exp iy =cosy + i sin y, (24.17)has been met already in equation (3.23). Its immediate extension isexp z =(expx)(cos y + i sin y). (24.18)As z varies over the complex plane, the modulus of exp z takes all real positivevalues, except that of 0. However, two values of z that differ by 2πki, for anyinteger k, produce the same value of exp z, as given by (24.18), and so exp z isperiodic with period 2πi. If we denote exp z by t, then the strip −π