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Question

sq.3 + i/ sq. 3 -i

Answer 1

By sq. do you mean that you are squaring (3 + i) in the numerator and (3-i) in the denominator?

If so, then your fraction equals
(6i + 8)/(-6i + 8) or (3i +4)/(-3i + 4).

Most likely, you mean take the square root of 3 in which case you need to multiply the numerator and denominator of the fraction by the conjugate of sqrt(3) - i. When you multiply a complex number (a + bi) by its conjugate (a - bi), you get a real number, a^2 + b^2.

sqrt(3) + i/ sqrt(3) - i
= (2sqrt(3)i +2)/4
= (sqrt(3)i +1)/2.

Answer 2

(SQRT(3) + i) / (SQRT (3) - i)

First Multiply numerator and denominator by conjugate of the denominator (which is SQRT(3) + i)

[(SQRT(3) + i) (SQRT(3) + i) ] / [ (SQRT (3) - i)(SQRT(3) + i) ]

Simplify

[ 3 + 2i SQRT(3) - 1 ] / (3 + 1)

[2 + 2i SQRT(3) ] / 4

1/2 + (i SQRT(3)) / 2

Answer 3

you multiply numerator and denominator by sq3+1

so (sq3+i)^2/ (sq3+i)(sq3-i)

numerator is 3+2i*sq3+i^2 = 2(1-i*sq3)
denominator= 3-i^2 = 3 -(-1)=4

and fraction (1-1* sq3)/2