# Provee the identity (4-i*square root6)/(2-i*square root6)=(square root3+2i*square root2)/(square root3+i*square root2)

*print*Print*list*Cite

We have to prove that (4 - i*sqrt 6)/(2 - i*sqrt 6)=(sqrt 3 + 2i*sqrt 2)/(sqrt 3 + i*sqrt 2)

The left hand side:

(4 - i*sqrt 6)/(2 - i*sqrt 6)

multiply the numerator and denominator by (2 + i*sqrt 6)

=> (4 - i*sqrt 6)*(2 + i*sqrt 6)/(2 - i*sqrt 6)*(2 + i*sqrt 6)

=> (8 + 4*i*sqtr 6 - 2*i*sqrt 6 + 6)/(4 + 6)

=> (14 + 2*i*sqtr 6)/10

=> (7 + i*sqrt 6)/5...(1)

The right hand side:

(sqrt 3 + 2i*sqrt 2)/(sqrt 3 + i*sqrt 2)

multiply the numerator and denominator by (sqrt 3 - i*sqrt 2)

=> (sqrt 3 + 2i*sqrt 2)*(sqrt 3 - i*sqrt 2)/(sqrt 3 + i*sqrt 2)*(sqrt 3 - i*sqrt 2)

=> (3 + 2i*sqrt 6 - i*sqrt 6 + 4) / (3 + 2)

=> (7 + i*sqrt 6)/5 ...(2)

As (1) and (2) are the same the identity is proved.

**The identity is proved a****s both the sides are equal to (7 + i*sqrt 6)/5.**

(4-i*sqrt6)/(2-i*sqrt6) = (sqrt3 + 2i*sqrt2)/(sqrt3 + i*sqrt2)

We'll manage the left side and we'll multiply the fraction by the conjugate of denominator:

(4-i*sqrt6)/(2-i*sqrt6) = (4-i*sqrt6)(2+i*sqrt6)/(2-i*sqrt6)(2+i*sqrt6)

(8 + 4i*sqrt6 - 2i*sqrt6 + 6)/(4 + 6) = (14 + 2i*sqrt6)/10

(14 + 2i*sqrt6)/10 = 2(7 + i*sqrt6)/10 = (7 + i*sqrt6)/5

We'll manage the right side and we'll multiply the fraction by the conjugate of denominator:

(sqrt3 + 2i*sqrt2)/(sqrt3 + i*sqrt2) = (sqrt3 + 2i*sqrt2)*(sqrt3 - i*sqrt2)/(3+2)

(sqrt3 + 2i*sqrt2)*(sqrt3 - i*sqrt2)/(5) = (3 - i*sqrt6 + 2i*sqrt6 + 4)/(5)

(3 - i*sqrt6 + 2i*sqrt6 + 4)/(5) = (7 + i*sqrt6)/5

**We notice that both sides we've get: (7 + i*sqrt6)/5 = (7 + i*sqrt6)/5.**