Question

if a+ib/c+id=x+iy prove that X²+y²=a²+b²/c²+d²​

Answer 1

Answer:

LHS = RHS

Step-by-step explanation:

 Here,

⇒ ( a + bi ) / ( c + id ) = x + iy

 Using the properties of complex numbers, we know :  

  if z = a / d  then  | z | = | a | / | d |    where z, a , d are  complex numbers.

Using the same property;

 Modulus of LHS = Modulus of RHS

$\implies \mathrm{\bigg| \dfrac{a+bi}{c+di} \bigg| =|x+iy|}$

$\implies \mathrm{\dfrac{\sqrt{a^2+b^2}}{\sqrt{c^2+d^2 } }=\sqrt{x^2 +y^2}}\\\\\\\implies \mathrm{\dfrac{a^2 +b^2}{c^2+d^2}=x^2+y^2 }$

LHS = RHS

Answer 2

$\bigstar$ Question :

If a+ib/c+id=x+iy prove that X²+y²=a²+b²/c²+d²​

$\bigstar$ Answer :

Let , z = x+iy , z1 = a+ib , z2 = c+id

So,Magnitudes are as follows ;

=> |z|=√(x²+y²)

=> |z1|=√(a²+b²)

=>|z2|=√(c²+d²)

$\bigstar$ Now , use properties of complex numbers to prove.

$Since,z=\frac{z_1}{z_2}$ [Given]

$\implies |z|=|\frac{z_1}{z_2} |\\\\=>|z|=\frac{|z_1|} {|z_2|} \\\\=>\sqrt{x^2+y^2} =\frac{\sqrt{a^2+b^2} }{\sqrt{c^2+d^2} } \\\\=>x^2+y^2=(\frac{\sqrt{a^2+b^2} }{\sqrt{c^2+d^2} } )^2\\\\=>x^2+y^2=(\sqrt{\frac{a^2+b^2}{c^2+d^2} })^2\\\\ \implies x^2+y^2=\frac{a^2+b^2}{c^2+d^2}$

Hence Proved.