Let $A = \{ z \in \mathbb{C} : |z - 2 - i| = 3 \}$, $B = \{ z \in \mathbb{C} : \text{Re}(z - iz) = 2 \}$, and $S = A \cap B$. Then $\sum_{z \in S} |z|^2$ is equal to
We have:
A: |z − (2+i)| = 3 ⇒ (x−2)^2 + (y−1)^2 = 9 (circle)
B: Re(z − iz) = 2 ⇒ Re((1−i)(x+iy)) = x + y = 2 (line)
Step 1: Intersect the circle with the line x + y = 2. Put y = 2 − x into the circle:
\[ (x-2)^2 + (1 - x)^2 = 9 \;\Longrightarrow\; 2x^2 - 6x + 5 = 9 \] \[ \Longrightarrow\; 2x^2 - 6x - 4 = 0 \;\Longrightarrow\; x^2 - 3x - 2 = 0. \]
Thus roots \(x_1, x_2\) satisfy \(x_1 + x_2 = 3,\; x_1 x_2 = -2\), and \(y_i = 2 - x_i\).
Step 2: For each point \(z_i = x_i + i y_i\),
\[ |z_i|^2 = x_i^2 + y_i^2 = (x_i + y_i)^2 - 2x_i y_i = 4 - 2x_i y_i \] (since \(x_i + y_i = 2\)).
Step 3: Sum over the two intersection points:
\[ \sum |z_i|^2 = \sum (4 - 2x_i y_i) = 8 - 2\sum x_i y_i. \] \[ x_i y_i = x_i (2 - x_i) = 2x_i - x_i^2 \Rightarrow \sum x_i y_i = 2(x_1+x_2) - (x_1^2 + x_2^2). \] \[ x_1^2 + x_2^2 = (x_1+x_2)^2 - 2x_1x_2 = 3^2 - 2(-2) = 9 + 4 = 13. \] \[ \sum x_i y_i = 2\cdot 3 - 13 = -7. \] \[ \therefore\ \sum |z|^2 = 8 - 2(-7) = 8 + 14 = \boxed{22}. \]
If $ A = \begin{pmatrix} 2 & 2 + p & 2 + p + q \\ 4 & 6 + 2p & 8 + 3p + 2q \\ 6 & 12 + 3p & 20 + 6p + 3q \end{pmatrix} $, then the value of $ \det(\text{adj}(\text{adj}(3A))) = 2^m \cdot 3^n $, then $ m + n $ is equal to:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)