Question

if x+iy=a+ib/a-ib then prove x sq. +y sq.=1​

Answer 1

Answer:

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Answer 2

Answer:

Given that :

$\tt{x+iy=\dfrac{a+ib}{a-ib}}$

$\longrightarrow\tt{\dfrac{a+ib}{a-ib}\times\dfrac{a+ib}{a+ib}}$

$\longrightarrow\tt{\dfrac{a^{2}+ab i +ab i + b^{2}i^{2}}{a^{2}+ab i - ab i - b^{2}i^{2}}}$

$\longrightarrow\tt{\dfrac{a^{2}+2ab i - b^{2}}{a^{2}+b^{2}}}$

$\longrightarrow\tt{\dfrac{a^{2}-b^{2}}{a^{2}+b^{2}}+\dfrac{2ab}{a^{2}+b^{2}}i}$

$\tt{Now,\:x-iy = \dfrac{a^{2}-b^{2}}{a^{2}+b^{2}}-\dfrac{2ab}{a^{2}+b^{2}}i}$

So,

$\tt{x^2 + y^2 = (x + iy)(x - iy)}$

Hence,

$\longrightarrow{\tt{\bigg[\dfrac{a^{2}-b^{2}}{a^{2}+b^{2}}+\dfrac{2ab}{a^{2}+b^{2}}i\bigg]\bigg[\dfrac{a^{2}-b^{2}}{a^{2}+b^{2}}-\dfrac{2ab}{a^{2}+b^{2}}i\bigg]}}$

$\longrightarrow\tt{\dfrac{{(a^{2}-b^{2})}^{2}}{(a^{2}+b^{2})^{2}}-\dfrac{4a^{2}b^{2}i^{2}}{(a^{2}+b^{2})^{2}}}$

$\longrightarrow\tt{\dfrac{{(a^{2}-b^{2})}^{2}}{(a^{2}+b^{2})^{2}}+\dfrac{4a^{2}b^{2}}{(a^{2}+b^{2})^{2}}}$

$\longrightarrow\tt{\dfrac{{(a^{2}-b^{2})}^{2}+4a^{2}b^{2}}{(a^{2}+b^{2})^{2}}}$

$\longrightarrow\tt{\dfrac{(a^{2})^{2}+(b^{2})^{2}-2a^{2}b^{2}+4a^{2}b^{2}}{(a^{2}+b^{2})^{2}}}$

$\longrightarrow\bf{\dfrac{a^{4}+b^{4}+2a^{2}b^{2}}{a^{4}+b^{4}+2a^{2}b^{2}}} = 1$