JEE Main 2022 — Mathematics Algebra

Given,

|z-1+i|\geq |z|,|z|<2&|z+i|=|z-1|

Now solving $|z-1+i|\geq |z|$∣z−1+i∣≥∣z∣ put $z=x+iy$z=x+iy we get, $x-y\leq 1$x−y≤1 so plotting we get,

Now solving $|z|<2$∣z∣<2 we get, ${x}^{2}+{y}^{2}<4$x2+y2<4 on plotting the diagram we get,

Now solving $|z+i|=|z-1|$∣z+i∣=∣z−1∣ we get $x+y=0$x+y=0

Now combine diagram of x+y=0&x-y\leq 1 will be,

Now combining all diagram we get,

Hence

$$w=2x+iy\in S$$w=2x+iy∈S

Now for $w\in S$w∈S, $2x\leq \frac{1}{2}$2x≤21​ so $x\leq \frac{1}{4}$x≤41​

Now $2x+y=0\Rightarrow y=-2x$2x+y=0⇒y=−2x putting the value in ${x}^{2}+{y}^{2}<4$x2+y2<4

$$\Rightarrow {(2x)}^{2}+{(2x)}^{2}<4$$⇒(2x)2+(2x)2<4

$$\Rightarrow {x}^{2}<\frac{1}{2}\Rightarrow x\in (\frac{-1}{\sqrt{2}},\frac{1}{\sqrt{2}})$$⇒x2<21​⇒x∈(2​−1​,2​1​)

So, $x\in (\frac{-1}{\sqrt{2}},\frac{1}{4}]$x∈(2​−1​,41​]