If \[\frac{\left( a^2 + 1 \right)^2}{2a - i} = x + iy\] find the value of \[x^2 + y^2\].

\[\frac{\left( a^2 + 1 \right)^2}{2a - i} = x + iy . . . . (1)\] \[ \Rightarrow \left[ \bar{\frac{\left( a^2 + 1 \right)^2}{2a - i}} \right] = \bar{{x + iy}}\] \[ \Rightarrow \frac{\left( a^2 + 1 \right)^2}{2a + i} = x - iy . . . . (2)\] \[\text { On multiplying (1) and (2), we get }\] \[\frac{\left( a^2 + 1 \right)^2}{2a - i} \times \frac{\left( a^2 + 1 \right)^2}{2a + i} = \left( x + iy \right)\left( x - iy \right)\] \[ \Rightarrow \frac{\left( a^2 + 1 \right)^4}{\left( 2a \right)^2 - i^2} = x^2 - i^2 y^2 \] \[ \Rightarrow \frac{\left( a^2 + 1 \right)^4}{\left( 2a \right)^2 + 1} = x^2 + y^2\] Hence, \[x^2 + y^2 = \frac{\left( a^2 + 1 \right)^4}{4 a^2 + 1}\].

If ( a 2 + 1 ) 2 2 a − I = X + I Y Find the Value of X 2 + Y 2 . - Mathematics

प्रश्न

If $\frac{{(a^{2} + 1)}^{2}}{2 a - i} = x + i y$ find the value of $x^{2} + y^{2}$ .

उत्तर

$\frac{{(a^{2} + 1)}^{2}}{2 a - i} = x + i y . . . . (1)$

$\Rightarrow [\bar{\frac{{(a^{2} + 1)}^{2}}{2 a - i}}] = \bar{x + i y}$

$\Rightarrow \frac{{(a^{2} + 1)}^{2}}{2 a + i} = x - i y . . . . (2)$

$On multiplying (1) and (2), we get$

$\frac{{(a^{2} + 1)}^{2}}{2 a - i} \times \frac{{(a^{2} + 1)}^{2}}{2 a + i} = (x + i y) (x - i y)$

$\Rightarrow \frac{{(a^{2} + 1)}^{4}}{{(2 a)}^{2} - i^{2}} = x^{2} - i^{2} y^{2}$

$\Rightarrow \frac{{(a^{2} + 1)}^{4}}{{(2 a)}^{2} + 1} = x^{2} + y^{2}$

Hence,

$x^{2} + y^{2} = \frac{{(a^{2} + 1)}^{4}}{4 a^{2} + 1}$ .

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Concept of Complex Numbers

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Q 14 Q 13 Q 15

पाठ 13: Complex Numbers - Exercise 13.5 [पृष्ठ ६३]

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आरडी शर्मा Mathematics [English] Class 11

पाठ 13 Complex Numbers
Exercise 13.5 | Q 14 | पृष्ठ ६३

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Concept of Complex Numbers

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If ( a 2 + 1 ) 2 2 a − I = X + I Y Find the Value of X 2 + Y 2 . - Mathematics

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If ( a 2 + 1 ) 2 2 a − I = X + I Y Find the Value of X 2 + Y 2 . - Mathematics

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