Find the mirror image of the curve z - z 1 z - z 2 = a , a ∈ R + a ≠ 1 about the line z - z 1 = z - z 2 .

z − z 1 z − z 2 = a ; line | z − z 1 | = | z − z 2 | We can assume the locus of the final image to be h , k of the image of h , k on the given line, the image of the locus lies on the original curve. ⇒ z − z 1 z − z 2 = a Hence, the answer is z − z 1 z − z 2 = a .

Find the mirror image of the curve z - z 1 z - z 2 = a , a ∈ R + a ≠ 1 about the line z - z 1 = z - z 2 .

z − z 1 z − z 2 = a ; line | z − z 1 | = | z − z 2 | We can assume the locus of the final image to be h , k of the image of h , k on the given line, the image of the locus lies on the original curve. ⇒ z − z 1 z − z 2 = a Hence, the answer is z − z 1 z − z 2 = a .

Let z 1 and z 2 are the two complex numbers satisfying z - 3 - 4 i = 3 Such that arg z 1 z 2 is maximum then find the value of z 1 - z 2 .

It is given that z - 3 - 4 i = 3 is a circle whose centre is 3 , 4 and radius is 3 if arg z 1 z 2 is maximum then z 1 & z 2 are extremities of chord of contact of a pair of tangents drawn from origin. Then z 1 - z 2 is length of chord of contact. Now as per given diagram, tan θ = 3 4 and O A = 4 then O D = O A cos θ . and A D = O A sin θ Hence, A B = 2 O A · sin θ . Here sin θ = 3 5 . hence A B = 2 4 · 3 5 . ⇒ A B = z 1 - z 2 = 24 5

If z 1 and z 2 are the two complex numbers satisfying | z - 3 - 4 i | = 8 and a r g z 1 z 2 = π 2 then find the range of the values of z 1 - z 2

It is given that z - 3 - 4 i = 8 We know that for two complex numbers a and b , a - b ≤ a + b ≤ a + b , ⇒ z - 3 - 4 i - 3 + 4 i ≤ z - 3 - 4 i + 3 + 4 i ≤ z - 3 - 4 i + 3 + 4 i ⇒ 8 - 5 ≤ z ≤ 8 + 5 = 3 ≤ z ≤ 13 . . . 1 Given a r g z 1 z 2 = π 2 ⇒ z 1 z 2 = i ⇒ z 1 = i z 2 Now, z 1 - z 2 = i z 2 - z 2 = z 2 1 - i = z 2 · 1 - i = 2 z 2 . . . 2 As z 1 and z 2 satisfies the condition in i , hence, we have z 2 ∈ 3 , 13 . Now, from equation 2 , we get z 1 - z 2 ∈ 3 2 , 13 2 . Thus, the range of z 1 - z 2 is 3 2 , 13 2 .

If z 2 + i z 1 = z 1 + z 2 and z 1 = 3 & z 2 = 4 , if affix of A , B , C are z 1 , z 2 , z 2 - i z 1 1 - i respectively. Then find the area of Δ A B C .

Given, z 2 + i z 1 = z 1 + z 2 ⇒ z 2 + i z 1 = i z 1 + z 2 And hence, i z 1 , origin 0 + i 0 and z 2 are collinear. (Assume points A and B are being represented by complex numbers z 1 and z 2 .) ⇒ arg iz 1 = arg z 2 ⇒ arg z 1 + π 2 = arg z 2 ⇒ arg z 2 - arg z 1 = π 2 So, O A ⊥ O B ⇒ A B = O A 2 + O B 2 = z 1 2 + z 2 2 = 3 2 + 4 2 = 5 . . . 1 Now consider, z 3 = z 2 - i z 1 1 - i (Point C ) So, z 3 1 - i = z 2 - i z 1 ⇒ z 3 - i z 3 = z 2 - i z 1 ⇒ z 2 - z 3 = i z 1 - z 3 ⇒ A C ⊥ B C As z 2 - z 3 = z 1 - z 3 ⇒ A C = B C . . . 2 A C 2 + B C 2 = A B 2 = 5 2 ⇒ 2 A C 2 = A B 2 ⇒ 2 A C 2 = 5 2 ⇒ A C = 5 2 = B C . . . 3 So, Area of ∆ A B C = 1 2 × A C × B C = 1 2 × 5 2 × 5 2 = 25 4 square units .(From 1 , 2 and 3 ).

