> 
> If C denotes the counterclockwise unit circle, then the value of the contour integral
> $$\frac{1}{2\pi j} \oint\_c Re\{ z\} \, dz \quad \text{is \_\_\_\_\_ }?$$
> 
> 
> 

Method $1$:  

$$Re\{ z\} = \frac{z + \bar{z}}{2} \quad \dots (1)$$
$$\implies \frac{1}{2\pi j} \oint\_c Re\{ z\} \, dz = \frac{1}{2\pi j} \oint\_c \frac{z + \bar{z}}{2} \, dz$$
$$=\frac{1}{2\pi j} \oint\_c \frac{z }{2} \, dz + \frac{1}{2\pi j} \oint\_c \frac{\bar{z}}{2} \, dz \quad \dots (2)$$
Now using Cauchy-Integral theorem, we get:
$$\frac{1}{2\pi j} \oint\_c \frac{z }{2} \, dz=0 \quad \dots (3)$$
as $f(z)=z$ is differentiable at every point inside C ($|z|=1$)  

and $\bar{z}=\frac{1}{z}$ for C( $|z|=1$),  

so, $eq(2)$ can be rewritten as:
$$\frac{1}{2\pi j} \oint\_c \frac{z }{2} \, dz + \frac{1}{2\pi j} \oint\_c \frac{\bar{z}}{2} \, dz = 0 + \frac{1}{2}\frac{1}{2\pi j} \oint\_c \frac{1}{z} \, dz$$
$$\frac{1}{2} .1=\frac{1}{2}$$

Method $2$:  

Let $z=re^{j \theta}$  

$$\implies \frac{1}{2\pi j} \oint\_c Re\{ z\} \, dz = \frac{1}{2\pi j} \oint\_c r cos (\theta) \, d(re^{j \theta}) \quad \dots (4)$$
Now, for C : $r=1 $ and $ \theta \to 0 $ to $2\pi $
$$=\frac{1}{2\pi j} \int\_0^{2\pi} cos (\theta) e^{j \theta}j \, d\theta$$
$$=\frac{1}{2\pi } \int\_0^{2\pi} cos (\theta) e^{j \theta} \, d\theta \quad \dots (5)$$
Now,
$$\int e^{ax}cos(bx) dx = \frac{e^{ax}}{a^2 +b^2}\{ a cos(bx) + b sin (bx)\} +C \quad \dots (6)$$
$$\implies \int\_0^{2\pi} cos (\theta) e^{j \theta} \, d\theta = [\frac{e^{jx}}{j^2 +1^2}\{ j cos(x) + sin (x)\}]\_0^{2\pi}$$
Here, denominator is coming to $0$ , so we can't use the formula described in $eq(6)$;  

Hence how to evaluate $eq(5)$ ?  

Any help or suggestions please...