Question

f(z) = {*3:2+1 (1:3); z = 1 Given the complex function: 1)Isf continuous on the complex plane? 2) Is fanalytic at z = (1; 0)?

Answer 1

fiz) = 3 2 1 1+31 : 2=1 - = (2-1) (2 + 1+ 2) 271 #2) Itzi z=1 for - S ₃²+ z+1 : ztl fre) } +3i z= lim fiz) = lim z²+ 2 + 1 Now 2 " 21 = f(1) 3 = bermit → limit doesn't exist at 2=1 so of is not countiniaus at z=1. elsewhere it is cauntiniaus because elsewhou it a polynomial in z . & palynomial in z is both countiniaus & differentable at an c. elsewhole it in z .& palable at 6 Now it t is analytic at z= Hoi - tis diffountable at alto, f is cantinlaus at altio out in port we show that fie Not camtiniaus f is Not diff at zzi t is not analytic at z oitoi