If a 2-1 22 a - i - x - iy- then x 2- y 2 is equal toa a 2-1 44 a 2-1b a-1 24 a 2-1c a2-124a2-12d none of these

(a)a2+144a2+1 x+iy #160;= #160;a2+122a i Taking modulus on both the sides, we get: x2+y2 #160;= #160;a2+124a2+1 Squaring #160;both #160;sides, #1 ...

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Byju's Answer

Standard XII

Mathematics

Integration by Partial Fractions

If a 2+1 22 a...

Question

If $\frac{(a^{2} + 1)^{2}}{2 a - i} = x + i y, then x^{2} + y^{2}$ is equal to
(a) $\frac{(a^{2} + 1)^{4}}{4 a^{2} + 1}$
(b) $\frac{(a + 1)^{2}}{4 a^{2} + 1}$
(c) $\frac{(a^{2} - 1)^{2}}{(4 a^{2} - 1)^{2}}$
(d) none of these

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Solution

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

Taking modulus on both the sides, we get:
$\sqrt{x^{2} + y^{2}} = \frac{{(a^{2} + 1)}^{2}}{\sqrt{4 a^{2} + 1}}$
$Squaring both sides, we get, x^{2} + y^{2} = \frac{{(a^{2} + 1)}^{4}}{4 a^{2} + 1}$

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If (a2+1)22a-i=x+iy, then x2+y2 is equal to (a) (a2+1)44a2+1 (b) (a+1)24a2+1 (c) (a2-1)2(4a2-1)2 (d) none of these

(a)a2+144a2+1 x+iy = a2+122a-i Taking modulus on both the sides, we get: x2+y2 = a2+124a2+1 Squaring both sides, we get,x2+y2=a2+144a2+1

If $\frac{(a^{2} + 1)^{2}}{2 a - i} = x + i y, then x^{2} + y^{2}$ is equal to
(a) $\frac{(a^{2} + 1)^{4}}{4 a^{2} + 1}$
(b) $\frac{(a + 1)^{2}}{4 a^{2} + 1}$
(c) $\frac{(a^{2} - 1)^{2}}{(4 a^{2} - 1)^{2}}$
(d) none of these

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

Taking modulus on both the sides, we get:
$\sqrt{x^{2} + y^{2}} = \frac{{(a^{2} + 1)}^{2}}{\sqrt{4 a^{2} + 1}}$
$Squaring both sides, we get, x^{2} + y^{2} = \frac{{(a^{2} + 1)}^{4}}{4 a^{2} + 1}$