Question

Evaluate the integration of (2z+3)dz from 1-2i to 3+i. a. along the path x=2t+1, y=4t^2-t-2, 0<=t<=1. b.along the strightline joining 1-2i and 3+i

Answer 1

$a) \newline f(z)=2z+3. \newline z=x+iy=(2t+1)+i(4t^2 -t-2), 0\le t \le 1. \newline z'(t)=2+i(8t-1), 0\le t \le 1.$

$\newline \int_{1-2i}^{3+i}f(z)dz=\int_{0}^{1}f(z(t))z'(t)dt= \newline$

$=\int_0^1(2(2t+1+i(4t^2-t-2)))(2+i(8t-1))dt= \newline =\int_0^1(4t+2+i(8t^2-2t-4))(2+i(8t-1))dt= \newline$

$=-\frac{64}{4}+\frac{24}{3}+\frac{38}{2}+ \newline +i(\frac{48}{3}+\frac{8}{2}-10)= \newline =-16+8+19+i(16+4-10)=11+10i. \newline b) \newline$

$\newline 1\le t\le 3, y(t)=kt+b \newline y(1)=k+b=-2,y(3)=3k+b=1, \newline 2k=3, k=\frac{3}{2} \newline b=-2-\frac{3}{2}=-\frac{7}{2} \newline$

$\int_{1-2i}^{3+i}2z+3 dz= \newline =\int_1^3((2(t+(\frac{3}{2}t-\frac{7}{2})i)+3)(1+\frac{3}{2}i)dt= \newline =\int_1^3(2t+3t-7i+3)(1+\frac{3}{2}i)dt= \newline =\int_1^3(5t+3-7i)(1+\frac{3}{2}i)dt= \newline =\int_1^3(5t+3+\frac{21}{2})+i(-7+\frac{15}{2}t+\frac{9}{2})dt= \newline$

$=(\frac{5}{2}t^2+\frac{27}{2}t+ \newline +i(\frac{15}{4}t^2-\frac{5}{2}t))_1^3= \newline =\frac{5}{2}3^2+\frac{27}{2}*3+ \newline +i(\frac{15}{4}3^2-\frac{5}{2}*3)- \newline -\frac{5}{2}-\frac{27}{2}- \newline -i(\frac{15}{4}-\frac{5}{2})= \newline =\frac{5}{2}*8+\frac{27}{2}*2+ +\newline i(\frac{15}{4}*8-\frac{5}{2}*2)=20+27+i(30-5)=47+25i$