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Integrala de la 1 la 1+i din z^2 unde z= nr complex

Răspuns :

z=a+bi z²=(a+bi)²=(a²-b²+2abi)
I=∫(a²-b²+2abi)dz=∫(a²-b²)dz+∫2abidz   z∈[1,1+i
notam  cu   I1=prima  integrala   si  cu   I2  a 2-a  imyegrala
I1=(a²-b²)*z/1↑(1+i)=(a²-b²)*(1+i-(a²-b²)*1=(a²-b²)(1=i-1)=(a²-b²)i
I2=2abi∫dz=2abi*z/1↑1+i=2abi(1+i-1)=2abi²=-2ab
I=I1+I2
I=(a²-b²)i+2ab  sau
I=2ab+(a²-b²)i