Adamm:
I = ∫
γ1 (Re(z))
2 dz + ∫
γ2 (Re(z))
2 dz = I
1 + I
2
z = 1(1−t)+t(2+i) = 1+t+ti, t∊[0, 1]
dz = (1+i) dt
| 7(1+i) | |
I1 = (1+i)∫01 (Re(1+t+ti))2 dt = (1+i)∫01 (t2+2t+1) dt = |
| |
| 3 | |
z = (2+i)(1−t)+t(−1+i) = 2−3t+i
dz = −3 dt
I
2 = −3 ∫
01 (Re(2−3t+i))
2 dt = −3 ∫
01 (4−12t+9t
2) dt = −3
duża szansa że to było dobrze, ale lepiej by ktoś sprawdził