If a2-12-2 a-1-x- iy find the value of x2-y2-

Given that : (a^2+1)^2/2a 1=x+iy ⇒ (a^2+1)^2/(2a i)=x+iy ⇒ (a^2+1)^2(2a+i)/(2a i)(2a+i)=x+iy ⇒ (a^2+1)^2(2a+i)/4a^2+1=x+iy ⇒ 2a(a^2+1)^2/4a^2+1 and y=(a^2+1)^2/ ...

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Byju's Answer

Standard XII

Mathematics

Conjugate of a Complex Number

If a2+12/2 a-...

Question

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

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Solution

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

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

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If (a2+1)22a−1=x+iy find the value of x2+y2.

Given that : (a2+1)22a−1=x+iy ⇒ (a2+1)2(2a−i)=x+iy⇒ (a2+1)2(2a+i)(2a−i)(2a+i)=x+iy⇒ (a2+1)2(2a+i)4a2+1=x+iy⇒ 2a(a2+1)24a2+1 and y=(a2+1)24a2+1∴ x+y2=4a2[(a2+1)24a2+1]2+[(a2+1)24a2+1]2=(4a2+1)(a2+1)4(4a2+1)2=(a2+1)4(4a2+1)

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