If a-i b-x-i2-2 x2-1- prove that a2-b2-x2-12-2 x2-12

Here a+ib=(x+i)^2/(2x^2+1)=x^2+i^2+2ix/2x^2+1 =x^2 1/2x^2+1+i2x/2x^2+1 Comparing both sides, we have a=x^2 1/2x^2+1 and b=2x/2x^2+1 ∴ a^2+b^2= ( x^2 1/2x^2+1 ...

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Byju's Answer

Standard XII

Mathematics

Iota

If a+i b=x+i2...

Question

If $a + i b = \frac{(x + i)^{2}}{(2 x^{2} + 1)}$ , prove that $a^{2} + b^{2} = \frac{(x^{2} + 1)^{2}}{(2 x^{2} + 1)^{2}}$

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Solution

Here $a + i b = \frac{(x + i)^{2}}{(2 x^{2} + 1)} = \frac{x^{2} + i^{2} + 2 i x}{2 x^{2} + 1}$

$= \frac{x^{2} - 1}{2 x^{2} + 1} + i \frac{2 x}{2 x^{2} + 1}$

Comparing both sides, we have

$a = \frac{x^{2} - 1}{2 x^{2} + 1} a n d b = \frac{2 x}{2 x^{2} + 1}$

$∴ a^{2} + b^{2} = {(\frac{x^{2} - 1}{2 x^{2} + 1})}^{2} + {(\frac{2 x}{2 x^{2} + 1})}^{2}$

$= \frac{(x^{2} - 1)^{2}}{(2 x^{2} + 1)^{2}} + \frac{(2 x)^{2}}{(2 x^{2} + 1)^{2}}$

$= \frac{(x^{2} - 1)^{2} + (2 x)^{2}}{(2 x^{2} + 1)^{2}}$

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

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

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If a+ib=(x+i)2(2x2+1), prove that a2+b2=(x2+1)2(2x2+1)2

Here a+ib=(x+i)2(2x2+1)=x2+i2+2ix2x2+1 =x2−12x2+1+i2x2x2+1 Comparing both sides, we have a=x2−12x2+1 and b=2x2x2+1 ∴ a2+b2=(x2−12x2+1)2+(2x2x2+1)2 =(x2−1)2(2x2+1)2+(2x)2(2x2+1)2 =(x2−1)2+(2x)2(2x2+1)2 =x4+1−2x2+4x2(2x2+1)2 =x4+1+2x2(2x2+1)2=(x2+1)2(2x2+1)2

If $a + i b = \frac{(x + i)^{2}}{(2 x^{2} + 1)}$ , prove that $a^{2} + b^{2} = \frac{(x^{2} + 1)^{2}}{(2 x^{2} + 1)^{2}}$