Interpret the loci 2 ≤ | z | ≤ 3 in z ∈ C . Also indicate on complex plane.

Given, 2 ≤ | z | ≤ 3 Now, z = 3 , represents the circle with radius 3 and centre is origin. So | z | ≤ 3 , represents all the points on this circle or inside this circle. Similarly, | z | = 2 , represents the circle with radius 2 and centre is origin. So | z | ≥ 2 , represents all the points on or outside this circle. Hence, 2 ≤ | z | ≤ 3 , represents all the points within the circular annulus of two concentric circles of radius 2 and 3 with centre at origin. All points on circumference of two circles are included.

Interpret the loci | z - 2 - 3 i | > 2 in z ∈ C . Also indicate on complex plane.

Given, | z - 2 - 3 i | > 2 Putting z = x + i y , we get | x + i y - 2 - 3 i | > 2 ⇒ | ( x - 2 ) + ( y - 3 ) i | > 2 ⇒ ( x - 2 ) 2 + ( y - 3 ) 2 > 2 ∵ x + i y = x 2 + y 2 ⇒ ( x - 2 ) 2 + ( y - 3 ) 2 > 4 , which represents all points are outside the circle with centre 2 , 3 and radius 2 .

Interpret the loci R e ( z ) > 2 in z ∈ C . Also indicate on complex plane.

Let z = x + i y , x , y ∈ R Here, R e z = x and I m z = y Since, R e ( z ) > 2 , which represents all the points to the right side of line x = 2 . This can be seen in an Argand plane.

Interpret the loci arg z - 1 - i = π 4 in z ∈ C . Also indicate on complex plane.

Given, arg ( z - 1 - i ) = π 4 Putting z = x + i y , we get ⇒ arg ( x + i y - 1 - i ) = π 4 ⇒ arg ( x - 1 + i y - 1 ) = π 4 ⇒ arg ( x - 1 + i y - 1 ) = π 4 ⇒ tan - 1 y - 1 x - 1 = π 4 ⇒ y - 1 x - 1 = tan π 4 ⇒ y - 1 x - 1 = 1 ⇒ y - 1 = x - 1 ⇒ y = x , it represents a half ray of a line of inclination 45 0 passing through point 1 + i i.e, of the line x = y , x > 1 .

Interpret the loci | z - 1 | 2 + | z + 1 | 2 = 4 in z ∈ C . Also indicate on complex plane.

Given, | z - 1 | 2 + | z + 1 | 2 = 4 Putting z = x + i y , we get ⇒ | x + i y - 1 | 2 + | x + i y + 1 | 2 = 4 ⇒ | x - 1 + i y | 2 + | x + 1 + i y | 2 = 4 ⇒ x - 1 2 + y 2 2 + x + 1 2 + y 2 2 = 4 ∵ x + i y = x 2 + y 2 ⇒ x - 1 2 + y 2 + x + 1 2 + y 2 = 4 ⇒ x 2 - 2 x + 1 + y 2 + x 2 + 2 x + 1 + y 2 = 4 ⇒ x 2 + 1 + y 2 + x 2 + 1 + y 2 = 4 ⇒ 2 x 2 + 2 y 2 + 2 = 4 ⇒ 2 x 2 + 2 y 2 = 2 ⇒ x 2 + y 2 = 1 , which represents a circle of radius 1 with centered at 0 , 0 . This can be seen an argand plane.

Interpret the loci | z + i | ≥ | z + 2 | in z ∈ C . Also indicate on complex plane.

Given, | z + i | ≥ | z + 2 | Putting z = x + i y , we get ⇒ | x + i y + i | ≥ | x + i y + 2 | ⇒ | x + i y + 1 | ≥ | x + 2 + i y | ⇒ | x + i y + 1 | 2 ≥ | x + 2 + i y | 2 (Modulus is always non -negative ) ⇒ x 2 + y + 1 2 2 ≥ x + 2 2 + y 2 2 ∵ | x + i y | = x 2 + y 2 ⇒ x 2 + y + 1 2 ≥ x + 2 2 + y 2 ⇒ x 2 + y 2 + 2 y + 1 ≥ x 2 + 4 x + 4 + y 2 (Subtract x 2 and y 2 on each side of inequality) ⇒ 2 y + 1 ≥ 4 x + 4 ⇒ 0 ≥ 4 x + 4 - 2 y - 1 (Subtract 1 and 2 y on each side of inequality) ⇒ 4 x + 4 - 2 y - 1 ≤ 0 ⇒ 4 x - 2 y + 3 ≤ 0 , it represents half of the plane that lies to the left of the line 4 x - 2 y + 3 = 0 . (including the points on the line) This can ben seen in a figure.