Write as a complex number in standard form?

Question

4 - i sqrt(3)
--------------
4 + i sqrt(3)

Thanks in advance. Please show work.

Answer 1

Multiply the top and bottom by: 4 - i sqrt(3)
(4 - i sqrt(3))(4 - i sqrt(3))
---------------------------------
(4 + i sqrt(3))(4 - i sqrt(3))

The denominator becomes a difference of squares:
(4 - i sqrt(3))²
-----------------
4² - (i sqrt(3))²

Expand the denominator:
(4 - i sqrt(3))²
-----------------
16 - (-1)(3)

(4 - i sqrt(3))²
-----------------
19

Expand the numerator:
4² - 8i sqrt(3) - (i sqrt(3))²
---------------------------------
19

Simplify:
16 - 8i sqrt(3) - (-1)(3)
----------------------------
19

19 - 8i sqrt(3)
-----------------
19

Answer:
1 - i (8/19)*sqrt(3)

Answer 2

When doing products and quotients, it's easier to work with polar form:

However, you dividing a number by its complex conjugate. So if
theta = arctan (sqrt(3)/4) (use radians)

then ratio is

e^(i*(-theta-theta)) = e^(i*2*theta)

Answer 3

rationalising by multiplying by 4-irt3
(4-irt3)^2/(4^2-i^2*3)
=4^2-8irt3+i^2*3/7
=16-3-8rt3i/7
=13/7 -(8rt3/7)i

Answer 4

You must multiply both top and bottom by conjugate of bottom.
The conjugate of a+ib is a-ib
4-isqrt 3 (4-isqrt3)(4-isqrt3)
----------- = --------------------------
4+isqrt3 (4+isqrt3)(4-isqrt3)

= (4-isqrt3)^2/(16 +3) = (4-isqrt3)^2/19
=(13-i8sqrt3)/29