PQ and PR are two infinite rays.QAR is an arc . point lying in the shaded region excluding the boundry satisfies

∣ z - 1 ∣ > 2; a r g (z - 1) < \frac{π}{4}

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∣ z - 1 ∣ > 2; a r g (z - 1) < \frac{π}{2}

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∣ z - 1 ∣ > 2; a r g (z - 1) < \frac{π}{4}

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∣ z - 1 ∣ > 2; a r g (z - 1) < \frac{π}{2}

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Solution

The correct option is C $∣ z - 1 ∣ > 2; a r g (z - 1) < \frac{π}{4}$
Equation of ray PQ $a r g (z + 1) = \frac{π}{4}$

Equation of ray PR $a r g (z + 1) = - \frac{π}{4}$

Shaded region is $\frac{- π}{4} < a r g (z + 1) < \frac{π}{4}$

$∣ a r g (z + 1) ∣ < \frac{π}{4}; ∣ P Q ∣= \sqrt{(\sqrt{2})^{2} + (\sqrt{2})^{2}} = 2$

$∣ P A ∣= 2; ∣ P R ∣ = 2$

So, arc QAR is of a circle of radius 2 unit with centre at P(-1,0).All the points in the shaded region are exterior to this circle $∣ z + 1 ∣= 2.$

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