FINDING ZEROS OF COMPLEX FUNCTIONS It is well kow sice the time of Newto that the zeros of a real fuctio f(x) ca be foud b carrig out the iterative procedure- f ( x[ x[ 1] x[ ] subject to x[0] x f '( x[ 0 Here x[0] represets a value lig withi the eighborhood of the root at x[]. The s i the square brackets represet subscripts. If we have f(x)=x -1 ad x[0]=1/, oe fids the zero at x[]=1. Had we started the iteratio with x[0]=- 3/, oe would fid the secod zero at x[]=-1. Thus the iterative procedure will coverge toward that zero of f(x) lig closest to the startig value x[0]. It is possible to exted this iterative procedure to complex fuctios f(z) where z=x+i. We do this i the preset article ad also show how other methods such as makig cotour maps of the absolute value of a complex fuctio to quickl locate its zeros.. Our startig poit will be a complex fuctio f(z) whose derivative f (z) exists.to fid its zeros we start with z[0] ad write- df(z[0])/dz={f(z[0])-0}/(z[0]-z[1]) Now if z[0] lies close eough to oe of the zeros of f(z), oe ca deduce form this last expressio that the followig iterative procedure will be valid- f ( z[ z[ 1] z[ ] subject to z[0] z f '( z[ 0 To get some idea of where to place z[0] it is a good idea to first obtai a rough cotourplot of the absolute value of f(z). That is, locate the eighborhood of the expected zero b otig where the cotours make small diameter circles i the z=x+i plae. The absolute value fuctio of f(z) ca be writte as- Abs{ } u( v( with u( iv( It eeds to be remembered that iterative solutios for complex fuctios will ofte ot coverge whe the startig poit lies outside of the immediate eighborhood of a zero. The boudar betwee coverget solutios ad those where the Newto method fails ca lead to ver itricate patter associated with fractals. Let us demostrate the iterative ad cotour map procedure for the special case of the complex fuctio-
z 1 i( z 1) Here we have u ( x 1 ad v( x( 1) 1 The cotour map for Abs(f(z) idicates small closed circles about z=1+i ad z=-1.so ruig the iterative process usig either z[0]=1++i or z[0]=-1+ as a startig poit, produces z[]=1+i ad -1, respectivel. Epsilo is take as a small icremet such as =0.1. I this case we could also have used a much shorter approach to get the aswer b otig that f(z) is a quadratic i z ad thus ca be writte as- f(z)=(z+1)(z-1-i) From this form the aswer ca be read off directl. Wheever f(z) is a quadratic, cubic, or fourth order polomial i z, the aswer for the zeros of f(z) are expressible i closed form ad the aswer ca be read off directel.thus the four zeros of- z 4 1 ( z 1)( z 1) ( z 1)( z 1)( z i)( z i) are see to be z[]=1, -1, i, -i.. The cotour plot for the absolute value of z 4-1 looks like this-
For quitic ad higher power polomial fuctios oe must(with a few exceptios) resort to iterative or cotour plot procedures. If the polomial form of f(z) cotais ol the highest power of z plus a costat such as f(z)=z -1, the solutios ca be give i closed form for a positive iteger. For example, take the complex fuctio- z 1 r exp( i ) 1 whe z is expressed i polar form. Solvig we have r=1 ad k / with k 1,,3,... So the zeros of f(z) all lie o a uit circle at / itervals. Expressed i Cartesia coordiates, the roots are located at- x 1 1 ta ad ta( ) 1 ta with k Aother complex fuctio isf ( z) si( z) si( x)cosh( i cos( x)sih( I this case- Abs{ } si cosh ( x)cosh ( cos ( x) ( cos ( x)sih ( A cotour plot of this fuctio follows-
It looks ver much like the phase plae plot for a simple pedulum. Here the zeros clearl lie alog the x axis at x={0,1,,3, etc.}..there are o zeros possible elsewhere i the z plae. This last result brigs to mid the as et uprove cojecture b Riema that the complex Riema Zeta Fuctio Z(+i) ca have zeros i the - plae ol alog the lie =1/. It certail looks like a valid statemet as show b the followig cotour map I have geerated
It appears that a defiitive proof that all roots of Zeta(x+i lie alog the x=1/ lie will be based a a topological argumet i which oe otes that all cotours of Abs(Zeta) for C<<1 form small radius circles about the zeros of f(z). As a fial example of fidig the zeros of a complex fuctio cosider- exp( z ) 1 u( iv( where u exp( x )cos(x 1 ad v exp( x )si(x Oe sees at oce b ispectio that z=0 is the ol possible zero of f(z)=exp(z )- 1. To cofirm this poit oe could carr out a Newto iteratio o this complex fuctio. Eve simpler is to just look at a cotour map of the absolute value of f(z). Doig so we fid the followig-
This icel cofirms that the ol zero lies withi the small circle cetered o the origi i the z plae.. September 1, 015