The first condition is:
\( |z - 1| \leq 1 \).
Substitute \( z = x + iy \), where \( x, y \in \mathbb{R} \):
\( |z - 1| = \sqrt{(x - 1)^2 + y^2} \leq 1 \).
Squaring both sides:
\( (x - 1)^2 + y^2 \leq 1. \) (1)
The second condition is:
\( |z - 5| \leq |z - 5i|. \)
Substitute \( z = x + iy \):
\( \sqrt{(x - 5)^2 + y^2} \leq \sqrt{x^2 + (y - 5)^2}. \)
Squaring both sides:
\( (x - 5)^2 + y^2 \leq x^2 + (y - 5)^2. \)
Simplify:
\( -10x - 10y \leq 0. \)
\( x + y \geq 0. \) (2)
From condition (1), \( (x - 1)^2 + y^2 \leq 1 \), the points lie within or on a circle centered at \( (1, 0) \) with radius 1.
From condition (2), \( x + y \geq 0 \), the points lie above or on the line \( y = -x \).
Since \( x, y \in \mathbb{Z} \), we identify the integer points satisfying both conditions. These points are:
\( (0, 0), (1, 0), (2, 0), (1, 1), (1, -1). \)
For each point \( z_k = x_k + iy_k \), the modulus squared is \( |z_k|^2 = x_k^2 + y_k^2 \). Calculate for each point:
\( \sum_{k=1}^{5} |z_k|^2 = 0 + 1 + 4 + 2 + 2 = 9. \)
Final Answer is \( 9 \).
Let \( z \) satisfy \( |z| = 1, \ z = 1 - \overline{z} \text{ and } \operatorname{Im}(z)>0 \)
Then consider:
Statement-I: \( z \) is a real number
Statement-II: Principal argument of \( z \) is \( \dfrac{\pi}{3} \)
Then: