Question

let a,b, and c denote complex constants. then use definition2, sec 15, of limit to show that :

let a,b, and c denote complex constants. then use definition2, sec 15, of limit to show that : b/lim z arrow z0 (z^2 + c) = z^2 0 + c c/lim[x + (2x + y)]z arrow (1 - i) = 1 FOR THE FIRST QUESTION IS THE LIMIE WHEN Z GOIS TO Z0AND THE SECOND QUESTION IS WHEN Z GOES TO (1-i). AND THIS IS THEDEFINITION: |f(Z) - W0| < epsilon whenever0 < |(z - z0)| < sigma

FOR THE FIRST QUESTION IS THE LIMIE WHEN Z GOIS TO Z0AND THE SECOND QUESTION IS WHEN Z GOES TO (1-i). AND THIS IS THEDEFINITION:

Answer 1

(a) let f(z) = z^2 + c. therefore f(z_0) = z^2_0 + c. now we have to find delta > 0 such that Vertical-bar f(z) - f(z_0) Vertical-bar < epsilon whenever Vertical-bar z - z_0 Vertical-bar < delta. We have Vertical-bar f(z) - f(z_0) Vertical-bar = Vertical-bar (z^2 + c) - (z^2_0 + c) Vertical-bar = Vertical-bar z^2 - z^2_0 Vertical-bar = Vertical-bar z - z_0 Vertical-bar Vertical-bar z + z_0 Vertical-bar now Vertical-bar f(z) - f(z_0) Vertical-bar < epsilon implies Vertical-bar z - z_0 Vertical-bar Vertical-bar z + z_0 Vertical-bar < epsilon implies Vertical-bar z - z_0 Vertical-bar < epsilon/ Vertical-bar z + z_0 Vertical-bar so we choose delta > 0 such that epsilon/ Vertical-bar z + z_0 Vertical-bar < delta. Thus for every epsilon > 0 there is delta > 0 such that Vertical-bar f(z) - f(z_0) Vertical-bar < epsilon whenever Vertical-bar z - z_0 Vertical-bar < delta. Hence we get lim_x rightarrow x_0 (z^2 + c) = z^2_0 + c. (b) Given problem lim_x rightarrow 1 - I [x + (2x + y)] = 1 is wrong because lim_x rightarrow 1 - I [x + (2x + y)] = lim_x + iy rightarrow 1 - I [x + (2x + y)]