If a + ib $$\dfrac{(x+i)^2}{2x^2+1}$$ Prove that $$a^2+b^2= \dfrac{(x^2+1)^2}{(2x^2+1)^2}$$

Asked by Pragya Singh | 1 year ago |  64

#### 1 Answer

##### Solution :-

Expression

$$a + ib = \dfrac{(x+i)^2}{2x^2+1}$$

$$\dfrac{x^2+i^2+2xi}{2x^2+1}$$

$$\dfrac{x^2-1+i2x}{2x^2+1}$$

$$\dfrac{x^2-1}{2x^2+1}+i (\dfrac{2x}{2x^2+1})$$

On comparing

$$a^2+b^2= \dfrac{(x^2+1)^2}{(2x^2+1)^2}+(\dfrac{2x}{2x^2+1})^2$$

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

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

$$\dfrac{(x^2+1)^2}{(2x^2+1)^2}$$

$$a + ib = \dfrac{(x^2+1)^2}{(2x^2+1)^2}$$

Hence, proved

Answered by Abhisek | 1 year ago

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