Question

Let p(z) be the principal branch of 21-i. Let D* = C\(-00,0) be all the complex numbers except for the non-positive real numbers. (a) (4 points) Find a function which is an antiderivative of p(x) on D". (b) (6 points) Let I be a contour such that (i) I is contained in D* and (ii) the initial point of I' is 1 and the terminal point of I is i. Compute (2)dr. Justify your answers.

Answer 1

solution : (1) Take HZ). 2-1 In(atu) - In (2:1) – In (x2-1) – 10 (2-1) 10 (912)) = (2-1) loz – 1n (2-1) By dit centiating the above equation, we get 2-1 Y(z) - 2 (Z) = (2-1) 9 (7) Z Z =6240 x 262-1) -2 (1-1) - PC) - q'(z) - P(Z). A principal brance of a(z) = antiderivative of Plz) on 0* is an 2- 6) z (2-1) q (Z) = (2-1) > 10 (9(Z)) = 2 (2-1) Inz- ln (2-1) because of (1) - In (Y(z)) =(2-1) (In(z)+i arg (z)) - (10(15)+i tas-'(12) Where arg (z) is the argument of *

(2) 9 (1) = 1 (2-1) 10(90) = (2-2)[/01+0) * -In (63) +itan"(112) =-love titan-1 (1/2) q (1) =é (lors pot-1) titan 1 (1/2) = 10 (55)- & citan-' (1/2) 96) - (costan! (12)) tisin (tan-'(12) 3 ; (2-1) qli) - (2-1) In 19 (1)) = (2-3) (In 1+1 t2) – 10 15 titan-(1/2) +(2-1) (0 + 7/2) – 1016 titan -|(12) = $xi Thy - ih/2 - 105 titan "(1/2) -(-1015) +ili+tan-'(120) + Ing(1) = (02-10v5 ) +/+tan-1(12]] Y() -e (/2-1015) e ilt+tan-'(120) e/2 Toro (cos(TT+ tan-1'/21) + isinkT+tan" (112) (-eas (ton="112) – isin (tan-11) el/2 V5

(3) The cos(tan-(02)) +isio (tan- (12) V5 Since r is contained in gt, it follows from the fundamental theorem of calculus for analytic function that P(z) dz= q (i) – 9 (1) 7/21 Cos (tan-|(122)) +isin(tan-|(12)] 5 eitan-1(1/2) --Le +/2+1) V5

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