Evaluate $\int_{C} log z~ dz$ where C is $|z|=y.$

Solution:

$I = \int_{C} \log z dz$ -------------------------(1)

Put $z=e^{j \theta}, \gamma = y$,

Differentiating z with respect to $\theta$.

$dz=j e^{j \theta} d \theta$ $\left[ \cfrac{d}{dx} e^{ax}=e^{ax} \cfrac{d}{dx}ax \right]$

and since it is unit circle, $\theta$ varies from $0 \ to \ 2\pi$.

Put z and dz value in equation (1)

$\begin{aligned} \therefore I &= \int_{0}^{2\pi}(\log e^{j \theta})j e^{j\theta} \cdot d \theta \\ I &= j \int_{0}^{2\pi}\log e^{j \theta} \cdot e^{j\theta} \cdot d \theta \\ I &= j \int_{0}^{2\pi} j \theta \cdot e^{j\theta} \cdot d \theta \\ I &= j \cdot j \int_{0}^{2\pi} \theta \cdot e^{j\theta} \cdot d \theta \\ \Longrightarrow I &= -1 \int_{0}^{2\pi} \theta \cdot e^{j\theta} \cdot d \theta \space \big[\because j^{2}=-1 \big] \end{aligned}$

Taking $u=\theta$ and $v = e^{j \theta}$,

Integration of $u \cdot v$,

$\int u \cdot v = u \int v - \int \left[ \cfrac{d}{dx}u \cdot \int v\right]$

OR

$\int u \cdot v = u \int v - u' \int \left[ \int v \right] +u'' \int \left[ \int \int v\right]$

$\begin{aligned} I &= - \left [ \theta \cdot \cfrac{e^{j \theta}}{j} - \cfrac{e^{j \theta}}{j \cdot j} \right]_{0}^{2\pi} \\ I &= - \left [ \theta \cdot \cfrac{e^{j \theta}}{j} - \cfrac{e^{j \theta}}{-1} \right]_{0}^{2\pi} \\ I &= - \left [ \theta \cdot \cfrac{e^{j \theta}}{j} + e^{j \theta} \right]_{0}^{2\pi} \\ I &= - \left [ \left ( 2\pi \cdot \cfrac{e^{j 2\pi}}{j} + e^{j 2\pi} \right) - \left( 0 \cdot \cfrac{e^{j0}}{j} + e^{j0}\right) \right] \\ I &= - \left [ 2\pi \cdot \cfrac{e^{j 2\pi}}{j} + e^{j 2\pi} - 0-1 \right] \\ I &= - \left [ 2\pi \cdot \cfrac{e^{j 2\pi}}{j} + e^{j 2\pi} -1 \right] \\ I &= - \left [ \cfrac{2\pi}{j} + 1 -1 \right] Since \quad[e^{2\pi \cdot j}=1] \\ I &= - \left [ \cfrac{2\pi}{j} \right] \end{aligned}$

OR

$I = -\left [2\pi \times \cfrac{1}{j} \right] = -\left [2\pi \times -j \right]=2\pi j \left[ \because \cfrac{1}{i} =-i\right]$