Complex Numbers
Complex Numbers
nta_abhyas_2025
Grade 11
Question:
For the circle with center $\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$, find the condition on $(Z-1)^n = Z^n \Rightarrow |Z-1| = |Z| \Rightarrow x = \frac{1}{2}$. Given $\frac{1}{2} + y^2 = 1 \Rightarrow y = \pm\frac{\sqrt{3}}{2}$. Which would exist when $n$ is a multiple of $6$: $\arg\left(\frac{1}{2} + \frac{\sqrt{3}}{2}i\right) = \tan^{-1}\sqrt{3} = \frac{\pi}{3}$; least value of $n$ is equal to $\frac{2\pi}{6}$
Step-by-Step Solution
Key Concept: Use modulus conditions to find loci, then apply argument conditions for the equation $(Z-1)^n = Z^n$
The condition $(Z-1)^n = Z^n$ gives $|Z-1| = |Z|$, which describes the perpendicular bisector $x = \frac{1}{2}$. Substituting into the circle equation $\frac{1}{4} + y^2 = 1$ yields $y = \pm\frac{\sqrt{3}}{2}$. The complex number $Z = \frac{1}{2} + \frac{\sqrt{3}}{2}i$ has argument $\frac{\pi}{3}$. For $(Z-1)^n = Z^n$, we need $n \cdot \arg(Z) = 2\pi k$, so $n \cdot \frac{\pi}{3} = 2\pi k$, giving $n = 6k$. The least positive value is $n = 6$.
Correct Answer: 6