If x - iy = $$\sqrt{\dfrac{a-ib}{c-id}}$$ prove that$$( x^2+y^2)^2=\dfrac{a^2+b^2}{c^2+d^2}$$

Asked by Pragya Singh | 1 year ago |  67

##### Solution :-

Expression

x - iy = $$\sqrt{\dfrac{a-ib}{c-id}}$$

$$\sqrt{\dfrac{a-ib}{c-id}}\times \dfrac{c+id}{c+id}$$

[On multiplying numerator and denominator by (c + id)]

$$\sqrt{\dfrac{(ac+bd)+i(ad-bc)}{c^2+d^2}}$$

$$( x - iy)^2 =\dfrac{(ac+bd)+i(ad-bc)}{c^2+d^2}$$

$$x^2 - y^2 -2ixy=\dfrac{(ac+bd)+i(ad-bc)}{c^2+d^2}$$

On comparing

$$x^2-y^2=\dfrac{ac+bd}{c^2+d^2}-2xy=\dfrac{ad-bc}{c^2+d^2}$$.......(1)

$$(x^2+y^2)^2=(x^2-y^2)^2+4x^2y^2$$

$$( \dfrac{ac+bd}{c^2+d^2})^2+\dfrac{ad-bc}{c^2+d^2}$$

$$\dfrac{a^2c^2+b^2d^2+2acbd+a^2d^2+b^2c^2-2adbc}{(c^2+d^2)^2}$$

$$\dfrac{a^2c^2+b^2d^2+a^2d^2+b^2c^2}{(c^2+d^2)^2}$$

$$\dfrac{a^2(c^2+d^2)b^2(c^2+d^2)}{(c^2+d^2)^2}$$

$$\dfrac{(c^2+d^2)(a^2+b^2)}{(c^2+d^2)^2}$$

$$\dfrac{a^2+b^2}{c^2+d^2}$$

Hence, proved

Answered by Abhisek | 1 year ago

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