Question

if x+iy=√{(a+iy)/(c+id)},prove that (x²+y²)²=(a²+b²)/(c²+d²)​

Answer 1

Answer:

Step-by-step explanation:

x - iy = √{(a -ib)/(c-id)}

concept :-

|z|ⁿ = |zⁿ| and also |z1/z2| = |z1|/|z2|

solution:-

x -iy = √{(a-ib)/(c-id)}

x + (-y)i = √{(a+ (-b)i)/(c+(-d)i)}

now, take modulus both sides,

|x + (-y)i| = |(a+(-b)i)/(c+(-d)i)|½

√(x² + y²) = |(a+(-ib))|½/|(c+(-d)i)|½

take square both sides,

(x² + y²) = |a +(-b)i|/|c+(-d)i|

(x²+y²) = √(a²+b²)/√(c²+d²) [ by using formula |(±a)+i(±b)| = √(a²+b²) ]

Take square both sides,

Hence,

(x² + y²)² = (a²+b²)/(c²+d²)