Complex Numbers
Complex Number in Iota Form
Complex Numbers_PYQ
Grade 11
Question:
Let $a$, $b$, $x$ and $y$ be real numbers such that $a - b = 1$ and $y \neq 0$. If the complex number $z = x + iy$ satisfies $\text{Im}\!\left(\dfrac{az + b}{z + 1}\right) = y$, then which of the following is(are) possible value(s) of $x$?
$1 - \sqrt{1 + y^2}$
$-1 - \sqrt{1 - y^2}$
$1 + \sqrt{1 + y^2}$
$-1 + \sqrt{1 - y^2}$
Step-by-Step Solution
Key Concept: The imaginary part of $\frac{az+b}{z+1}$ (with $a-b=1$) telescopes beautifully: the $a$-dependence cancels entirely, reducing the condition to a circle equation in $x$.
**Step 1: Substitute b = a − 1 and rationalise**
With $b = a-1$ and $z = x+iy$:
$$\dfrac{az+b}{z+1} = \dfrac{a(x+iy)+(a-1)}{(x+1)+iy}.$$
Multiply by conjugate $(x+1) - iy$.
**Step 2: Compute the imaginary part**
The imaginary part of the numerator after expansion:
$$\text{Im}[\text{numerator}] = ay(x+1) - (ax+a-1)y = y[a(x+1) - (ax+a-1)] = y[1] = y.$$
So $\text{Im}\!\left(\dfrac{az+b}{z+1}\right) = \dfrac{y}{(x+1)^2 + y^2}$.
**Step 3: Set equal to y and solve**
$\dfrac{y}{(x+1)^2+y^2} = y$. Since $y \neq 0$: $(x+1)^2 + y^2 = 1$, so $(x+1)^2 = 1 - y^2$, giving $x = -1 \pm \sqrt{1-y^2}$.
**Step 4: Match to options**
$x = -1 + \sqrt{1-y^2}$ matches option (4); $x = -1 - \sqrt{1-y^2}$ matches option (2).
Correct Answer: 2, 4