Complex Numbers
Properties of Complex Numbers
Grade 11
Question:
<p>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 \(x\)?</p>
<p>(1) \(-1 - \sqrt{1 - y^2}\)</p>
<p>(2) \(1 + \sqrt{1 + y^2}\)</p>
<p>(3) \(1 - \sqrt{1 + y^2}\)</p>
<p>(4) \(-1 + \sqrt{1 - y^2}\)</p>
Step-by-Step Solution
Key Concept: Expand the quotient using the conjugate method to separate real and imaginary parts, then equate the imaginary part to y. This creates a constraint equation that determines which values of x are possible regardless of the specific values of a, b.
<p><strong>Step 1:</strong> Write z = x + iy and compute the quotient using conjugate multiplication:</p><p>$$\frac{az + b}{z + 1} = \frac{(ax + b) + iay}{(x + 1) + iy} \cdot \frac{(x + 1) - iy}{(x + 1) - iy}$$</p><p><strong>Step 2:</strong> Expand the numerator:</p><p>$$= \frac{[(ax + b)(x + 1) + ay^2] + i[ay(x + 1) - y(ax + b)]}{(x + 1)^2 + y^2}$$</p><p><strong>Step 3:</strong> Extract the imaginary part:</p><p>$$\text{Im}\left(\frac{az + b}{z + 1}\right) = \frac{y[a(x + 1) - (ax + b)]}{(x + 1)^2 + y^2} = \frac{y(a - b)}{(x + 1)^2 + y^2}$$</p><p><strong>Step 4:</strong> Use the given condition that this equals y:</p><p>$$\frac{y(a - b)}{(x + 1)^2 + y^2} = y$$</p><p><strong>Step 5:</strong> Since y ≠ 0, divide both sides by y:</p><p>$$\frac{a - b}{(x + 1)^2 + y^2} = 1$$</p><p><strong>Step 6:</strong> Substitute a - b = 1:</p><p>$$(x + 1)^2 + y^2 = 1$$</p><p><strong>Step 7:</strong> This is a circle equation. For real y ≠ 0 to exist, we need $(x + 1)^2 < 1$, which gives $-2 < x < 0$. The possible values are typically x = -1 (when y = 0 is excluded) and boundary analysis shows x ∈ [-2, 0].</p><p>∴ Answer: AC</p>
Correct Answer: AC