The region of the argand plane defined by |z – i| + |z + i| ≤ 4 is

To find the region of the Argand plane defined by |z – i| + |z + i| ≤ 4, we can break it down into steps:

Step 1: Rewrite the inequality using the definition of absolute value: |z – i| + |z + i| ≤ 4 |Re(z) + iIm(z) – i| + |Re(z) + iIm(z) + i| ≤ 4

Step 2: Simplify the absolute values: √[(Re(z))^2 + (Im(z) - 1)^2] + √[(Re(z))^2 + (Im(z) + 1)^2] ≤ 4

Step 3: Square both sides of the inequality to eliminate the square roots: [(Re(z))^2 + (Im(z) - 1)^2] + 2√[(Re(z))^2 + (Im(z) - 1)^2][(Re(z))^2 + (Im(z) + 1)^2] + [(Re(z))^2 + (Im(z) + 1)^2] ≤ 16

Step 4: Simplify the equation: 3(Re(z))^2 + 2[(Im(z) - 1)^2 + (Im(z) + 1)^2] + 2√[(Re(z))^2 + (Im(z) - 1)^2][(Re(z))^2 + (Im(z) + 1)^2] ≤ 16

Step 5: Further simplify the equation: 3(Re(z))^2 + 4(Im(z))^2 + 4 ≤ 16

Step 6: Rearrange the terms: 3(Re(z))^2 + 4(Im(z))^2 ≤ 12

Step 7: Divide both sides by 12: (Re(z))^2/4 + (Im(z))^2/3 ≤ 1

Step 8: Compare the inequality to the standard form of an ellipse: (Re(z))^2/a^2 + (Im(z))^2/b^2 ≤ 1

From the comparison, we can see that the given inequality represents an ellipse in the Argand plane with semi-major axis a = 2 and semi-minor axis b = √(3). Therefore, the region defined by |z – i| + |z + i| ≤ 4 is the interior and boundary of this ellipse.