Complex Numbers
Rotation – Locus of a Complex Set
Complex Numbers_PYQ
Grade 11
Question:
Let $a,b\in\mathbb{R}$ and $a^2+b^2\neq0$. Suppose $S=\left\{z\in\mathbb{C}:z=\dfrac{1}{a+ibt},\;t\in\mathbb{R},\;t\neq0\right\}$, where $i=\sqrt{-1}$. If $z=x+iy$ and $z\in S$, then $(x,y)$ lies on
the circle with radius $\dfrac{1}{2a}$ and centre $\left(\dfrac{1}{2a},0\right)$ for $a>0,\;b\neq0$
the circle with radius $-\dfrac{1}{2a}$ and centre $\left(-\dfrac{1}{2a},0\right)$ for $a<0,\;b\neq0$
the $X$-axis for $a\neq0,\;b=0$
the $Y$-axis for $a=0,\;b\neq0$
Step-by-Step Solution
Key Concept: From $x=a/(a^2+b^2t^2)$ and $x^2+y^2=1/(a^2+b^2t^2)$, the key step is $x^2+y^2=x/a$, which is the circle $(x-1/(2a))^2+y^2=1/(4a^2)$ centred at $(1/(2a),0)$.
**Step 1: Write z in terms of x and y**
$z=\dfrac{1}{a+ibt}=\dfrac{a-ibt}{a^2+b^2t^2}$, so $x=\dfrac{a}{a^2+b^2t^2}$ and $y=\dfrac{-bt}{a^2+b^2t^2}$.
**Step 2: Derive the locus equation**
Compute $x^2+y^2=\dfrac{a^2+b^2t^2}{(a^2+b^2t^2)^2}=\dfrac{1}{a^2+b^2t^2}=\dfrac{x}{a}$. So $x^2+y^2-\dfrac{x}{a}=0$, i.e., $\left(x-\dfrac{1}{2a}\right)^2+y^2=\dfrac{1}{4a^2}$.
**Step 3: Check option (1): a>0, b≠0**
Centre $\left(\dfrac{1}{2a},0\right)$ with $\dfrac{1}{2a}>0$ and radius $\dfrac{1}{2a}$. This matches the derived circle. TRUE ✓
**Step 4: Check option (2): a<0, b≠0**
Derived centre is $\left(\dfrac{1}{2a},0\right)$ with $\dfrac{1}{2a}<0$ for $a<0$. Option (2) claims centre $\left(-\dfrac{1}{2a},0\right)>0$ — wrong sign. FALSE ✗
**Step 5: Check option (3): a≠0, b=0**
With $b=0$: $z=1/a$ (real constant), so $x=1/a$, $y=0$. The point lies on the $X$-axis. TRUE ✓
**Step 6: Check option (4): a=0, b≠0**
With $a=0$: $z=1/(ibt)=-i/(bt)$, purely imaginary (since $t\neq0$, $b\neq0$). So $x=0$, and $(x,y)$ lies on the $Y$-axis. TRUE ✓ — but wait, is option (4) listed? If so, it is also correct.
Correct Answer: 1, 3