Complex Numbers
Concentric Circles / Modulus Condition
nta_pyq_2024_jan
Grade 11
Question:
Let the complex numbers $\alpha$ and $\frac{1}{\bar{\alpha}}$ lie on the circles $|z-z_0|^2=4$ and $|z-z_0|^2=16$ respectively, where $z_0=1+i$. Then the value of $100|\alpha|^2$ is
Step-by-Step Solution
Key Concept: Write out the two circle conditions for $\alpha$ and $1/\bar{\alpha}$ explicitly. The condition for $1/\bar{\alpha}$ on the second circle can be manipulated using $|\alpha|^2$ to get a second equation. Subtract the two equations to find $|\alpha|^2$.
From $|\alpha-z_0|^2=4$: $|\alpha|^2-\alpha\bar{z}_0-\bar{\alpha}z_0=2-|z_0|^2=2-2=0$, so $|\alpha|^2-\alpha\bar{z}_0-\bar{\alpha}z_0=2\cdots(1)$.
From $|\frac{1}{\bar{\alpha}}-z_0|^2=16$: expand to get $1-\bar{\alpha}z_0-\alpha\bar{z}_0+|\alpha|^2|z_0|^2=16|\alpha|^2$, so $1-\bar{\alpha}z_0-\alpha\bar{z}_0=14|\alpha|^2\cdots(2)$.
$(1)-(2)$: $|\alpha|^2-1=2-14|\alpha|^2\Rightarrow 15|\alpha|^2=3\Rightarrow|\alpha|^2=\frac{1}{5}$.
$100|\alpha|^2=20$.
Correct Answer: 20