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

Solution:

Given |z| = 1.

and $\left|{\mathrm{z}}^2-\mathrm{az}+1\right|=1$

$\Rightarrow \left|\mathrm{z}\right|\left|\mathrm{z}-\mathrm{a}+\frac{1}{\mathrm{z}}\right|=1$

$\Rightarrow \left|\mathrm{a}-\left(\mathrm{z}+\frac{1}{\mathrm{z}}\right)\right|=1$                     (1)

Now, $\mathrm{z}=\cos \mathrm{\theta }+\mathrm{isin}\mathrm{\theta }$

$\Rightarrow \mathrm{z}+\frac{1}{\mathrm{z}}=2\cos \mathrm{\theta }$

So, (1) reduces to

$|\mathrm{a}-2\cos \mathrm{\theta }|=1$

So, locus of a is circle whose center is $(2\cos \mathrm{\theta }, 0)$ and radius is 1.

Since, $-2\le 2\cos \mathrm{\theta }\le 2$, 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 line segment with end points at -2 and 2 on the Argand plane. Therefore, the area S consists of the area of two semi circles and area of rectangle ABCD.

So, the required area is $\mathrm{\pi }+8$.