Wednesday, 18 August 2021
Consider the following two subsets of \(\mathbb{C}\) :
  \[A=\left\{\frac{1}{z}:|z|=2\right\} \text { and } B=\left\{\frac{1}{z}:|z-1|=2\right\} \text { . }\]
Then \(\qquad\)
(a) \(A\) is a circle, but \(B\) is not a circle.
\(\qquad\)
(b) \(B\) is a circle, but \(A\) is not a circle.
\(\qquad\qquad\)
(c) \(A\) and \(B\) are both circles.
\(\qquad\qquad\quad\)
(d) Neither \(A\) nor \(B\) is a circle.
Subscribe to:
Post Comments (Atom)
Define \(f: \mathbb{R} \rightarrow \mathbb{R}\) by \[f(x)= \begin{cases}(1-\cos x) \sin \left(\frac{1}{x}\right), & x \neq 0 \\ 0, ...
- 
16 programers are playing in a single elimination tournament. Each player has a diff erent skill level and when two play against each othe...
- 
Post by mathematicalcircles .
- 
Find all polynomial \(P(x)\) with degree \(\leq n\) and non negative coefficients such that \[P(x)P(\frac{1}{x})\leq P(1)^2\]for all positiv...
 
 
No comments:
Post a Comment