Question

Question 1) Find I = z +2 3z - 2 + 3i 22 + (2i - 2)2 - 4i ] dz, C:\z| = 3, CW

Answer 1

Date Page NO I-61 C:111=3 , Cw J.L2+2 + 32-2+3; de z²+(21-2)=-40 are af z=-2, - bon polon S x²+(21-2) Z-41=0 7²22+Riz-yito 2=2; 2=-2i (All are inside circle c) 7 and by Residue theoren n t f(3) dz = Srila, taat x = x, Z. - where a, a, az... an are residies at zu which are inside the contour C. Here we are having the integral in clockwire direction So we use are negative sign 9. f(z)dz 3 - 21 (4, +4,7 ton I = I t Iz I = & dz = -2 ri (residue at z=-2) = -2ri (1) Jc Z+2 - residue = lim (2+2). I i li a 2-2 (2+2) 326 32-2+3: dz = -ari residue at z=2, -22) 2+(21-2)=-4ion residue at z=2 line ( (37-2+31) = (303) - 2+371 7+2 (7-2)(2+ 2;) (2+R2) 7 → -ri residue at 7-21; lim (z+7 ( 32=2+3) = 3 (2)-2+3i (2-2)(2 + (-2;-2) Att hitrexat at I w-ari ( 4 +31 + 3i+2) Iz = -2ni (3) abri stri Pita

So I = I + In I = -2Ti - bril I - - 8ril Answer
The answer must be -8πi because on changing the direction of integral i.e. clockwise there must be a negative sign. In standard notation we consider direction of closed loop to be counter-clockwise direction (anticlockwise) as positive.

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