NIMCET Complex Number Previous Year Questions (PYQs)
Study Stuff for MCA Examinations
Solution
Sum of Complex Roots of Unity
Given:
\[ x_k = \cos\left(\frac{2\pi k}{n}\right) + i \sin\left(\frac{2\pi k}{n}\right) = e^{2\pi i k/n} \]
Required: Find: \[ \sum_{k=1}^{n} x_k \]
This is the sum of all \( n^\text{th} \) roots of unity (from \( k = 1 \) to \( n \)).
We know: \[ \sum_{k=0}^{n-1} e^{2\pi i k/n} = 0 \] So shifting index from \( k = 1 \) to \( n \) just cycles the same roots: \[ \sum_{k=1}^{n} e^{2\pi i k/n} = 0 \]
✅ Final Answer: \( \boxed{0} \)
Solution
Let \(r=|z|<\sqrt{3}-1\). Using triangle inequality, \[ |z^{2}+2z\cos\alpha|\le |z|^{2}+2|z||\cos\alpha|\le r^{2}+2r. \] Since \(r<\sqrt{3}-1\), \[ r^{2}+2r<(\sqrt{3}-1)^{2}+2(\sqrt{3}-1)= (3-2\sqrt{3}+1)+2\sqrt{3}-2=2. \] Hence, \[ \boxed{|z^{2}+2z\cos\alpha|<2}. \]
Solution
Solution
Solution
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