Consider the region S of complex numbers a such that $|z^2-az+1|=1,$ where $\left|z\right|=1.$ Then area of S in the Argand plane is  $\pi +k$  then  $k=$  ___________.

(1) Given  $\left|z\right|=1$
And  $|z^2-az+1|=1 \Rightarrow \left|z\right||z-a+\frac{1}{z}|=1 \Rightarrow |a-(z+\frac{1}{z})|=1$                
Now,  $z=\cos \theta +i\sin \theta \Rightarrow z+\frac{1}{z}=2\cos \theta$ 
So, (1) reduces to $|a-2\cos \theta |=1$  
So, locus of  a  is circle whose center is $(2\cos \theta ,0)$ and radius is 1. 
Since,  $-2\le 2\cos \theta \le 2,$ we have following region traced by the locus of  a.
 
As shown in the figure the set S consists of the locus of units circles centered at a point On the lines segment with end points at -2 and 2 on the Agrand plane. Therefore, the area S consists of the area of two semi circles and area of rectangle  ABCD.
So, the required area is $\pi \text{\hspace{0.17em}+\hspace{0.17em}8.}$