Find the locus of mid-point of line segment intercepted between real and imaginary axes, by the line a z ¯ + a ¯ z + b = 0 , where ' b ' is real parameter and ' a ' is a fixed complex number such that Re ( a ) ≠ 0 , I m ( a ) ≠ 0

Here, we have the equation of a line segment a z ¯ + a ¯ z + b = 0 . . . 1 , where ' b ' is real parameter, and a is a fixed complex number such that Re a ≠ 0 , Im a ≠ 0 . Now, let z = x + i y ⇒ z ¯ = x - i y and a = c + i d ⇒ a ¯ = c - i d Substituting in 1 , a z ¯ + a ¯ z + b = 0 ⇒ c + i d x - i y + c - i d x + i y + b = 0 ⇒ c x + i d x - i y c + y d + c x + i y c - i d x + y d + b = 0 2 c x + 2 y d + b = 0 . . . 2 To find x and y intercepts, put, x = 0 in 2 ⇒ y = - b 2 d and put, y = 0 in 2 ⇒ x = - b 2 c So, x - intercept is - b 2 c , 0 and y - intercept is 0 , - b 2 d So, co-ordinate of mid point is (say h , k ) = 0 - b 2 c 2 , 0 - b 2 d 2 , , ∴ h = - b 4 c . . . 3 and k = - b 4 d . . . 4 Solving 3 and 4 we get, c h = k d Replacing h → x and k → y c x - k y = 0 Now, replace x → z + z ¯ 2 , y → z - z ¯ 2 i , c → a + a ¯ 2 and d → a - a ¯ 2 i c x - d y = 0 ⇒ a + a ¯ 2 z + z ¯ 2 - a - a ¯ 2 i z - z ¯ 2 i = 0 ⇒ a z + a z ¯ + a ¯ z + a ¯ z ¯ 4 - a z - a z ¯ - a ¯ z + a ¯ z ¯ 4 i 2 = 0 a z + a z ¯ + a ¯ z + a ¯ z ¯ + a z - a z ¯ - a ¯ z + a ¯ z ¯ 4 = 0 ⇒ a z + a ¯ z ¯ = 0

If w ≠ 1 is n th root of unity, then find the value of ∑ k = 0 n - 1 z 1 + w k z 2 2 .

We have to find the value of ∑ k = 0 n - 1 ∣ z 1 + w k z 2 l 2 For a complex number z , we know that, z 2 = z z ¯ , and z i + z j ¯ = z i ¯ + z j ¯ , Using the properties mentioned above ∑ k = 0 n - 1 ∣ z 1 + w k z 2 l 2 = ∑ k = 0 n - 1 z 1 + w k z 2 ∣ z 1 + w k z 2 | = ∑ k = 0 n - 1 z 1 + w k z 2 z 1 ¯ + w k z 2 ¯ = ∑ k = 0 n - 1 z 1 + w k z 2 z 1 ¯ + w k ¯ z 2 ¯ ∑ k = 0 n - 1 z 1 z 1 ¯ + w k z 2 z 1 ¯ + w k ¯ z 2 ¯ z 1 + w k z 2 w k ¯ z 2 ¯ = ∑ k = 0 n - 1 z 1 z 1 ¯ + ∑ k = 0 n - 1 w k z 2 z 1 ¯ + ∑ k = 0 n - 1 w k ¯ z 2 ¯ z 1 + ∑ k = 0 n - 1 w k z 2 w k ¯ z 2 ¯ = ∑ k = 0 n - 1 z 1 2 + z 2 z 1 ¯ ∑ k = 0 n - 1 w k + z 2 ¯ z ∑ k = 0 n - 1 w k ¯ 1 + z 2 z 2 ¯ ∑ k = 0 n - 1 w k w k ¯ = ∑ k = 0 n - 1 z 1 2 + z 2 z 1 ¯ ∑ k = 0 n - 1 w k + z 2 ¯ z 1 ∑ k = 0 n - 1 w k ¯ + z 2 z 2 ¯ ∑ k = 0 n - 1 w 2 k As, w is nth root of unity, ∑ k = 0 n - 1 w k = ∑ k = 0 n - 1 w k ¯ = 0 and w 2 = 1 Now, substituting these values we get, ∑ k = 0 n - 1 z 1 2 + z 2 z 1 ¯ ∑ k = 0 n - 1 w k + z 2 ¯ z 1 ∑ k = 0 n - 1 w k ¯ + z 2 z 2 ¯ ∑ k = 0 n - 1 w 2 = n z 1 2 + z 2 z 1 ¯ 0 + z 2 ¯ z 1 0 + z 2 z 2 ¯ n = n z 1 2 + z 2 2