If a+ib= (x+i)^2 2x^2+1 , prove that a^2+b^2= (x^2+1)^2 (2x^2+1)^2 .

a+ib= (x+i)^2 2x^2+1 = x^2+i^2+2xi 2x^2+1 = x^2-1+i2x 2x^2+1 = x^2-1 2x^2+1 +i ( 2x 2x^2+1 ) On comparing real and imaginary parts, we obtain a= x^2-1 2x^2+1 and b = 2x 2x^2+1 a^2+b^2= ( x^2-1 2x^2+1 )^2+ ( 2x 2x^2+1 )^2 = x^4+1-2x^2+4x^2 (2x^2+1)^2 = x^4+1+2x^2 (2x^2+1)^2 = (x^2+1)^2 (2x^2+1)^2 a^2+b^2= (x^2+1)^2 (2x^2+1)^2 Hence, proved.

If a+ib= (x+i)^2 2x^2+1 , prove that a^2+b^2= (x^2+1)^2 (2x^2+1)^…

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

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$\text{a+ib}=\frac{(\text{x+i})^2}{2\text{x}^2+1}$ $=\frac{\text{x}^2+\text{i}^2+2\text{xi}}{2\text{x}^2+1}$ $=\frac{\text{x}^2-1+\text{i2x}}{2\text{x}^2+1}$ $=\frac{\text{x}^2-1}{2\text{x}^2+1}+\text{i}\Big(\frac{2\text{x}}{2\text{x}^2+1}\Big)$ On comparing real and imaginary parts, we obtain $\text{a}=\frac{\text{x}^2-1}{2\text{x}^2+1} \text{ and b} =\frac{2\text{x}}{2\text{x}^2+1}$ $\therefore\ \text{a}^2+\text{b}^2=\Big(\frac{\text{x}^2-1}{2\text{x}^2+1}\Big)^2+\Big(\frac{2\text{x}}{2\text{x}^2+1}\Big)^2$ $=\frac{\text{x}^4+1-2\text{x}^2+4\text{x}^2}{(2\text{x}^2+1)^2}$ $=\frac{\text{x}^4+1+2\text{x}^2}{(2\text{x}^2+1)^2}$ $=\frac{(\text{x}^2+1)^2}{(2\text{x}^2+1)^2}$$\therefore\ \text{a}^2+\text{b}^2=\frac{(\text{x}^2+1)^2}{(2\text{x}^2+1)^2}$
Hence, proved.

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