Use the principal branch of log z $\int_{-1}^{1} log z\, dz$
My attempt was we see how -1 to 1 became pi to 0? That should make the e^whatever terms go away. Then we can do it by parts. $\int_{-1}^{1} log z\, dz$ =$\int_{\pi}^{0} log (e^{i\theta})ie^{i\theta}\, dz=-2+i\pi$
Is this correct result? Could please show me another method of resolution?
Could someone help me through this problem?