Question

a+ib/c+id=x+iy, then prove that a²+b²/c²+d²=x²+y²

Answer 1

$\large\underline{\sf{Solution-}}$

Given complex number is

$\rm :\longmapsto\:\dfrac{a + ib}{c + id} = x + iy$

Taking modulus on both sides,

$\rm :\longmapsto\:\bigg |\dfrac{a + ib}{c + id}\bigg | = |x + iy|$

We know,

$\boxed{ \tt{ \: \bigg| \frac{z_1}{z_2} \bigg| = \frac{ |z_1| }{ |z_2| } \: }}$

and

$\boxed{ \tt{ \: z = a + ib \: \: \rm \implies\: |z| = \sqrt{ {a}^{2} + {b}^{2} } \: }}$

So, using these results, we get

$\rm :\longmapsto\:\dfrac{ |a + ib| }{ |c + id| } = \sqrt{ {x}^{2} + {y}^{2} }$

$\rm :\longmapsto\:\dfrac{ \sqrt{ {a}^{2} + {b}^{2}}}{ \sqrt{ {c}^{2}+{d}^{2}}} = \sqrt{ {x}^{2}+{y}^{2}}$

On squaring both sides, we get

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

Hence, Proved

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Additional Information :-

$\boxed{ \tt{ \: z \: \overline{z} = { |z| }^{2} \: }}$

$\boxed{ \tt{ \: |z| \: = \: |\overline{z}| \: }}$

$\boxed{ \tt{ \: \overline{z_1 + z_2} = \overline{z_1} + \overline{z_2} \: }}$

$\boxed{ \tt{ \: \overline{z_1 \times z_2} = \overline{z_1} \times \overline{z_2} \: }}$

$\boxed{ \tt{ \: \overline{z_1 - z_2} = \overline{z_1} - \overline{z_2} \: }}$

$\boxed{ \tt{ \: \overline{z_1 \div z_2} = \overline{z_1} \div \overline{z_2} \: }}$