[bit.listserv.hellas] Quiz Answer

P5047G@TEMPLEVM.BITNET (Apostolis Apostolopoulos) (02/11/90)

   Geia sas kai pali.
Loipon i apantisi sto quizaki pou esteila prin kana dio meres einai i akolouthi
kai den exei na kanei tipota me topologia, alla me migadikes sinartiseis.

Theoroume ti migadiki sinartisi sqrt(z):
sqrt(z)=exp(1/2(logz)),alla logz=log|z|+it opou t=gwnia theta
dialegoume to depleted xy plane afou dioksoume to branch arnitiko y, opote
exoume ena simlpy connected domain kai to sqrt(z) orizetai kai einai analitiko
Dialegoume to t na ikanopoin to: -pi/2<t<3(pi)/2
Twra, sqrt(1)=exp(1/2(log1+0))=1
      sqrt(-1)=exp(1/2(log1+(pi)i))=exp((pi/2)i)=i
      Sqrt(-1)*sqrt(-1)=exp((pi)i)=-1
etsi loipon  sqrt((-1)(-1)) not eq. sqrt(-1)*sqrt(-1)


Twra to an ena polionimo exei rizes i oxi exei na kanei me to apo poio
pedio dialegontai oi sintelestes tou, p.x.
x**2+1 den exei rizes sto pragmatiko pedio alla exei poli kales rizes sto
migadiko. Idiaitera gia rizes sto pedio twn ritwn iparxei to Eisenstein
Criterion pou mas leei ti ginetai (I.N. Herstein "Topics in Algebra).

Afta kai sto epanaepeikoinwnein.