Question

for all real number x.let mapping f(x)=1/x-t.if there exists real number a,bc,d for which f(a),f(b),f(c) and f(d) on complex plane.find area of square​

Answer 1

The area of the square is AB^2 = a-b / (a^2 + 1)^2 (b^2 + 1)^2 { (a - b) (a^2b^2-2ab) + 2a}

Step-by-step explanation:

f(x)=1/x-i = x + i / x - i

f(x) = x + i / x^2 - i^2 = x + i / x  - i

     = x / x^2 + 1 + i . 1 / x^2 + 1

Now  

f(x) = [ x / x^2 + 1  + 1 / x^2 + 1]

Now AB = BC = CD = DA

Let's suppose the length of the side of square is AB.

Now  

AB^2 = (x2 - 4)2 + (y2 - y1)^2

After solving we get;

AB^2 = a-b / (a^2 + 1)^2 (b^2 + 1)^2 { (a - b) (a^2b^2-2ab) + 2a}

Thus the area of the square is AB^2 = a-b / (a^2 + 1)^2 (b^2 + 1)^2 { (a - b) (a^2b^2-2ab) + 2a}

Answer 2

Answer:

The area of the square is AB^2 = a-b / (a^2 + 1)^2 (b^2 + 1)^2 { (a - b) (a^2b^2-2ab) + 2a}

Step-by-step explanation:

f(x)=1/x-i = x + i / x - i

f(x) = x + i / x^2 - i^2 = x + i / x  - i

     = x / x^2 + 1 + i . 1 / x^2 + 1

Now  

f(x) = [ x / x^2 + 1  + 1 / x^2 + 1]

Now AB = BC = CD = DA

Let's suppose the length of the side of square is AB.

Now  

AB^2 = (x2 - 4)2 + (y2 - y1)^2

After solving we get;

AB^2 = a-b / (a^2 + 1)^2 (b^2 + 1)^2 { (a - b) (a^2b^2-2ab) + 2a}

Thus the area of the square is AB^2 = a-b / (a^2 + 1)^2 (b^2 + 1)^2 { (a - b) (a^2b^2-2ab) + 2a}

Step-by-step explanation:

Hope this answer will help you.✌️