We are given two circles in the complex plane:
The problem states that $C_2$ lies entirely within $C_1$. This geometric condition implies $R_1 > d + R_2$, where $d$ is the distance between the centers.
Calculate the distance $d$ between the centers of $C_1$ (origin $0$) and $C_2$ ($3+4i$):
$ d = |0 - (3 + 4i)| = |-(3 + 4i)| = \sqrt{(-3)^2 + (-4)^2} $
$ d = \sqrt{9 + 16} = \sqrt{25} = 5 $
When one circle ($C_2$) is inside another ($C_1$), the minimum distance between points $z_1$ on $C_1$ and $z_2$ on $C_2$ is given by the difference between the radii adjusted by the distance between centers:
$ \min|z_1 - z_2| = R_1 - (d + R_2) $
We are given $\min|z_1 - z_2| = 2$. Substitute the known values:
$ 2 = r - (5 + 5) $
$ 2 = r - 10 $
Solving for the radius $r$ of $C_1$:
$ r = 2 + 10 = 12 $
The radius of $C_1$ is $R_1 = 12$. Note that $12 > 5 + 5$, consistent with $C_2$ being inside $C_1$.
The maximum distance between points $z_1$ on $C_1$ and $z_2$ on $C_2$ occurs along the line connecting their centers, extending outwards:
$ \max|z_1 - z_2| = R_1 + d + R_2 $
Substitute the values $R_1 = 12$, $d = 5$, and $R_2 = 5$:
$ \max|z_1 - z_2| = 12 + 5 + 5 $
$ \max|z_1 - z_2| = 22 $
Let f and g be functions satisfying $f(x+y) = f(x)f(y), f(1) = 7$ and $g(x+y) = g(xy), g(1) = 1$, for all $x, y \in \mathbb{N}$. If $\sum_{x=1}^{n} \left(\frac{f(x)}{g(x)}\right) = 19607$, then n is equal to :
Let S be the set of the first 11 natural numbers. Then the number of elements in $A = \{B \subseteq S : n(B) \geq 2$ and the product of all elements of B is even is ________.