JEE Main 2021 — Mathematics Algebra

We know that the complex number $z$z satisfying $|z-{z}_{0}|=r$∣z−z0​∣=r represents a circle with centre ${z}_{0}$z0​ and radius $r$r units.

Hence, for ${S}_{1}=|z-1|\leq \sqrt{2},z$S1​=∣z−1∣≤2​,z lies on and inside the circle

of radius $\sqrt{2}$2​ units and centre $(1,0).$(1,0).

For ${S}_{2},$S2​, let $z=x+iy$z=x+iy

Now, $(1-i)(z)=(1-i)(x+iy)$(1−i)(z)=(1−i)(x+iy)

$$\Rightarrow (1-i)(z)=x+iy-ix-{i}^{2}y$$⇒(1−i)(z)=x+iy−ix−i2y

$$\Rightarrow (1-i)(z)=x+iy-ix+y$$⇒(1−i)(z)=x+iy−ix+y

$$\Rightarrow Re((1-i)z)=x+y$$⇒Re((1−i)z)=x+y

$$\Rightarrow x+y\geq 1.$$⇒x+y≥1.

And, for ${S}_{3},$S3​, again let $z=x+iy,$z=x+iy,

$$\Rightarrow y\leq 1.$$⇒y≤1.

Plotting all the inequalities on the graph, we get

Now, the common part is shown in the shaded part, hence, we get infinite points in the shaded region.

$\Rightarrow {S}_{1}\cap {S}_{2}\cap {S}_{3}$⇒S1​∩S2​∩S3​ has infinitely many elements.