Answer to Question #236287 in Chemical Engineering for Lokika

Question

3)Evaluate∫C z(ReZ)dz,Where C:z(t)=t, 0≤t≤1. ; 1+i(t-1), 1≤t≤2.

Explain the problem with step by step process?

Answer 1

Set z1 = −1 = z2. Then Arg(z1z2) = Arg(1) = 0, whereas Arg(z1) + Arg(z2) = π + π = 2π. Thus Arg(z1z2) 6= Arg(z1) + Arg(z2). (b) Set z1 = 0 and z2 = −1. Then Arg  z1 z2  = Arg(0) = 0, whereas Arg(z1) − Arg(z2) = 0 − π = −π. Thus Arg  z1 z2  6= Arg(z1) − Arg(z2). (c) Set z = −1. Then Arg(¯z) = Arg(1) = 0, whereas −Arg(z) = −Arg(−1) = −π. Thus Arg(¯z) 6= −Arg(z). (d) Set z = −1. Then Arg(−z) = Arg(1) = 0, whereas Arg(z)+π = Arg(−1)+π = π +π = 2π. Thus Arg(−z) 6= Arg(z) + π.