Complex Numbers
Complex Plane / Geometry
Grade Class 11
Question:
<p>If \(\alpha\) lies on circle of radius \(r\) and \(1/\bar{\alpha}\) lies on circle of radius \(2r\), both centred at \(z_0=x_0+iy_0\) with \(|z_0|^2=2\), then \(|\alpha|^2\) is:</p>
Step-by-Step Solution
Key Concept: |\alpha-z_0|^2=r^2 and |1/ᾱ-z_0|^2=4r^2. From first: |\alpha|^2-z_0ᾱ-z̄_0\alpha+|z_0|^2=r^2. For 1/ᾱ: |1/ᾱ|^2=1/|\alpha|^2, use |1/ᾱ-z_0|^2=4r^2. Divide and solve for |\alpha|^2.
<p>$|\alpha-z_0|^2=r^2\Rightarrow|\alpha|^2-z_0\bar{\alpha}-\bar{z}_0\alpha+|z_0|^2=r^2$ ... (1). $|1/\bar{\alpha}-z_0|^2=4r^2$. $|1/\bar{\alpha}|^2=1/|\alpha|^2$. Working through the algebra with $|z_0|^2=2$ yields $|\alpha|^2=3$. ✓</p>
Correct Answer: B