Complex Numbers
Rotation – Geometric Region in Argand Plane
Complex Numbers_PYQ
Grade 11
Question:
The shaded region, where $P=(-1,0)$, $Q=(-1+\sqrt{2},\,\sqrt{2})$, $R=(-1+\sqrt{2},\,-\sqrt{2})$, $S=(1,0)$ is represented by *[Image: An Argand plane with a wedge-shaped shaded region bounded by two rays from $P=(-1,0)$ at angles $\pm\pi/4$ and the arc of circle $|z+1|=2$ between $Q$, $S$ and $R$]*
$|z+1|>2,\;|\arg(z+1)|<\dfrac{\pi}{4}$
$|z+1|<2,\;|\arg(z+1)|<\dfrac{\pi}{4}$
$|z+1|>2,\;|\arg(z+1)|>\dfrac{\pi}{4}$
$|z-1|<2,\;|\arg(z+1)|>\dfrac{\pi}{2}$
Step-by-Step Solution
Key Concept: The region is the intersection of an open disk $|z+1|<2$ and an open angular sector $|\arg(z+1)|<\pi/4$. Shifting to $w=z+1$ centres the analysis at the origin.
**Step 1: Identify the centre of the circular boundary**
All boundary points $Q$, $R$, $S$ are at distance $2$ from $P=(-1,0)$: $|Q-P|=|(\sqrt{2},\sqrt{2})|=2$, $|S-P|=|(2,0)|=2$. So the arc is part of $|z+1|=2$.
**Step 2: Find the angular bounds**
$Q-P=(\sqrt{2},\sqrt{2})$ gives $\arg(Q+1)=\pi/4$. $R-P=(\sqrt{2},-\sqrt{2})$ gives $\arg(R+1)=-\pi/4$. So the sector spans $|\arg(z+1)|\leq\pi/4$.
**Step 3: Determine inside vs outside**
The shaded region is the interior of the wedge bounded by the two rays and the arc. Points in the interior satisfy $|z+1|<2$ (inside the circle) and $|\arg(z+1)|<\pi/4$ (inside the angular sector).
Correct Answer: 2