Question

if x+iy=root3-iroot2 /2root3-iroot2 find x and y​

Answer 1

Answer:

• x = 1/4
• y = (3√6)/16

Given,

$\implies x + iy = \dfrac{ \sqrt{3} + \sqrt{2}i}{2 \sqrt{3} - \sqrt{2}i}$

To find,

• x & y = ?

Solution:

Rationalising the given information we get:

$\implies x + iy = \dfrac{ \sqrt{3} + \sqrt{2}i}{2 \sqrt{3} - \sqrt{2}i} \times \dfrac{2 \sqrt{3} + \sqrt{2}i}{2 \sqrt{3} + \sqrt{2}i}\\\\\\\implies x + iy = \dfrac{6 + \sqrt{6}i + 2\sqrt{6}i - 2}{ (2\sqrt{3})^2 - (\sqrt{2}i)^2}\\\\\\\implies x + iy = \dfrac{4 + 3\sqrt{6}i}{12 + 4}\\\\\\\implies x + iy = \dfrac{ 4 + 3\sqrt{6}i}{16} \implies \dfrac{4}{16} + \dfrac{3\sqrt{6}i}{16}\\\\\\\implies \boxed{x + iy = \dfrac{1}{4} + \dfrac{3\sqrt{6}i}{16}}$

Comparing the coefficients of real and imaginary parts, we get:

→ x = 1/4 ; y = (3√6)/16

Hence the value of x is 1/4 and the value of y is (3√6)/16.