PQ and PR are two infinite rays. QAR is an ar | vedclass

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(C) The vertex of the rays $PQ$ and $PR$ is at $P(-1, 0)$,which corresponds to the complex number $z_0 = -1$.The ray $PQ$ passes through $(-1 + \sqrt{

Vedclass · Accepted Answer

$|z + 1| > 2; |\arg (z + 1)| < \frac{\pi }{4}$

Vedclass · Answer

(C) The vertex of the rays $PQ$ and $PR$ is at $P(-1, 0)$,which corresponds to the complex number $z_0 = -1$.The ray $PQ$ passes through $(-1 + \sqrt{2}, \sqrt{2}i)$,so its angle with the real axis is $\arg(z - (-1)) = \arg(z + 1) = 	an^{-1}(\frac{\sqrt{2}}{\sqrt{2}}) = \frac{\pi}{4}$.The ray $PR$ passes through $(-1 + \sqrt{2}, -\sqrt{2}i)$,so its angle is $\arg(z + 1) = -\frac{\pi}{4}$.The shaded region lies between these rays,so $|\arg(z + 1)| The arc $QAR$ is part of a circle centered at $P(-1, 0)$. The distance from $P(-1, 0)$ to $A(1, 0)$ is $|1 - (-1)| = 2$. Thus,the radius of the circle is $2$.The shaded region is outside this circle,so $|z - (-1)| > 2$,which is $|z + 1| > 2$.Therefore,the conditions are $|z + 1| > 2$ and $|\arg(z + 1)| < \frac{\pi}{4}$.

$PQ$ and $PR$ are two infinite rays. $QAR$ is an arc. $A$ point lying in the shaded region,excluding the boundary,satisfies:

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